Review Problems for the Final

Math 345

These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won't appear on the test.

1. Prove that if n is an integer, then $(4 n + 6, 3 n + 4)$ is either 1 or 2. Give specific examples which show that both cases can occur.

2. Find the greatest common divisor of 847 and 133 and write it as a linear combination with integer coefficients of 847 and 133.

3. Show that the following set is a subgroup of $GL(2, \real)$ :

$$H = \left\{\left[\matrix{x & 0 \cr 0 & y \cr}\right] \Bigm| x, y \in \real, \quad x y \ne 0\right\}$$

However, show that it is not a normal subgroup of $GL(2, \real)$ .

4. Consider the map $\phi: \integer_8 \rightarrow \integer_8$ given by $\phi(n) = n + 2 \mod{8}$ . Is $\phi$ a group homomorphism? Why or why not?

5. Consider the map $\phi: \integer \rightarrow \integer$ given by $\phi(n) = n^2$ . Is $\phi$ a group homomorphism? Why or why not?

6. $\integer \times \integer$ is a group under componentwise addition and $\integer$ is a group under addition. Prove that

$$\dfrac{\integer \times \integer}{\langle (7, 25) \rangle} \approx \integer.$$

7. $\real^2$ is a group under componentwise addition and $\real$ is a group under addition. Let

$$H = \left\{x \cdot (19, -\sqrt{7}) \Bigm| x \in \real\right\}.$$

Prove that $\dfrac{\real^2}{H} \approx \real$ .

8. $\integer \times \integer$ and $\integer \times \integer \times \integer$ are groups under componentwise addition. Let

$$H = \{t \cdot (-4, 3, 1) \mid t \in \integer\}.$$

Show that

$$\dfrac{\integer \times \integer \times \integer}{H} \approx \integer \times \integer.$$

9. Here is the multiplication table for the Klein 4-group V:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & a & & 1 & & c & & b & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & b & & b & & c & & 1 & & a & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & c & & c & & b & & a & & 1 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Write down all the subgroups of V.

10. Find all integer solutions $(x, y)$ to

$$x^2 - y^2 + 2 y = 12.$$

11. Find an element of order 30 in $\integer_{25} \times \integer_{12}$ .

12. Find the primary decomposition of $U_{16}$ .

13. (a) What is the order of the element $a^4$ in the cyclic group

$$\{a^k \mid a^{22} = 1\}?$$

(b) What is the order of the element 10 in $\integer_{45}$ ?

(c) What elements generate the cyclic group $\integer_{12}$ ?

14. Subgroups of cyclic groups are cyclic. Give an example of an abelian group which is not cyclic, but in which every proper subgroup is cyclic.

15. (a) Prove that a group cannot be the union of two proper subgroups.

(b) Find a group which is a union of three proper subgroups.

16. Let $\phi: G \to H$ be a group homomorphism. Prove that $\phi$ is injective if and only if $\ker \phi = \{1\}$ .

17. Is there a group homomorphism $\phi: \integer_6 \rightarrow \integer_{12}$ such that $\ker \phi = \{0\}$ ? Construct such a homomorphism, or show that such a homomorphism cannot exist.

18. (a) Give an example of a finite group which is not abelian.

(b) Give an example of an abelian group which is not finite.

(c) Give an example of a group which is neither finite nor abelian.

19. Let $SL(2, \real)$ denote the subgroup of $GL(2, \real)$ consisting of matrices of determinant 1. Show that the following matrices lie in the same left coset of $SL(2, \real)$ :

$$\left[\matrix{1 & 1 \cr 1 & 4 \cr}\right] \quad\hbox{and}\quad \left[\matrix{5 & 2 \cr 11 & 5 \cr}\right].$$

20. Give an example of a finite commutative ring with 1 which is not an integral domain.

21. (a) Define $f: \real \times \real \to \real$ by

$$f(x, y) = x^3 + y^3.$$

Show that f is surjective.

(b) Define $g: \real \times \real \times \real \to \real$ by

$$g(x, y,z) = x - 2 y + 3 z.$$

Show that g is surjective.

(c) Define $h: M(2, \real) \to M(2, \real)$ by

$$h\left(\left[\matrix{a & b \cr c & d \cr}\right]\right) = \left[\matrix{a & 2 b \cr 3 c & 4 d \cr}\right].$$

Show that h is surjective.

(d) Define $k: \real \to \real \times \real$ by

$$k(x) = (x, x).$$

Show that k is not surjective.

(e) Give an example of a group map $p: \integer \to \integer$ which is not surjective, and a surjective function $q: \integer \to \integer$ which is not a group map.

22. (a) Explain why $\rational$ is not a group under multiplication.

(b) Do the nonzero elements of $\integer_6$ form a group under multiplication mod 6?

(c) Show that the nonzero elements of $\integer_5$ form a group under multiplication mod 5. What group?

23. Reduce $32^{2011} \mod{41}$ to an integer in the set $\{0, 1, \ldots, 40\}$ .

24. Reduce $\dfrac{148!}{3 \cdot 75} \mod{149}$ to an integer in the set $\{0, 1, \ldots, 148\}$ . (Note: 149 is prime.)

25. The definition of a subring of a ring does not require that you check associativity for addition or multiplication. Explain why.

26. Prove that if I is an ideal in a ring R with identity and $1 \in I$ , then $I = R$ .

27. Show that the only (two-sided) ideals in $M(2, \real)$ are the zero ideal and the whole ring.

28. Consider the following subset of the ring $\integer \times \integer$ :

$$S = \{(m + n, m - n) \mid m, n \in \integer\}.$$

Check each axiom for an ideal. If the axiom holds, prove it. If the axiom does not hold, give a specific counterexample.

29. (a) Show that $x^2 + x + 1$ is irreducible in $\integer_5[x]$ .

(b) Find $[(2 x + 3) + \langle x^2 + x + 1 \rangle]^{-1}$ in $\dfrac{\integer_5[x]}{\langle x^2 + x + 1 \rangle}$ .

(c) Compute the product of the cosets $\left((x^2 + 2) + \langle x^2 + x + 1\rangle\right) \cdot \left((3 x + 4) + \langle x^2 + x + 1\rangle\right)$ in the quotient ring $\dfrac{\integer_5[x]}{\langle x^2 + x + 1\rangle}$ . Write your answer in the form $(ax + b) + \langle x^2 + x + 1\rangle$ , where $a, b \in \integer_5$ .

30. Factor $x^4 + 64$ in $\rational[x]$ .

31. (a) Show that $x^4 + 1$ has no roots in $\integer_5$ .

(b) Show that $x^4 + 1$ factors in $\integer_5[x]$ .

32. In the ring $\real[x]$ , consider the subset

$$\langle x^2 - x - 2, x^2 - 1 \rangle = \{a(x)(x^2 - x - 2) + b(x)(x^2 - 1) \mid a(x), b(x) \in \real[x]\}.$$

(a) Show that $\langle x^2 - x - 2, x^2 - 1 \rangle$ is an ideal.

(b) Is $x^2 + x + 3$ in $\langle x^2 - x - 2, x^2 - 1 \rangle$ ?

33. $x^2 + 2 = (x + 1)(x + 2)$ is a factorization of $x^2 + 2$ into irreducibles in $\integer_3[x]$ . Find a different factorization of $x^2 + 2$ into irreducibles in $\integer_3[x]$ .

34. Compute the product of the cycles $(2\ 4\ 6\ 3)(1\ 3\ 4)$ (right to left) and write the result as a product of disjoint cycles.

35. Define $\phi: \integer[x] \to \integer[x]$ by

$$\phi(f(x)) = f(x)^2.$$

Determine which of the axioms for a ring map are satisfied by $\phi$ . If an axiom is not satisfied, give a specific example which shows that the axiom is violated.

36. Define $\phi: \integer_2[x] \to \integer_2[x]$ by

$$\phi(f(x)) = f(x)^2.$$

(a) Show that $\phi$ is a ring map.

(b) Determine the kernel of $\phi$ .

(c) Show that $x^4 + 1 \in \im \phi$ . Is $\phi$ surjective?

37. Find the quotient and the remainder when $2 x^4 + 3 x^3 + x + 1$ is divided by $3 x^2 + 1$ in $\integer_5[x]$ .

38. (a) Explain why $x^4 + 1$ has no roots in $\real$ .

(b) Is $x^4 + 1$ irreducible in $\real[x]$ ?

39. List the zero divisors and the units in $\integer_2 \times \integer_3$ .

40. Prove that if I is a left ideal in a division ring R, then either $I = \{0\}$ or $I = R$ .

41. Let R be a ring, and let $r \in R$ . The centralizer $C(r)$ of r is the set of elements of R which commute with r:

$$C(r) = \{a \in R \mid ra = ar\}.$$

Prove that $C(r)$ is a subring of R.

