Quotient Rings of Polynomial Rings


 Nathaniel Wade
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1 Quotient Rings of Polynomial Rings Let F be a field. is a field if and only if p(x) is irreducible. In this section, I ll look at quotient rings of polynomial rings. Let F be a field, and suppose p(x). is the set of all multiples (by polynomials) of p(x), the (principal) ideal generated by p(x). When you form the quotient ring, it is as if you ve set p(x) multiples of p(x) equal to 0. If a(x), then a(x) + is the coset of represented by a(x). Define a(x) = b(x) (mod p(x)) (a(x) is congruent to b(x) mod p(x)) to mean that p(x) a(x) b(x). In words, this means that a(x) and b(x) are congruent mod p(x) if they differ by a multiple of p(x). In equation form, this says a(x) b(x) = k(x) p(x) for some k(x), or a(x) = b(x) + k(x) p(x) for some k(x). Lemma. Let R be a commutative ring, and suppose a(x), b(x), p(x) R[x]. Then a(x) = b(x) (mod p(x)) if and only if a(x) + = b(x) +. Proof. Suppose a(x) = b(x) (mod p(x)). Then a(x) = b(x) + k(x) p(x) for some k(x) R[x]. a(x) + = b(x) + k(x) p(x) + = b(x) +. Conversely, suppose a(x) + = b(x) +. Then This means that a(x) = b(x) (mod p(x)). a(x) a(x) + = b(x) +. a(x) = b(x) + k(x) p(x) for some k(x) R[x]. Depending on the situation, I may write a(x) = b(x) (mod p(x)) or a(x) + = b(x) +. Example. (A quotient ring of the rational polynomial ring) Take p(x) = x in. Then two polynomials are congruent mod x if they differ by a multiple of x. For example, (x + 3x + ) (x + x + 7) = x x = (x + )(x ), so x + 3x + = x + x + 7 (mod x ). If I take an arbitrary rational polynomial and apply the Division Algorithm to divide it by x, I ll get a rational number as a remainder. For example, ( x 7x + x + = x3 + x 7 x 9 ) ( (x ) + 3 ). The equation shows that x 7x + x + and 3 x 7x + x + = 3 differ by a multiple of x. So (mod x ).
2 Thus, mod (x ) every rational polynomial is congruent to a rational number. This makes the following isomorphism reasonable: x Q. To prove this, I ll use the First Isomorphism Theorem. Define φ : Q by φ(f(x)) = f(). That is, φ evaluates a polynomial at x =. Since φ(f(x) + g(x)) = f() + g() = φ(f(x)) + φ(g(x)) and φ(f(x)g(x)) = f()g() = φ(f(x)) φ(g(x)), it follows that φ is a ring map. I claim that kerφ = x. First, let p(x)(x ) x, where p(x). Then φ(p(x)(x )) = p()( ) = 0. p(x)(x ) kerφ, so x kerφ. Conversely, suppose that f(x) kerφ. Then φ(f(x)) = 0, or f() = 0. This means that x = is a root of f(x), so x is a factor of f(x), by the Root Theorem. So f(x) = p(x)(x ) for some p(x), and f(x) x. This proves that kerφ x, and hence kerφ = x. Next, I ll show that φ is surjective. Let q Q. I can think of q as a constant polynomial, and doing so, φ(q) = q. φ is surjective. Using these results, x = im φ = Q. kerφ The first equality follows from the fact that x = kerφ. The isomorphism follows from the First Isomorphism Theorem. The second equality follows from the fact that φ is surjective. In the last example, irreducible. Theorem. was a field. The next result says that this is the case exactly when p(x) is is a field if and only if p(x) is irreducible. Proof. Since is a commutative ring with identity, so is. Suppose p(x) 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 a(x) +, for a(x). What does it mean for a(x) + to be nonzero? It means that a(x) /, so p(x) a(x). Now what is the greatest common divisor of a(x) and p(x)? Well, (a(x), p(x)) p(x), but p(x) is irreducible its only factors are units and unit multiples of p(x). Suppose (a(x), p(x)) = k p(x), where k F and k 0. Then k p(x) a(x), i.e. k p(x)b(x) = a(x) for some b(x). But then p(x)[k b(x)] = a(x) shows that p(x) a(x), contrary to assumption. The only other possibility is that (a(x), p(x)) = k, where k F and k 0. So I can find polynomials m(x), n(x), such that a(x)m(x) + p(x)n(x) = k.
