Solutions to Problem Set 28

Math 345/504

5-2-2018

1. Let R be a ring and let I be an ideal in R. Prove using the definition of coset addition and properties of R that the quotient ring $\dfrac{R}{I}$ satisfies the associative axiom for addition.

Let $a, b, c \in R$ . Then

$$[(a + I) + (b + I)] + (c + I) = [(a + b) + I] + (c + I) = [(a + b) + c] + I = [a + (b + c)] + I =$$

$$(a + I) + [(b + c) + I] = (a + I) + [(b + I) + (c + I)].\quad\halmos$$


2. (a) Find the number of elements in the quotient ring $\dfrac{\integer_5[x]}{\langle 4 x^3 +
   x^2 + 3 \rangle}$ .

(b) Find all the roots of $4 x^3
   + x^2 + 3$ in $\integer_5$ . Prove or disprove: $\dfrac{\integer_5[x]}{\langle 4 x^3 + x^2 + 3 \rangle}$ is a field.

(a) Every coset can be simplified to the form $(a x^2 + b x + c) + \langle 4 x^3 +
   x^2 + 3 \rangle$ , where $a, b, c \in \integer_5$ . Hence, the quotient ring has $5^3 = 125$ elements.

(b)

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

$4 x^3 + x^2 + 3$ has roots, so it's not irreducible, and $\dfrac{\integer_5[x]}{\langle 4
   x^3 + x^2 + 3 \rangle}$ is not a field.

In fact,

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


3. Consider the quotient ring $\dfrac{\integer_5[x]}{\langle x^2 + 2 \rangle}$ .

(a) How many elements are there in $\dfrac{\integer_5[x]}{\langle x^2 + 2 \rangle}$ ?

(b) Compute $[(x + 1)(x + 4) +
   \langle x^2 + 2 \rangle] + [(x^3 + 4x^2 + 2) + \langle x^2 + 2
   \rangle$ . Simplify your answer to the form $(ax + b) + \langle
   x^2 + 2 \rangle$ , where $a, b \in \integer_5$ .

(c) Compute $[(x + 3) + \langle
   x^2 + 2 \rangle]^2$ . Simplify your answer to the form $(ax +
   b) + \langle x^2 + 2 \rangle$ , where $a, b \in \integer_5$ .

(d) Compute $[(3x + 2) + \langle
   x^2 + 2 \rangle]^{-1}$ . Simplify your answer to the form $(ax + b) + \langle x^2 + 2 \rangle$ , where $a, b
   \in \integer_5$ .

(a) Since every element can be written as $(ax + b) + \langle x^2 + 2 \rangle$ , where $a, b \in \integer_5$ , there are $5^2 = 25$ elements.

(b)

$$[(x + 1)(x + 4) + \langle x^2 + 2 \rangle] + [(x^3 + 4x^2 + 2) + \langle x^2 + 2 \rangle = (x^3 + 1) + \langle x^2 + 2 \rangle = (x^3 + 1) + 4x(x^2 + 2) + \langle x^2 + 2 \rangle =$$

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

(c)

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

(d)

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

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

Therefore, $[(3x + 2) + \langle
   x^2 + 2 \rangle]^{-1} = (x + 1) + \langle x^2 + 2 \rangle$ .


[Math 504]

4. A ring R (not necessarily commutative or having an identity element) is Boolean if $x^2 = x$ for all $x \in
   R$ .

(a) Prove that in a Boolean ring, $x = -x$ for all $x \in R$ .

(b) Prove that a Boolean ring is commutative.

(a) Let R be a Boolean ring and let $x \in R$ . By assumption, $(x + x)^2 = x + x$ . Expanding the left side, I have

$$x^2 + 2 x^2 + x^2 = x + x, \quad 4 x^2 = 2 x, \quad 4x = 2 x, \quad 2 x = 0, \quad x = -x.\quad\halmos$$

(b) Let R be a Boolean ring and let $x, y \in R$ . By assumption, $(x + y)^2 = x + y$ . Expanding the left side, I have

$$x^2 + x y + y x + y^2 = x + y, \quad x + x y + y x + y = x + y, \quad x y + y x = 0, \quad x y = -y x, \quad x y = y x.$$

The last step used the result of part (a) (which shows that $-y x = y x$ ).


Mingle some brief folly with your wisdom. - Horace


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