42. Let

$$I = \left\{\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] \Bigm| x, y, z \in \real\right\}.$$

Prove that I is a left ideal, but not a right ideal, in the ring $M(3, \real)$ .

43. (a) List the elements of $U_{42}$ .

(b) List the elements of the subgroup $\langle 25 \rangle$ in $U_{42}$ .

(c) List the cosets of the subgroup $\langle 25 \rangle$ in $U_{42}$ .

(d) Is the quotient group isomorphic to $\integer_2 \times \integer_2$ or $\integer_4$ ?

44. Find the primary decomposition and the invariant factor decomposition for $\integer_{24} \times \integer_{28} \times \integer_{21}$ .

45. What is the largest possible order of an element of $\integer_{30} \times \integer_{45} \times \integer_{60}$ ?

46. Let $f, g: G \to H$ be group maps. Let

$$E = \left\{x \in G \Bigm| f(x) = g(x)\right\}.$$

Prove that E is a subgroup of G. (E is called the equalizer of f and g.)

47. Let R be a ring such that for each $r \in R$ , there is a unique element $s \in R$ such that $r s r = r$ . Prove that R has no zero divisors.

48. Suppose $f: R \to S$ is a ring homomorphism and R and S are rings with identity, but do not assume that $f(1_R) = 1_S$ . Prove that if f is surjective, then $f(1_R) = 1_S$ .

49. Factor $3 x^3 + 2 x^2 + 3 x + 2$ in $\integer_5[x]$ .

50. Find the remainder when $x^{41} + 3 x^{39} + 4 x^{11} + 2 x^9 + 5 x + 3$ is divided by $x + 4$ in $\integer_5[x]$ .

51. Calvin Butterball thinks $x^2 + 1 \in \integer_2[x]$ is irreducible, based on the fact that solving $x^2 + 1 = 0$ gives $x = \pm i$ , which are complex numbers. Is he right?

52. Find the greatest common divisor of $x^4 + x^3 + x^2 + 2 x + 3$ and $x^3 + 4 x^2 + 2 x + 3$ in $\integer_5[x]$ and express the greatest common divisor as a linear combination (with coefficients in $\integer_5[x]$ ) of the two polynomials.

53. The following set is an ideal in the ring $\integer_2 \times \integer_8$ :

$$I = \{(0, 0), (0, 4), (1, 0), (1, 4)\}.$$

(a) List the cosets of I in $\integer_2 \times \integer_8$ .

(b) Construct addition and multiplication tables for the quotient ring $\dfrac{\integer_2 \times \integer_8}{I}$ .

(c) Is $\dfrac{\integer_2 \times \integer_8}{I}$ an integral domain?

Solutions to the Review Problems for the Final

1. Prove that if n is an integer, then $(4 n + 6, 3 n + 4)$ is either 1 or 2. Give specific examples which show that both cases can occur.

Note that

$$3 (4 n + 6) - 4 (3 n + 4) = 2.$$

Now $(4 n + 6, 3 n + 4)$ divides $4 n + 6$ and $3 n + 4$ , so it divides $3 (4 n + 6) - 4 (3 n + 4)$ , and hence it divides 2. The only positive integers that divide 2 are 1 and 2. Hence, $(4 n + 6, 3 n + 4)$ is either 1 or 2.

If $n = 1$ , I have $4 n + 6 = 10$ and $3 n + 4 = 7$ , and $(10, 7) = 1$ .

If $n = 2$ , I have $4 n + 6 = 14$ and $3 n + 4 = 10$ , and $(14, 10) = 2$ .

This shows that both cases can occur.

2. Find the greatest common divisor of 847 and 133 and write it as a linear combination with integer coefficients of 847 and 133.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 847 & & - & & 51 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 133 & & 6 & & 8 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 49 & & 2 & & 3 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 35 & & 1 & & 2 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 14 & & 2 & & 1 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 7 & & 2 & & 0 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The greatest common divisor is 7, and

$$7 = (-8)(847) + (51)(133).\quad\halmos$$

3. Show that the following set is a subgroup of $GL(2, \real)$ :

$$H = \left\{\left[\matrix{x & 0 \cr 0 & y \cr}\right] \Bigm| x, y \in \real, \quad x y \ne 0\right\}$$

However, show that it is not a normal subgroup of $GL(2, \real)$ .

Since $\displaystyle \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] \in H$ , H contains the identity.

If $\displaystyle \left[\matrix{x & 0 \cr 0 & y \cr}\right] \in H$ , then

$$\left[\matrix{x & 0 \cr 0 & y \cr}\right] = \left[\matrix{x^{-1} & 0 \cr 0 & y^{-1} \cr}\right] \in H.$$

(Note that $x y \ne 0$ implies $x \ne 0$ and $y \ne 0$ , so $x^{-1}$ and $y^{-1}$ are defined.) Therefore, H is closed under taking inverses.

Finally,

$$\left[\matrix{x & 0 \cr 0 & y \cr}\right] \left[\matrix{x' & 0 \cr 0 & y' \cr}\right] = \left[\matrix{x x' & 0 \cr 0 & y y' \cr}\right] \in H.$$

(If $x y \ne 0$ and $x' y' \ne 0$ , then $x, x', y, y' \ne 0$ , so $x x' \ne 0$ and $y y' \ne 0$ .) Thus, H is closed under products. Hence, H is a subgroup.

However,

$$\left[\matrix{1 & 2 \cr 1 & 3 \cr}\right] \left[\matrix{1 & 0 \cr 0 & 2 \cr}\right] \left[\matrix{1 & 2 \cr 1 & 3 \cr}\right]^{-1} = \left[\matrix{1 & 2 \cr 1 & 3 \cr}\right] \left[\matrix{1 & 0 \cr 0 & 2 \cr}\right] \left[\matrix{3 & -2 \cr -1 & 1 \cr}\right] = \left[\matrix{-1 & 2 \cr -3 & 4 \cr}\right] \notin H.$$

Therefore, H is not a normal subgroup.

4. Consider the map $\phi: \integer_8 \rightarrow \integer_8$ given by $\phi(n) = n + 2 \mod{8}$ . Is $\phi$ a group homomorphism? Why or why not?

A group homomorphism must map the identity in the domain to the identity in the range. The identity in $\integer_8$ is 0. However, $\phi(0) = 2$ . Therefore, $\phi$ is not a homomorphism.

5. Consider the map $\phi: \integer \rightarrow \integer$ given by $\phi(n) = n^2$ . Is $\phi$ a group homomorphism? Why or why not?

In this case, $\phi(0) = 0$ , so $\phi$ does map the identity to the identity. However,

$$\phi(1 + 1) = \phi(2) = 2^2 = 4, \quad\hbox{but}\quad \phi(1) + \phi(1) = 1 + 1 = 2.$$

Since $\phi(a + b) \ne \phi(a) + \phi(b)$ for all a and b, $\phi$ is not a homomorphism.

6. $\integer \times \integer$ is a group under componentwise addition and $\integer$ is a group under addition. Prove that

$$\dfrac{\integer \times \integer}{\langle (7, 25) \rangle} \approx \integer.$$

Define $f: \integer \times \integer \to \integer$ by

$$f(x, y) = 25 x - 7 y.$$

f can be represented by matrix multiplication:

$$\left(\left[\matrix{x \cr y \cr}\right]\right) = \left[\matrix{25 & -7 \cr}\right] \left[\matrix{x \cr y \cr}\right].$$

Hence, it's a group map.

Let $n (7, 25) = (7 n, 25 n) \in \langle (7, 25) \rangle$ . Then

$$f((7 n, 25 n) = 25 (7 n) - 7(25 n) = 0.$$

Thus, $\langle (7, 25) \rangle \subset \ker f$ .

Let $(x, y) \in \ker f$ . Then

$$\eqalign{ f(x, y) & = 0 \cr 25 x - 7 y & = 0 \cr 25 x & = 7 y \cr}$$

Now $25 \mid 7 y$ but $(7, 25) = 1$ . By Euclid's lemma, $25 \mid y$ . Say $y = 25 n$ . Then

$$25 x = 7 (25 n), \quad\hbox{so}\quad x = 7 n.$$

Therefore,

$$(x, y) = (7 n, 25 n) = n (7, 25) \in \langle (7, 25) \rangle.$$

Thus, $\ker f \subset \langle (7, 25) \rangle$ .

Hence, $\langle (7, 25) \rangle = \ker f$ .