3 Then ( ) ( ) a(x) k m(x) + p(x) k n(x) =. ( ) ( ) a(x) k m(x) + p(x) k n(x) + = +, ( ) a(x) k m(x) + = +, ( ) (a(x) + ) m(x) + = +. k This shows that m(x) + is the multiplicative inverse of a(x) +. a(x) + k is invertible, and is a field. Going the other way, suppose that p(x) is not irreducible. Then I can find polynomials c(x), d(x) such that p(x) = c(x)d(x), where c(x) and d(x) both have smaller degree than p(x). Because c(x) and d(x) have smaller degree than p(x), they re not divisible by p(x). In particular, But p(x) = c(x)d(x) gives c(x) + 0 and d(x) + 0. p(x) + = c(x)d(x) +, 0 = (c(x) + ) (d(x) + ). This shows that has zero divisors. 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) In divisors, because (x )(x + ) = x = 0 ( mod x ). This is a replay of the last part of the proof, and demonstrates in this case that hence not a field. x, x and x + are zero x is not a domain, Example. (A quotient ring which is a field) Consider x + x +. x + x + is irreducible in. x is a field. + x + Now if this is really true, I ought to be able to take a nonzero element and find a multiplicative inverse. For example, I ll find the inverse of (x 3 + ) + x + x +. Apply the Extended Euclidean algorithm to x 3 + and x + x + : x x + x + x x + 3 x 8x x x + 3 x 0
4 ( 3 x = x + 3 ) ( x (x + x + ) ) (x 3 + ). = ( x 3 x + 3 ) (x + x + ) ( x 3 ) (x 3 + ). Going mod x + x +, I get + x + x + = ( x 3 ) (x 3 + ) + x + x +, ( + x + x + = ( x 3 ) ((x + x + x + ) 3 + ) + x + x + ). Thus, ( x 3 ) + x + x + is the reciprocal of (x 3 + ) + x + x +. Example. (A field with elements) Consider x + x + in Z [x]. You can check that it has no roots, so Z [x] it s irreducible. x is a field. + x + By the Division Algorithm, I can reduce every polynomial in Z [x] to a polynomial ax+b mod x +x+. For example, take x + x 3 + in Z [x]. By the Division Algorithm, x + x 3 + = (x + x + )(x + ) + x. This equation says that x + x 3 + and x differ by a multiple of x + x +, so they represent the same coset mod x + x +. (x + x 3 + ) + x + x + = x + x + x +. There are two possibilities for a and two for b, a total of. It follows that elements. The elements are Z [x] x + x x + x +, + x + x +, x + x + x +, (x + ) + x + x +. Here are the addition and multiplication tables for Z [x] x + x + : + 0 x x x x + 0 x + x x x x + 0 x + x + x 0 0 x x x x + x 0 x x + x + 0 x + x is a field with
5 The addition table is fairly easy to understand: For example, x + (x + ) =, because x = 0 (mod ). For the multiplication table, take x x as an example. x x = x ; I apply the Division Algorithm to get x = (x + x + ) + (x + ). So x x = x + ( mod x + x + ). In the same way, you can construct a field of order p n for any prime n and any n. Just take Z p [x] and form the quotient ring Z p[x], where f(x) is an irreducible polynomial of degree n. f(x) Example. (Computations in a quotient ring) (a) Show that x 3 + x + has no roots in Z 3 : x 0 x 3 + x + (mod 3) Since x 3 + x + is a cubic, it follows that it s irreducible. (b) How many elements are there in x 3 + x +? By the Division Algorithm, every element of x 3 + x + x 3 is a field. + x + x 3 + x + can be written in the form (ax + bx + c) + x 3 + x +, where a, b, c Z 3. x 3 + x + has 33 = 7 elements. (c) Express [ (x + x + ) + x 3 + x + ] [ (x + ) + x 3 + x + ] is a field. in the form (ax + bx + c) + x 3 + x +, where a, b, c Z 3. [ (x + x + ) + x 3 + x + ] [ (x + ) + x 3 + x + ] = (x + x 3 + x + x + ) + x 3 + x +. By the Division Algorithm, x + x 3 + x + x + = (x + )(x 3 + x + ) + x. (x + x 3 + x + x + ) + x 3 + x + = x + x 3 + x +. (d) Find [ (x + ) + x 3 + x + ]. Apply the Extended Euclidean algorithm: x 3 + x +  x + x + x + x x + x + x + x + 0
6 (x + x + )(x + ) (x + )(x 3 + x + ) =, (x + x + )(x + ) (x + )(x 3 + x + ) =. [ (x + x + ) + x 3 + x + ][ (x + ) + x 3 + x + ] = + x 3 + x +. [ (x + ) + x 3 + x + ] = (x + x + ) + x 3 + x +. c 009 by Bruce Ikenaga 6
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