In this section, I'll look at quotient rings of polynomial rings.
Let F be a field, and suppose . is the set of all multiples (by polynomials) of , the (principal) ideal generated by . When you form the quotient ring , it is as if you've set multiples of equal to 0.
If , then is the coset of represented by .
Define ( is congruent to mod ) to mean that
In words, this means that and are congruent mod if they differ by a multiple of . In equation form, this says for some , or for some .
Lemma. Let R be a commutative ring, and suppose . Then if and only if .
Proof. Suppose . Then for some . Hence,
Conversely, suppose . Then
This means that .
Depending on the situation, I may write or .
Example. ( A quotient ring of the rational polynomial ring) Take in . Then two polynomials are congruent mod if they differ by a multiple of .
(a) Show that .
(b) Find a rational number r such that .
(c) Prove that .
(b) By the Remainder Theorem, when is divided by , the remainder is
(c) I'll use the First Isomorphism Theorem. Define by
That is, evaluates a polynomial at . Note that
It follows that is a ring map.
I claim that . Now if and only if
That is, if and only if 2 is a root of f. By the Root Theorem, this is equivalent to , which is equivalent to .
Next, I'll show that is surjective. Let . I can think of q as a constant polynomial, and doing so, . Therefore, is surjective.
Using these results,
The first equality follows from the fact that . The isomorphism follows from the First Isomorphism Theorem. The second equality follows from the fact that is surjective.
In the last example, was a field. The next result says that this is the case exactly when is irreducible.
Theorem. is a field if and only if is irreducible.
Proof. Since is a commutative ring with identity, so is .
Suppose is irreducible. I need to show that is a field. I need to show that nonzero elements are invertible.
Take a nonzero element of --- say , for . What does it mean for to be nonzero? It means that , so .
Now what is the greatest common divisor of and ? Well, , but is irreducible --- its only factors are units and unit multiples of .
Suppose , where and . Then , i.e. for some . But then shows that , contrary to assumption.
The only other possibility is that , where and . So I can find polynomials , , such that
This shows that is the multiplicative inverse of . Therefore, is invertible, and is a field.
Going the other way, suppose that is not irreducible. Then I can find polynomials , such that , where and both have smaller degree than .
Because and have smaller degree than , they're not divisible by . In particular,
This shows that has zero divisors. Therefore, it's not an integral domain --- and since fields are integral domains, it can't be a field, either.
Example. ( A quotient ring which is not an integral domain) Prove that is not an integral domain by exhibiting a pair of zero divisors.
and are zero divisors, because
Example. ( A quotient ring which is a field) (a) Show that is a field.
(b) Find the inverse of in .
(a) Since for all , it follows that has no rational roots. Hence, it's irreducible, and the quotient ring is a field.
(b) Apply the Extended Euclidean algorithm to and :
Reducing mod , I get
Thus, is the inverse of .
Example. ( A field with 4 elements) (a) Prove that is a field.
(b) Find so that
(c) Construct addition and multiplication tables for .
(a) Let . Then and . Since f has no roots in , it's irreducible. Hence, is a field.
(b) By the Division Algorithm,
This equation says that and x differ by a multiple of , so they represent the same coset mod .
(c) By the Division Algorithm, if , then
There are two possibilities for a and two for b, a total of 4. It follows that is a field with 4 elements. The elements are
Here are the addition and multiplication tables for :
The addition table is fairly easy to understand: For example, , because .
For the multiplication table, take as an example. ; I apply the Division Algorithm to get
Alternatively, you can use the fact that in the quotient ring (omitting the coset notation), so (remember that in ).
Remark. In the same way, you can construct a field of order for any prime n and any . Just take and form the quotient ring , where is an irreducible polynomial of degree n.
Example. ( Computations in a quotient ring) (a) Show that is a field.
(b) How many elements are there in ?
Express your answer in the form , where .
(d) Find .
(a) has no roots in :
Since is a cubic, it follows that it's irreducible. Hence, is a field.
(b) By the Division Algorithm, every element of can be written in the form
There are 3 choices each for a, b, and c. Therefore, has elements.
By the Division Algorithm,
(d) Apply the Extended Euclidean algorithm:
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