Let $z \in \integer$ . Note that

$$1 = (25, -7) = 2 \cdot 25 + 7 \cdot (-7).$$

Multiplying by z, I get

$$z = 25 (2 z) - 7 (7 z).$$

Then

$$f(2 z, 7 z) = 25 (2 z) - 7 (7 z) = z.$$

This proves that $\im f = \integer$ .

Hence,

$$\dfrac{\integer \times \integer}{\langle (7, 25) \rangle} = \dfrac{\integer \times \integer}{\ker f} \approx \im f = \integer.\quad\halmos$$

7. $\real^2$ is a group under componentwise addition and $\real$ is a group under addition. Let

$$H = \left\{x \cdot (19, -\sqrt{7}) \Bigm| x \in \real\right\}.$$

Prove that $\dfrac{\real^2}{H} \approx \real$ .

Define $f: \real^2 \to \real$ by

$$f(x, y) = \sqrt{7} x + 19 y.$$

Note that

$$f\left(\left[\matrix{x \cr y \cr}\right]\right) = \left[\matrix{\sqrt{7} & 19 \cr}\right] \left[\matrix{x \cr y \cr}\right].$$

Since f can be expressed as multiplication by a constant matrix, it's a linear transformation, and hence a group map.

Let $x \cdot (19, -\sqrt{7}) \in H$ . Then

$$f[x \cdot (19, -\sqrt{7})] = f(19 x, -\sqrt{7} x) = \sqrt{7}(19 x) + 19(-\sqrt{7} x) = 0.$$

Therefore, $x \cdot (19, -\sqrt{7}) \in \ker f$ , and hence $H \subset \ker f$ .

Let $(x, y) \in \ker f$ . Then

$$\eqalign{ f(x, y) & = 0 \cr \sqrt{7} x + 19 y & = 0 \cr 19 y & = -\sqrt{7} x \cr \noalign{\vskip2pt} y & = -\dfrac{\sqrt{7}}{19} x \cr}$$

Hence,

$$(x, y) = \left(x, -\dfrac{\sqrt{7}}{19} x\right) = \dfrac{1}{19} x \cdot (19, -\sqrt{7}) \in H.$$

Therefore, $\ker f \subset H$ . Hence, $\ker f = H$ .

Let $z \in \real$ . Note that

$$f\left(\dfrac{1}{\sqrt{7}} z, 0\right) = \sqrt{7} \cdot \dfrac{1}{\sqrt{7}} z + 19 \cdot 0 = z.$$

Hence, $\im f = \real$ .

Thus,

$$\dfrac{\real^2}{H} = \dfrac{\real^2}{\ker f} \approx \im f = \real.\quad\halmos$$

8. $\integer \times \integer$ and $\integer \times \integer \times \integer$ are groups under componentwise addition. Let

$$H = \{t \cdot (-4, 3, 1) \mid t \in \integer\}.$$

Show that

$$\dfrac{\integer \times \integer \times \integer}{H} \approx \integer \times \integer.$$

Define $f: \integer \times \integer \times \integer \to \integer \times \integer$ by

$$f(x, y, z) = (x + 4 z, y - 3 z).$$

Note that

$$f\left(\left[\matrix{x \cr y \cr z \cr}\right]\right) = \left[\matrix{ 1 & 0 & 4 \cr 0 & 1 & -3 \cr}\right] \left[\matrix{x \cr y \cr z \cr}\right].$$

Since f can be written as multiplication by a constant matrix, it is a group map.

Let $t \cdot (-4, 3, 1) = (-4 t, 3 t, t) \in H$ . Then

$$f(-4 t, 3 t, t) = (-4 t + 4 t, 3 t - 3 t) = (0, 0).$$

Hence, $(-4 t, 3 t, t) \in \ker f$ , so $H \subset \ker f$ .

Let $(x, y, z) \in \ker f$ . Then

$$\eqalign{ f(x, y, z) & = (0, 0) \cr (x + 4 z, y - 3 z) & = (0, 0) \cr}$$

This gives $x + 4 z = 0$ and $y - 3 z = 0$ . The first equation gives $x = -4 z$ and the second equation gives $y = 3 z$ . Hence,

$$(x, y, z) = (-4 z, 3 z, z) \in H.$$

Therefore, $\ker f \subset H$ , and hence $\ker f = H$ .

Let $(a, b) \in \integer \times \integer$ . Then

$$f(a, b, 0) = (a, b).$$

Hence, f is surjective, and $\im f = \integer \times \integer$ .

Therefore,

$$\dfrac{\integer \times \integer \times \integer}{H} = \dfrac{\integer \times \integer \times \integer}{\ker f} \approx \im f = \integer \times \integer.\quad\halmos$$

9. Here is the multiplication table for the Klein 4-group V:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & a & & 1 & & c & & b & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & b & & b & & c & & 1 & & a & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & c & & c & & b & & a & & 1 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Write down all the subgroups of V.

By Lagrange's theorem, the order of a subgroup must divide the order of the group. Hence, there could be subgroups of order 1, 2, or 4.

The subgroup of order 1 is $\{1\}$ ; the subgroup of order 4 is the whole group. A subgroup of order 2 must contain the identity and another element; by closure under inverses, the other element must be its own inverse. Hence, the subgroups of V are:

$$V, \{1, a\}, \{1, b\}, \{1, c\}, \{1\}.\quad\halmos$$

10. Find all integer solutions $(x, y)$ to

$$x^2 - y^2 + 2 y = 12.$$

$$\eqalign{ x^2 - y^2 + 2 y & = 12 \cr x^2 - y^2 + 2 y - 1 & = 12 - 1 \cr x^2 - (y - 1)^2 & = 11 \cr (x - (y - 1))(x + (y - 1)) & = 11 \cr (x - y + 1)(x + y - 1) & = 11 \cr}$$

This equation expresses 11 as a product of two integers $x - y + 1$ and $x + y - 1$ . There are four ways to do this.

Case 1.

$$\eqalign{ x - y + 1 & = 11 \cr x + y - 1 & = 1 \cr}$$

Solving simultaneously, I get $x = 6$ and $y = -4$ .

Case 2.

$$\eqalign{ x - y + 1 & = 1 \cr x + y - 1 & = 11 \cr}$$

Solving simultaneously, I get $x = 6$ and $y = 6$ .

Case 3.

$$\eqalign{ x - y + 1 & = -1 \cr x + y - 1 & = -11 \cr}$$

Solving simultaneously, I get $x = -6$ and $y = -4$ .

Case 4.

$$\eqalign{ x - y + 1 & = -11 \cr x + y - 1 & = -1 \cr}$$

Solving simultaneously, I get $x = -6$ and $y = 6$ .

The solutions are $(6, -4)$ , $(6, 6)$ , $(-6, -4)$ , and $(-6, 6)$ .

11. Find an element of order 30 in $\integer_{25} \times \integer_{12}$ .

5 has order 5 in $\integer_{25}$ .

2 has order 6 in $\integer_{12}$ .

Hence, $(5, 2)$ has order $[5, 6] = 30$ in $\integer_{25} \times \integer_{12}$ .

12. Find the primary decomposition of $U_{16}$ .

$$U_{16} = \{1, 3, 5, 7, 9, 11, 13, 15\}.$$

The operation is multiplication mod 16. The possibilities are

$$\integer_2 \times \integer_2 \times \integer_2, \quad \integer_2 \times \integer_4, \quad \integer_8.$$

I start computing the orders of elements. The order of an element can be 1, 2, 4, 8, or 16, so I can repeatedly square until I get the identity.

$$3^2 = 9, \quad 3^4 = 1.$$

Since 3 has order 4, and since every element of $\integer_2 \times \integer_2 \times \integer_2$ has order 2 or less, $\integer_2 \times \integer_2 \times \integer_2$ is ruled out.

$$5^2 = 9, \quad 5^4 = 1.$$

$$7^2 = 1.$$

$$9^2 = 1.$$

$$11^2 = 9, \quad 11^4 = 1.$$

$$13^2 = 9, \quad 13^4 = 1.$$

$$15^2 = 1.$$

Since there are no elements of order 8, the group can't be $\integer_8$ . Hence, $U_{16} \approx \integer_2 \times \integer_4$ .

13. (a) What is the order of the element $a^4$ in the cyclic group

$$\{a^k \mid a^{22} = 1\}?$$

(b) What is the order of the element 10 in $\integer_{45}$ ?

(c) What elements generate the cyclic group $\integer_{12}$ ?

(a) The order of $a^k$ in the cyclic group of order n with generator a is $\dfrac{n}{(n, k)}$ . So the order of $a^4$ in $\{a^k \mid a^{22} = 1\}$ is

$$\dfrac{22}{(4, 22)} = \dfrac{22}{2} = 11.\quad\halmos$$

(b) The order of 10 in $\integer_{45}$ is

$$\dfrac{45}{(45, 10)} = \dfrac{45}{5} = 9.\quad\halmos$$

(c) The order of the element $m \in \integer_{12}$ is $\dfrac{12}{(m, 12)}$ . If m generates $\integer_{12}$ , it must have order 12, so

$$\dfrac{12}{(m, 12)} = 12.$$

This implies that $(m, 12) = 1$ ; that is, m is relatively prime to 12. Therefore, the generators are $\{1, 5, 7, 11\}$ .

14. Subgroups of cyclic groups are cyclic. Give an example of an abelian group which is not cyclic, but in which every proper subgroup is cyclic.

V is not cyclic, since there are no elements of order 4. However, every subgroup of V is cyclic.

15. (a) Prove that a group cannot be the union of two proper subgroups.

(b) Find a group which is a union of three proper subgroups.

(a) Suppose G is a group, H and K are proper subgroups, and $G = H \cup K$ . Since H is not all of G, I can find an element $k \in K$ such that $k \notin H$ . Likewise, I can find an element $h \in H$ such that $h \notin K$ .

Now consider the element $h k$ . It's in G, so it's either in H or K. But $h k = h' \in H$ gives $k = h^{-1} h' \in H$ , contradicting the assumption that $k \notin H$ . And $h k = k' \in K$ gives $h = k' k^{-1} \in K$ , which contradicts the assumption that $h \notin K$ .

Therefore, G cannot be the union of H and K.

(b) Consider the Klein 4-group V:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & a & & b & & c & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & a & & 1 & & c & & b & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & b & & b & & c & & 1 & & a & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & c & & c & & b & & a & & 1 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

V is the union of the proper subgroups $\{1, a\}$ , $\{1, b\}$ , and $\{1, c\}$ .

16. Let $\phi: G \to H$ be a group homomorphism. Prove that $\phi$ is injective if and only if $\ker \phi = \{1\}$ .

Suppose that $\phi$ is injective. (This means that different inputs go to different outputs, or alternatively, that $\phi(x) = \phi(y)$ implies $x = y$ .) I want to show that $\ker \phi = \{1\}$ .

Since $\{1\} \subset \ker \phi$ , I need to show $x \in \ker \phi$ implies $x = 1$ . Therefore, take $x \in \ker \phi$ , so $\phi(x) = 1$ . Now $\phi(1) = 1$ , so $\phi(x) = 1 = \phi(1)$ . Since $\phi$ is injective, this implies that $x = 1$ , which is what I wanted to show.

Conversely, suppose $\ker \phi = \{1\}$ . I want to show that $\phi$ is injective. To do this, suppose $\phi(x) = \phi(y)$ . I need to show $x = y$ . Rearrange the equation:

$$\phi(x) = \phi(y), \quad \phi(x)^{-1} \phi(x) = \phi(x)^{-1} \phi(y), \quad 1 = \phi(x)^{-1} \phi(y), \quad 1 = \phi(x^{-1}) \phi(y), \quad 1 = \phi(x^{-1} y).$$

But this means that $x^{-1}y \in \ker \phi = \{1\}$ , i.e.

$$x^{-1}y = 1, \quad\hbox{so}\quad x = y.$$

Therefore, $\phi$ is injective.

17. Is there a group homomorphism $\phi: \integer_6 \rightarrow \integer_{12}$ such that $\ker \phi = \{0\}$ ? Construct such a homomorphism, or show that such a homomorphism cannot exist.

If $\phi: \integer_6 \rightarrow \integer_{12}$ is a homomorphism such that $\ker \phi = \{0\}$ , then $\phi$ is 1-1. Since the image of $\phi$ will be isomorphic to $\integer_6/\{0\} \approx \integer_6$ , the image of such a map must be a cyclic subgroup of order 6.

The only subgroup of order 6 in $\integer_{12}$ is

$$\{0, 2, 4, 6, 8, 10\}.$$

The only possibility is that $\phi$ maps $\integer_6$ isomorphically onto this subgroup. Such an isomorphism must send the generator $1 \in \integer_6$ to a generator of $\{0, 2, 4, 6, 8, 10\}$ . Since 2 generates $\{0, 2, 4, 6, 8, 10\}$ , I will try $\phi(1) = 2$ .

Since $\phi$ is supposed to be a group map, this forces $\phi(x) = 2 x \mod{12}$ for $x \in \integer_6$ . Then if $x, y \in \integer_6$ ,

$$\phi(x + y) = 2(x + y) \mod{12} = (2 x + 2 y) \mod{12} = 2 x \mod{12} + 2 y \mod{12} = \phi(x) + \phi(y).$$

Hence, $\phi$ is a group map.

Finally, the only element of $\integer_6$ that maps to 0 is 0, by inspection. Thus, $\ker \phi = \{0\}$ , and $\phi$ satisfies the conditions of the problem.

18. (a) Give an example of a finite group which is not abelian.

(b) Give an example of an abelian group which is not finite.

(c) Give an example of a group which is neither finite nor abelian.

(a) $S_3$ is finite, but not abelian.

(b) $\integer$ is abelian, but not finite.

(c) $GL(2, \real)$ is an infinite group which is not abelian. For example,

$$\left[\matrix{1 & 2 & 1 & 0 \cr}\right] \left[\matrix{0 & 1 \cr 0 & 1 \cr}\right] = \left[\matrix{0 & 3 \cr 0 & 1 \cr}\right] \quad\hbox{but}\quad \left[\matrix{0 & 1 \cr 0 & 1 \cr}\right] \left[\matrix{1 & 2 & 1 & 0 \cr}\right] = \left[\matrix{1 & 0 \cr 1 & 0 \cr}\right].\quad\halmos$$

19. Let $SL(2, \real)$ denote the subgroup of $GL(2, \real)$ consisting of matrices of determinant 1. Show that the following matrices lie in the same left coset of $SL(2, \real)$ :

$$\left[\matrix{1 & 1 \cr 1 & 4 \cr}\right] \quad\hbox{and}\quad \left[\matrix{5 & 2 \cr 11 & 5 \cr}\right].$$

If H is a subgroup of a group G, then $a H = b H$ if and only if $b^{-1} a \in H$ . In this case,

$$\left[\matrix{5 & 2 \cr 11 & 5 \cr}\right]^{-1} \left[\matrix{1 & 1 \cr 1 & 4 \cr}\right] = \dfrac{1}{3} \left[\matrix{5 & -2 \cr -11 & 5 \cr}\right] \left[\matrix{1 & 1 \cr 1 & 4 \cr}\right] = \dfrac{1}{3} \left[\matrix{3 & -3 \cr -6 & 9 \cr}\right] = \left[\matrix{1 & -1 \cr -2 & 3 \cr}\right].$$

Now

$$\det \left[\matrix{1 & -1 \cr -2 & 3 \cr}\right] = 3 - 2 = 1.$$

Hence,

$$\left[\matrix{5 & 2 \cr 11 & 5 \cr}\right]^{-1} \left[\matrix{1 & 1 \cr 1 & 4 \cr}\right] = \left[\matrix{1 & -1 \cr -2 & 3 \cr}\right] \in SL(2, \real).$$

This shows that the matrices lie in the same left coset of $SL(2, \real)$ .

20. Give an example of a finite commutative ring with 1 which is not an integral domain.

$\integer_4$ is finite, commutative, and has a multiplicative identity 1. But $2\cdot 2 = 0$ , so it's not a domain.

21. (a) Define $f: \real \times \real \to \real$ by

$$f(x, y) = x^3 + y^3.$$

Show that f is surjective.

(b) Define $g: \real \times \real \times \real \to \real$ by

$$g(x, y,z) = x - 2 y + 3 z.$$

Show that g is surjective.

(c) Define $h: M(2, \real) \to M(2, \real)$ by

$$h\left(\left[\matrix{a & b \cr c & d \cr}\right]\right) = \left[\matrix{a & 2 b \cr 3 c & 4 d \cr}\right].$$

Show that h is surjective.

(d) Define $k: \real \to \real \times \real$ by

$$k(x) = (x, x).$$

Show that k is not surjective.

(e) Give an example of a group map $p: \integer \to \integer$ which is not surjective, and a surjective function $q: \integer \to \integer$ which is not a group map.

(a) Let $z \in \real$ . Then

$$f(\root 3 \of z, 0) = (\root 3 \of z)^3 + 0^3 = z.$$

Therefore, f is surjective.

(b) Let $w \in \real$ . Then

$$g(w, 0, 0) = w - 2 \cdot 0 + 3 \cdot 0 = w.$$

Therefore, g is surjective.

(c) Let $\displaystyle \left[\matrix{w & x \cr y & z \cr}\right] \in M(2, \real)$ . Then

$$h\left(\left[\matrix{w & \dfrac{x}{2} \cr \noalign{\vskip2 pt} \dfrac{y}{3} & \dfrac{z}{4} \cr}\right]\right) = \left[\matrix{w & x \cr y & z \cr}\right].$$

Therefore, h is surjective.

(d) $(\sqrt{17}, \pi) \in \real \times \real$ . But if

$$k(x) = (\sqrt{17}, \pi) \quad\hbox{then}\quad x = \sqrt{17} \quad\hbox{and}\quad x = \pi.$$

This contradiction shows that there is no $x \in \real$ such that $k(x) = (\sqrt{17}, \pi)$ . Hence, k is not surjective.

(e) The function $p: \integer \to \integer$ defined by $p(x) = 2 x$ is a group map, since

$$p(a + b) = 2( a + b) = 2 a + 2 b = p(a) + p(b).$$

However, p is not surjective, since (for example) there is no $n \in \integer$ such that $p(n) = 1$ .

The function $q: \integer \to \integer$ given by $q(x) = x + 1$ is surjective: If $y \in \integer$ , then

$$q(y - 1) = (y - 1) + 1 = y.$$

But q is not a group map: $q(0) = 1$ , so q does not map the identity to the identity.

For that matter, the identity map $\hbox{id}: \integer \to \integer$ is a surjective group map, and the function $r: \integer \to \integer$ given by $r(x) = x^2$ is neither surjective nor a group map. The properties of surjectivity and being a group map are independent.

22. (a) Explain why $\rational$ is not a group under multiplication.

(b) Do the nonzero elements of $\integer_6$ form a group under multiplication mod 6?

(c) Show that the nonzero elements of $\integer_5$ form a group under multiplication mod 5. What group?

(a) $\rational$ is not a group under multiplication because not every element has a multiplicative inverse. To be specific, $0 \in \rational$ does not have a multiplicative inverse.

(b) $\{1, 2, 3, 4, 5\}$ is not a group under multiplication mod 6, because it is not closed under the operation: $2 \cdot 3 = 0 \notin \{1, 2, 3, 4, 5\}$ , for instance.

(c) Here is the operation table:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\cdot$ & & 1 & & 2 & & 3 & & 4 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & 2 & & 3 & & 4 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 2 & & 2 & & 4 & & 1 & & 3 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 3 & & 3 & & 1 & & 4 & & 2 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 4 & & 4 & & 3 & & 2 & & 1 & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The table shows that set is closed under the operation. Take for granted that multiplication mod 6 is associative (since $\integer_6$ is a ring under addition and multiplication mod 6). 1 is the identity element. The inverse of 2 is 3, the inverse of 3 is 2, and 4 is its own inverse. Therefore, this set is a group; it's usually denoted $\integer_5^*$ .

$\integer_5^*$ a group with 4 elements. and the table shows that not every element has order 2. Therefore, $\integer_5^*$ is not isomorphic to $\integer_2 \times \integer_2$ ; it must be isomorphic to $\integer_4$ .

23. Reduce $32^{2011} \mod{41}$ to an integer in the set $\{0, 1, \ldots, 40\}$ .

$41 \notdiv 32$ , so by Fermat's theorem, $32^{40} = 1 \mod {41}$ . Therefore,

$$32^{2011} = 32^{2000}\cdot 32^{11} = (32^{40})^{50}\cdot 32^{11} =$$

$$1^{50} \cdot (32^5)^2 \cdot 32 = 33554432^2 \cdot 32 \mod{41}.$$

Since $33554432 = 818400\cdot 41 + 32$ , $33554432 = 32 \mod{41}$ . Hence,

$$33554432^2 \cdot 32 = 32^3 = 9 \mod{41}.\quad\halmos$$

24. Reduce $\dfrac{148!}{3 \cdot 75} \mod{149}$ to an integer in the set $\{0, 1, \ldots, 148\}$ . (Note: 149 is prime.)

$$\eqalign{ x & = \dfrac{148!}{3 \cdot 75} \mod{149} \cr \noalign{\vskip2 pt} 3 \cdot 75 x & = 148! = -1 \mod{149} \cr}$$

At this point, you could use the Extended Euclidean Algorithm to find the inverses of 3 and 75 mod 149. But it's easier to note that

$$150 = 149 + 1 = 1 \mod{149}.$$

Since $2 \cdot 75 = 150$ and $50 \cdot 3 = 150$ , I have

$$\eqalign{ 2 \cdot 50 \cdot 3 \cdot 75 x & = 2 \cdot 50 \cdot (-1) \mod{149} \cr x & = -100 = 49 \mod{149} \quad\halmos \cr}$$

25. The definition of a subring of a ring does not require that you check associativity for addition or multiplication. Explain why.

When you consider a subset S of a ring R, addition and multiplication are associative as operations in R. In showing that S is a subring, you're confining the operations to a subset, so they must continue to be associative.

(People often say that associativity is inherited from R by S.) For similar reasons, the definition of a subgroup does not require that you check associativity.

26. Prove that if I is an ideal in a ring R with identity and $1 \in I$ , then $I = R$ .

Since $I \subset R$ by definition, I only need to prove the opposite containment. Let $r \in R$ . Now $1 \in I$ , so $r \cdot 1 \in I$ , i.e. $r \in I$ . Hence, $R \subset I$ , so $I = R$ .

27. Show that the only (two-sided) ideals in $M(2, \real)$ are the zero ideal and the whole ring.

Let S be an ideal in $M(2, \real)$ , and suppose S is nonzero. I'll show that $S = M(2, \real)$ .

S contains a nonzero matrix A. If A is invertible, then $A \in S$ implies $A^{-1}A \in S$ , i.e. $I \in S$ , where I is the identity matrix. By the last problem, this implies that $S = M(2, \real)$ .

Suppose then that A is not invertible. Any $2 \times 2$ matrix row reduces to one of the following:

$$\left[\matrix{0 & 0 \cr 0 & 0 \cr}\right], \quad \left[\matrix{1 & \ast \cr 0 & 0 \cr}\right], \quad \left[\matrix{0 & 1 \cr 0 & 0 \cr}\right], \quad \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right].$$

A is not invertible, so it doesn't row reduce to I; it's nonzero, so it doesn't row reduce to the zero matrix.

Suppose A row reduces to $\displaystyle \left[\matrix{1 & \ast \cr 0 & 0 \cr}\right]$ . There are elementary matrices $E_1$ , ..., $E_k$ such that

$$E_1\cdots E_k A = \left[\matrix{1 & \ast \cr 0 & 0 \cr}\right].$$

Since $A \in S$ , this equation shows that $\displaystyle \left[\matrix{1 & \ast \cr 0 & 0 \cr}\right] \in S$ .

Since S is an ideal,

$$\left[\matrix{1 & \ast \cr 0 & 0 \cr}\right] \left[\matrix{1 & -\ast \cr 0 & 1 \cr}\right] = \left[\matrix{1 & 0 \cr 0 & 0 \cr}\right] \in S.$$

Again, since S is an ideal,

$$\left[\matrix{0 & 1 \cr 1 & 0 \cr}\right] \left[\matrix{1 & 0 \cr 0 & 0 \cr}\right] = \left[\matrix{0 & 0 \cr 1 & 0 \cr}\right] \in S.$$

And again, since S is an ideal,

$$\left[\matrix{0 & 0 \cr 1 & 0 \cr}\right] \left[\matrix{0 & 1 \cr 1 & 0 \cr}\right] = \left[\matrix{0 & 0 \cr 0 & 1 \cr}\right] \in S.$$

Hence,

$$\left[\matrix{1 & 0 \cr 0 & 0 \cr}\right] + \left[\matrix{0 & 0 \cr 0 & 1 \cr}\right] = \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] \in S.$$

Hence, $S = M(2, \real)$ .

A similar argument shows that if A row reduces to $\displaystyle \left[\matrix{0 & 1 \cr 0 & 0 \cr}\right]$ , then $S = M(2, \real)$ .

Therefore, the only ideals in $M(2, \real)$ are the zero ideal and the whole ring.

28. Consider the following subset of the ring $\integer \times \integer$ :

$$S = \{(m + n, m - n) \mid m, n \in \integer\}.$$

Check each axiom for an ideal. If the axiom holds, prove it. If the axiom does not hold, give a specific counterexample.

The zero element is in S, since $(0, 0) = (0 + 0, 0 - 0) \in S$ .

Let $(m + n, m - n) \in S$ . Then

$$-(m + n, m - n) = (-m - n, -m + n) = ((-m) + (-n), (-m) - (-n)) \in S.$$

Let $(a + b, a - b), (c + d, c - d) \in S$ . Then

$$(a + b, a - b) + (c + d, c - d) = ((a + c) + (b + d), (a + c) - (b + d)) \in S.$$

I have $(5, 1) = (3 + 2, 3 - 2) \in S$ . Then

$$(1, 0) \cdot (5, 1) = (5, 0).$$

But $(5, 0) \notin S$ . For suppose $(5, 0) = (m + n, m - n)$ for $m, n \in \integer$ . Then

$$m + n = 5 \quad\hbox{and}\quad m - n = 0.$$

Adding the two equations gives $2 m = 5$ , but this equation has no integer solutions.

Thus, S is not an ideal in $\integer \times \integer$ .

29. (a) Show that $x^2 + x + 1$ is irreducible in $\integer_5[x]$ .

(b) Find $[(2 x + 3) + \langle x^2 + x + 1 \rangle]^{-1}$ in $\dfrac{\integer_5[x]}{\langle x^2 + x + 1 \rangle}$ .

(c) Compute the product of the cosets $\left((x^2 + 2) + \langle x^2 + x + 1\rangle\right) \cdot \left((3 x + 4) + \langle x^2 + x + 1\rangle\right)$ in the quotient ring $\dfrac{\integer_5[x]}{\langle x^2 + x + 1\rangle}$ . Write your answer in the form $(ax + b) + \langle x^2 + x + 1\rangle$ , where $a, b \in \integer_5$ .

(a) Since it's a quadratic, it suffices to show that it has no roots in $\integer_5$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & x & & $x^2 + x + 1 \mod{5}$ & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 0 & & 1 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 1 & & 3 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 2 & & 2 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 3 & & 3 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 4 & & 1 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

It has no roots in $\integer_5$ , so it's irreducible over $\integer_5$ .

(b) In general, you can find an inverse using the Extended Euclidean Algorithm. In this case, the coset representative $2 x + 3$ is linear, so I can just Apply the Division Algorithm:

$$\eqalign{ x^2 + x + 1 & = (3 x + 1)(2 x + 3) + 3 \cr (x^2 + x + 1) - (3 x + 1)(2 x + 3) & = 3 \cr 2(x^2 + x + 1) - 2(3 x + 1)(2 x + 3) & = 2 \cdot 3 \cr 2(x^2 + x + 1) - (x + 2)(2 x + 3) & = 1 \cr 2(x^2 + x + 1) + (4 x + 3)(2 x + 3) & = 1 \cr 2(x^2 + x + 1) + (4 x + 3)(2 x + 3) + \langle x^2 + x + 1 \rangle & = 1 + \langle x^2 + x + 1 \rangle \cr (4 x + 3)(2 x + 3) + \langle x^2 + x + 1 \rangle & = 1 + \langle x^2 + x + 1 \rangle \cr [(4 x + 3) + \langle x^2 + x + 1 \rangle] [(2 x + 3) + \langle x^2 + x + 1 \rangle] & = 1 + \langle x^2 + x + 1 \rangle \cr}$$

Hence,

$$[(2 x + 3) + \langle x^2 + x + 1 \rangle]^{-1} = (4 x + 3) + \langle x^2 + x + 1 \rangle.\quad\halmos$$

(c) First,

$$\left((x^2 + 2) + \langle x^2 + x + 1\rangle\right)\cdot \left((3 x + 4) + \langle x^2 + x + 1\rangle\right) = (x^2 + 2)(3 x + 4) + \langle x^2 + x + 1\rangle =$$

$$(3 x^3 + 4 x^2 + x + 3) + \langle x^2 + x + 1\rangle.$$

Apply the Division Algorithm:

$$3 x^3 + 4 x^2 + x + 3 = (x^2 + x + 1)(3 x + 1) + (2 x + 2).$$

Therefore, the product is

$$(3 x^3 + 4 x^2 + x + 3) + \langle x^2 + x + 1\rangle = (2 x + 2) + \langle x^2 + x + 1\rangle.\quad\halmos$$

30. Factor $x^4 + 64$ in $\rational[x]$ .

The idea is to add a middle term to complete the square, then subtract it back off:

$$x^4 + 64 = x^4 + 16 x^2 + 64 - 16 x^2 = (x^2 + 8)^2 - (4 x)^2 = (x^2 + 8 + 4 x)(x^2 + 8 - 4 x) = (x^2 + 4 x + 8)(x^2 - 4 x + 8).$$

You can check using the Quadratic Formula that $x^2 + 4 x + 8$ and $x^2 - 4 x + 8$ do not factor over $\rational$ .

31. (a) Show that $x^4 + 1$ has no roots in $\integer_5$ .

(b) Show that $x^4 + 1$ factors in $\integer_5[x]$ .

(a)

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & 0 & & 1 & & 2 & & 3 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $x^4 + 1$ & & 1 & & 2 & & 2 & & 2 & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} \quad\halmos $$

(b) $x^4 + 1 = (x^2 + 2)(x^2 + 3)$ .

32. In the ring $\real[x]$ , consider the subset

$$\langle x^2 - x - 2, x^2 - 1 \rangle = \{a(x)(x^2 - x - 2) + b(x)(x^2 - 1) \mid a(x), b(x) \in \real[x]\}.$$

(a) Show that $\langle x^2 - x - 2, x^2 - 1 \rangle$ is an ideal.

(b) Is $x^2 + x + 3$ in $\langle x^2 - x - 2, x^2 - 1 \rangle$ ?

(a) Suppose $a(x) (x^2 - x - 2) + b(x) (x^2 - 1), c(x) (x^2 - x - 2) + d(x) (x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle$ , where $a(x), b(x), c(x), d(x) \in \real[x]$ . Then

$$[a(x) (x^2 - x - 2) + b(x) (x^2 - 1)] + [c(x) (x^2 - x - 2) + d(x) (x^2 - 1)] =$$

$$[a(x) + c(x)](x^2 - x - 2) + [b(x) + d(x)](x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle.$$

I have

$$0 = 0 \cdot (x^2 - x - 2) + 0 \cdot (x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle.$$

Let $a(x) (x^2 - x - 2) + b(x) (x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle$ . Then

$$-[a(x) (x^2 - x - 2) + b(x) (x^2 - 1)] = (-a(x))(x^2 - x - 2) + (-b(x))(x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle.$$

Finally, let $f(x) \in \real[x]$ and let $a(x) (x^2 - x - 2) + b(x) (x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle$ . Then

$$f(x) [a(x) (x^2 - x - 2) + b(x) (x^2 - 1)] = [f(x) a(x)](x^2 - x - 2) + [f(x) b(x)](x^2 - 1) \in \langle x^2 - x - 2, x^2 - 1 \rangle.$$

(Note that $\real[x]$ is commutative, so I only need to check multiplication on the left.) Hence, $\langle x^2 - x - 2, x^2 - 1 \rangle$ is an ideal.

(b) The greatest common divisor of $x^2 - x - 2 = (x - 2)(x + 1)$ and $x^2 - 1 = (x - 1)(x + 1)$ is $x + 1$ , and it must divide any linear combination $a(x) (x^2 - x - 2) + b(x) (x^2 - 1)$ . Suppose then that

$$x^2 + x + 3 = a(x) (x^2 - x - 2) + b(x) (x^2 - 1).$$

Then $x + 1 \mid x^2 + x + 3$ . But in fact,

$$x^2 + x + 3 = x (x + 1) + 3.$$

Thus, $x + 1 \notdiv x^2 + x + 3$ , and so $x^2 + x + 3$ cannot be an element of $\langle x^2 - x - 2, x^2 - 1 \rangle$ .

33. $x^2 + 2 = (x + 1)(x + 2)$ is a factorization of $x^2 + 2$ into irreducibles in $\integer_3[x]$ . Find a different factorization of $x^2 + 2$ into irreducibles in $\integer_3[x]$ .

Since $2 \cdot 2 = 1$ in $\integer_3$ ,

$$x^2 + 2 = (x + 1)(x + 2) = 2 (x + 1) \cdot 2(x + 2) = (2 x + 2)(2 x + 1).\quad\halmos$$

34. Compute the product of the cycles $(2\ 4\ 6\ 3)(1\ 3\ 4)$ (right to left) and write the result as a product of disjoint cycles.

$$\matrix{1 & 2 & 3 & 4 & 5 & 6 & \cr & & & & & & (1 3 4) \cr 3 & 2 & 4 & 1 & 5 & 6 & \cr & & & & & & (2 4 6 3)\cr 2 & 4 & 6 & 1 & 5 & 3 & \cr}$$

The product is $(1 2 4)(3 6)$ .

35. Define $\phi: \integer[x] \to \integer[x]$ by

$$\phi(f(x)) = f(x)^2.$$

Determine which of the axioms for a ring map are satisfied by $\phi$ . If an axiom is not satisfied, give a specific example which shows that the axiom is violated.

First, $\phi(1) = 1^2 = 1$ .

If $f(x), g(x) \in \integer[x]$ ,

$$\phi\left(f(x) g(x))\right) = (f(x) g(x))^2 = f(x)^2 g(x)^2 = \phi(f(x)) \phi(g(x)).$$

However,

$$\phi(x + x) = \phi(2 x) = 4 x^2, \quad\hbox{but}\quad \phi(x) + \phi(x) = x^2 + x^2 = 2 x^2.$$

Thus, $\phi(x + x) \ne \phi(x) + \phi(x)$ .

36. Define $\phi: \integer_2[x] \to \integer_2[x]$ by

$$\phi(f(x)) = f(x)^2.$$

(a) Show that $\phi$ is a ring map.

(b) Determine the kernel of $\phi$ .

(c) Show that $x^4 + 1 \in \im \phi$ . Is $\phi$ surjective?

(a) If $f(x), g(x) \in \integer[x]$ ,

$$\phi\left(f(x)g(x))\right) = (f(x)g(x))^2 = f(x)^2 g(x)^2 = \phi(f(x))\phi(g(x)),$$

$$\phi\left(f(x) + g(x)\right) = \left(f(x) + g(x)\right)^2 = f(x)^2 + 2 f(x) g(x) + g(x)^2 = f(x)^2 + g(x)^2 = \phi(f(x)) + \phi(g(x)).$$

$$\phi(1) = 1^2 = 1.$$

Therefore, $\phi$ is a ring map.

(b) $\phi(f(x)) = 0$ means $f(x)^2 = 0$ , which is only possible if $f(x) = 0$ (since $\integer_2[x]$ is an integral domain). Therefore, $\ker \phi = \{0\}$ .

(c)

$$x^4 + 1 = x^4 + 2 x^2 + 1 = (x^2 + 1)^2 = \phi(x^2 + 1).$$

$\phi$ is not surjective. If $\phi(f(x)) = x$ , then $f(x)^2 = x$ . This implies $\deg f(x)^2 = \deg x = 1$ , or $2 \deg f(x) = 1$ . Obviously, $2 \notdiv 1$ . This contradiction shows that x is not in the image of $\phi$ , and $\phi$ is not surjective.

37. Find the quotient and the remainder when $2 x^4 + 3 x^3 + x + 1$ is divided by $3 x^2 + 1$ in $\integer_5[x]$ .

$$\hbox{\epsfysize=1.75 in \epsffile{final-review1.eps}}$$

The quotient is $4 x^2 + x + 2$ and the remainder is 4.

38. (a) Explain why $x^4 + 1$ has no roots in $\real$ .

(b) Is $x^4 + 1$ irreducible in $\real[x]$ ?

(a) Since $x^4 \ge 0$ for all x, it follows that $x^4 + 1 \ge 1 > 0$ . In particular, no real value of x makes it 0.

(b)

$$x^4 + 1 = (x^4 + 2 x^2 + 1) - 2 x^2 = (x^2 + 1)^2 - 2 x^2 = (x^2 + 1 - \sqrt{2} x)(x^2 + 1 + \sqrt{2} x).$$

Hence, $x^4 + 1$ is not irreducible in $\real[x]$ .

39. List the zero divisors and the units in $\integer_2 \times \integer_3$ .

$$(0, 1)(1, 0) = (0, 0), \quad (0, 2)(1, 0) = (0, 0), \quad (1, 0)(0, 1) = (0, 0).$$

The zero divisors are $(0, 1)$ , $(0, 2)$ , and $(1, 0)$ .

$$(1, 1)(1, 1) = (1, 1), \quad (1, 2)(1, 2) = (1, 1).$$

The units are $(1, 1)$ and $(1, 2)$ .

40. Prove that if I is a left ideal in a division ring R, then either $I = \{0\}$ or $I = R$ .

Suppose $I \ne \{0\}$ . Then I can find a nonzero element $x \in I$ . Since R is a division ring, x is invertible. Since I is a left ideal, $x^{-1}\cdot x \in I$ . But $x^{-1}\cdot x = 1$ , so $1 \in I$ . An ideal that contains 1 is the whole ring, so $I = R$ .

41. Let R be a ring, and let $r \in R$ . The centralizer $C(r)$ of r is the set of elements of R which commute with r:

$$C(r) = \{a \in R \mid ra = ar\}.$$

Prove that $C(r)$ is a subring of R.

Let $a, b \in C(r)$ , so $ar = ra$ and $br = rb$ . Then

$$(a + b)r = a r + b r = r a + r b = r(a + b).$$

Therefore, $a + b \in C(r)$ .

Since $0 \cdot r = 0 = r \cdot 0$ , I have $0 \in C(r)$ .

Let $a \in C(r)$ , so $a r = r a$ . Then

$$(-a)r = -(a r) = -(r a) = r(-a).$$

Hence, $-a \in C(r)$ .

Let $a, b \in C(r)$ , so $a r = r a$ and $b r = r b$ . Then

$$(a b) r = a (b r) = a (r b) = (a r) b = (r a) b = r (a b).$$

Therefore, $a b \in C(r)$ .

Hence, $C(r)$ is a subring.

42. Let

$$I = \left\{\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] \Bigm| x, y, z \in \real\right\}.$$

Prove that I is a left ideal, but not a right ideal, in the ring $M(3, \real)$ .

$$\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] + \left[\matrix{0 & x' & 0 \cr 0 & y' & 0 \cr 0 & z' & 0 \cr}\right] = \left[\matrix{0 & x + x' & 0 \cr 0 & y + y' & 0 \cr 0 & z + z' & 0 \cr}\right] \in I.$$

Hence, I is closed under sums.

Elements of I are exactly the $3 \times 3$ matrices with all-zero first and third columns. Thus,

$$\left[\matrix{0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr}\right] \in I.$$

If

$$\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] \in I, \quad\hbox{then}\quad -\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] = \left[\matrix{0 & -x & 0 \cr 0 & -y & 0 \cr 0 & -z & 0 \cr}\right] \in I.$$

Thus, I is closed under taking additive inverses.

Let

$$\left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] \in I \quad\hbox{and}\quad \left[\matrix{a & b & c \cr d & e & f \cr g & h & i \cr}\right] \in M(3, \real).$$

Then

$$\left[\matrix{a & b & c \cr d & e & f \cr g & h & i \cr}\right] \left[\matrix{0 & x & 0 \cr 0 & y & 0 \cr 0 & z & 0 \cr}\right] = \left[\matrix{ 0 & a x + b y + c z & 0 \cr 0 & d x + e y + f z & 0 \cr 0 & g x + h y + i z & 0 \cr}\right] \in I.$$

Hence, I is a left ideal.

However,

$$\left[\matrix{0 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 1 & 0 \cr}\right] \in I \quad\hbox{and}\quad \left[\matrix{0 & 0 & 0 \cr 1 & 1 & 1 \cr 0 & 0 & 0 \cr}\right] \in M(3, \real).$$

But

$$\left[\matrix{0 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 1 & 0 \cr}\right] \left[\matrix{0 & 0 & 0 \cr 1 & 1 & 1 \cr 0 & 0 & 0 \cr}\right] = \left[\matrix{1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr}\right] \notin I.$$

Hence, I is not a right ideal.

43. (a) List the elements of $U_{42}$ .

(b) List the elements of the subgroup $\langle 25 \rangle$ in $U_{42}$ .

(c) List the cosets of the subgroup $\langle 25 \rangle$ in $U_{42}$ .

(d) Is the quotient group isomorphic to $\integer_2 \times \integer_2$ or $\integer_4$ ?

(a)

$$U_{42} = \{1, 5, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41\}.\quad\halmos$$

(b)

$$\langle 25 \rangle = \{1, 25, 37\}.\quad\halmos$$

(c)

$$\eqalign{ \langle 25 \rangle & = \{1, 25, 37\} \cr 5 \cdot \langle 25 \rangle & = \{5, 41, 17\} \cr 11 \cdot \langle 25 \rangle & = \{11, 23, 29\} \cr 13 \cdot \langle 25 \rangle & = \{13, 31, 19\} \cr} \quad\halmos$$

(d) Note that

$$5^2 = 25, \quad 11^2 = 37, \quad 13^2 = 1.$$

The results are all elements of the identity coset $\{1, 25, 37\}$ .

So all three cosets have order 2.

$\dfrac{U_{42}}{\langle 25\rangle}$ is isomorphic to $\integer_2 \times \integer_2$ , since every element squares to the identity.

44. Find the primary decomposition and the invariant factor decomposition for $\integer_{24} \times \integer_{28} \times \integer_{21}$ .

$$\eqalign{ \integer_{24} & \approx \integer_3 \times \integer_8 \cr \integer_{28} & \approx \integer_4 \times \integer_7 \cr \integer_{21} & \approx \integer_3 \times \integer_7 \cr}$$

Therefore, the primary decomposition is

$$\integer_4 \times \integer_8 \times \integer_3 \times \integer_3 \times \integer_7 \times \integer_7.$$

Here's the work for the invariant factor decomposition:

$$\matrix{ 4 & 8 \cr 3 & 3 \cr 7 & 7 \cr \noalign{\vskip2 pt\hrule\vskip2 pt} 84 & 168 \cr}$$

The invariant factor decomposition is $\integer_{84} \times \integer_{168}$ .

45. What is the largest possible order of an element of $\integer_{30} \times \integer_{45} \times \integer_{60}$ ?

The primary decomposition is

$$\integer_{30} \times \integer_{45} \times \integer_{60} \approx \integer_2 \times \integer_3 \times \integer_5 \times \integer_9 \times \integer_5 \times \integer_4 \times \integer_3 \times \integer_5.$$

Compute the invariant factor decomposition:

$$\matrix{ & 2 & 4 \cr 3 & 3 & 9 \cr 5 & 5 & 5 \cr \noalign{\vskip2pt \hrule \vskip2pt} 15 & 30 & 180 \cr}$$

The invariant factor decomposition is $\integer_{15} \times \integer_{30} \times \integer_{180}$ . Hence, the largest possible order of an element is 180.

46. Let $f, g: G \to H$ be group maps. Let

$$E = \left\{x \in G \Bigm| f(x) = g(x)\right\}.$$

Prove that E is a subgroup of G. (E is called the equalizer of f and g.)

Let $x, y \in E$ , so $f(x) = g(x)$ and $f(y) = g(y)$ . Then

$$f(x) f(y) = g(x) g(y), \quad\hbox{so}\quad f(x y) = g(x y).$$

Therefore, $x y \in E$ .

Since $f(1) = 1 = g(1)$ , $1 \in E$ .

Let $x \in E$ , so $f(x) = g(x)$ . Then $f(x)^{-1} = g(x)^{-1}$ , so $f(x^{-1}) = g(x^{-1})$ . Hence, $x^{-1} \in E$ .

Therefore, E is a subgroup of G.

47. Let R be a ring such that for each $r \in R$ , there is a unique element $s \in R$ such that $r s r = r$ . Prove that R has no zero divisors.

Suppose that $r \in R$ is a zero divisor, so $r \ne 0$ and $r t = 0$ for some $t \ne 0$ . Let s be the unique element of R such that $r s r = r$ . Then

$$r(s + t)r = r s r + r t r = r + 0 = r.$$

But $x = s$ was the unique solution to $r x r = r$ , so $s = s + t$ , and $t = 0$ . This contradiction implies that there is no such t, so R has no zero divisors.

48. Suppose $f: R \to S$ is a ring homomorphism and R and S are rings with identity, but do not assume that $f(1_R) = 1_S$ . Prove that if f is surjective, then $f(1_R) = 1_S$ .

Since f is surjective, there is an element $e \in R$ such that $f(e) = 1_S$ . Then

$$1_S = f(e) = f(e \cdot 1_R) = f(e) \cdot f(1_R) = 1_S \cdot f(1_R) = f(1_R).\quad\halmos$$

49. Factor $3 x^3 + 2 x^2 + 3 x + 2$ in $\integer_5[x]$ .

If a cubic or quadratic polynomial over a field factors, it must have a linear factor, i.e. a root. Therefore, I'll try the elements of $\integer_5$ to find the roots.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & x & & $3 x^3 + 2 x^2 + 3 x + 2$ & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 0 & & 2 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 1 & & 0 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 2 & & 0 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 3 & & 0 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & \cr & 4 & & 4 & \cr height2 pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

1, 2, and 3 are roots, so $x - 1 = x + 4$ , $x - 2 = x + 3$ , and $x - 3 = x + 2$ are factors. Since the leading coefficient is 3, I must have

$$3 x^3 + 2 x^2 + 3 x + 2 = 3(x + 4)(x + 3)(x + 2).\quad\halmos$$

50. Find the remainder when $x^{41} + 3 x^{39} + 4 x^{11} + 2 x^9 + 5 x + 3$ is divided by $x + 4$ in $\integer_5[x]$ .

Notice that $x + 4 = x - 1$ in $\integer_5[x]$ . By the Remainder Theorem, the remainder is

$$1^{41} + 3 \cdot 1^{39} + 4 \cdot 1^{11} + 2 \cdot 1^9 + 5 \cdot 1 + 3 = 3.\quad\halmos$$

51. Calvin Butterball thinks $x^2 + 1 \in \integer_2[x]$ is irreducible, based on the fact that solving $x^2 + 1 = 0$ gives $x = \pm i$ , which are complex numbers. Is he right?

In fact, since $2 = 0$ in $\integer_2$ ,

$$(x + 1)^2 = x^2 + 2 x + 1 = x^2 + 1.$$

Thus, $x^2 + 1$ is not irreducible in $\integer_2[x]$ .

52. Find the greatest common divisor of $x^4 + x^3 + x^2 + 2 x + 3$ and $x^3 + 4 x^2 + 2 x + 3$ in $\integer_5[x]$ and express the greatest common divisor as a linear combination (with coefficients in $\integer_5[x]$ ) of the two polynomials.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & a & & q & & y & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & $x^4 + x^3 + x^2 + 2 x + 3$ & & & & $x + 2$ & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & $x^3 + 4 x^2 + 2 x + 3$ & & $x + 2$ & & 1 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & $x^2 + 2$ & & $x + 4$ & & 0 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The greatest common divisor is $x + 2$ , and

$$1 \cdot (x^4 + x^3 + x^2 + 2 x + 3) - (x + 2)(x^3 + 4 x^2 + 2 x + 3) = x + 2.\quad\halmos$$

53. The following set is an ideal in the ring $\integer_2 \times \integer_8$ :

$$I = \{(0, 0), (0, 4), (1, 0), (1, 4)\}.$$

(a) List the cosets of I in $\integer_2 \times \integer_8$ .

(b) Construct addition and multiplication tables for the quotient ring $\dfrac{\integer_2 \times \integer_8}{I}$ .

(c) Is $\dfrac{\integer_2 \times \integer_8}{I}$ an integral domain?

(a)

$$\eqalign{ I & = \{(0, 0), (0, 4), (1, 0), (1, 4)\} \cr (0, 1) + I & = \{(0, 1), (0, 5), (1, 1), (1, 5)\} \cr (0, 2) + I & = \{(0, 2), (0, 6), (1, 2), (1, 6)\} \cr (0, 3) + I & = \{(0, 3), (0, 7), (1, 3), (1, 7)\} \cr}\quad\halmos$$

(b) I will let $(0, 0)$ , $(0, 1)$ , $(0, 2)$ , and $(0, 3)$ stand for their respective cosets.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & + & & $(0, 0)$ & & $(0, 1)$ & & $(0, 2)$ & & $(0, 3)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 0)$ & & $(0, 0)$ & & $(0, 1)$ & & $(0, 2)$ & & $(0, 3)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 1)$ & & $(0, 1)$ & & $(0, 2)$ & & $(0, 3)$ & & $(0, 0)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 2)$ & & $(0, 2)$ & & $(0, 3)$ & & $(0, 0)$ & & $(0, 1)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 3)$ & & $(0, 3)$ & & $(0, 0)$ & & $(0, 1)$ & & $(0, 2)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\cdot$ & & $(0, 0)$ & & $(0, 1)$ & & $(0, 2)$ & & $(0, 3)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 0)$ & & $(0, 0)$ & & $(0, 0)$ & & $(0, 0)$ & & $(0, 0)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 1)$ & & $(0, 0)$ & & $(0, 1)$ & & $(0, 2)$ & & $(0, 3)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 2)$ & & $(0, 0)$ & & $(0, 2)$ & & $(0, 0)$ & & $(0, 2)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $(0, 3)$ & & $(0, 0)$ & & $(0, 3)$ & & $(0, 2)$ & & $(0, 1)$ & \cr height2 pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$

(c) Since $((0, 2) + I)((0, 2) + I) = I$ (which is the zero element in $\dfrac{\integer_2 \times \integer_8}{I}$ ), the quotient ring $\dfrac{\integer_2 \times \integer_8}{I}$ is not an integral domain.

The best thing for being sad is to learn something. - Merlyn, in T. H. White's The Once and Future King

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