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Fields as Quotients of Polynomial Rings

Fields as Quotients of Polynomial Rings. In the preceding slides we have constructed fields by starting with the ring GF 2 [x] and “mod-ing off” an irreducible polynomial

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Fields as Quotients of Polynomial Rings

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  1. Fields as Quotients of Polynomial Rings • In the preceding slides we have constructed fields by starting with the ring GF2[x] and “mod-ing off” an irreducible polynomial • We were able to show that the results were fields by explicitly constructing their addition and multiplication tables and verifying the field properties from those tables • Our first goal is to prove that this construction works for an arbitrary base field F

  2. F[x] mod an Irreducible Polynomial is a Field Theorem 1Let F be a field and q(x) F[x] be an irreducible polynomial of degree n over F. Let K be the set of polynomials in F[x] of degree less than n and define  on K be the usual addition of polynomials in F[x]. Define the operation  on K by f(x)g(x) = (f(x)g(x)) mod q(x). Then (K, , ) is a field containing F and a root of q(x). Moreover, if F is finite, |K| = |F|n. The field constructed in the theorem is often denoted by F[x]/q(x).

  3. Extension Fields Containing Polynomial Roots Definition 1: If F is a subfield of field K, we say that K is an extension of F. Corollary 1Let F be a field. Then for every nonconstant polynomial f(x)  F[x], there is an extension field of F that contains a root of f. proof Since f(x) can be written as a product of irreducible polynomials, we can write f(x) = q(x)s(x) with q(x) irreducible. If we let K = F[x]/q(x), then K is an extension of F containing a root  of q(x). Since f(x) = q(x)s(x),  is also a root of f(x). Corollary 2Let F be a field. Then for every nonconstant polynomial f(x)  F[x], there is an extension field K of F, an element   K and a polynomial g(x) in K[x] such that f(x) = (x-)g(x) over K.

  4. Splitting Field of a Polynomial Theorem 2Given a field F and a monic polynomial f(x)  F[x] of degree n, there is an extension K of F and elements 1, . . . , n  K such that, as a polynomial over K, f(x) = (x- 1)(x- n).

  5. The Splitting Field of a Polynomial • If a monic polynomial f(x) F[x], F a field, can be written as a product of linear polynomials in an extension K of F, we say that f(x) splits completely over K. • The smallest such field is called the splitting field of f(x). • Theorem 2 shows that every nonconstant polynomial splits completely over some extension of F • We are interested in the case where a polynomial of degree n has n distinct roots in its splitting field.

  6. Formal Derivatives • Given a polynomial f(x) = a0 + a1x + a2x2 +  + anxn over a field F, we define the formal derivative of f to be the polynomial f(x) = a1 + 2a2x +  + nanxn-1 • Formal derivatives satisfy the usual product and power rules: • (f(x)g(x)) = f(x)g(x) + f(x)g(x) • ((f(x))n) = n(f(x))n-1f(x) • We will have occasion to consider the formal derivative of a product of powers of linear terms • Suppose f(x) = (x-)a(x-b)b(x-g)c • Then f(x) = [(x-)a](x-b)b(x-g)c + (x-)a[(x-b)b(x-g)c] = a(x-)a-1 (x-b)b(x-g)c +(x-)a[b(x-b)b-1(x-g)c + (x-b)bc(x-g)c-1 ] = a(x-)a-1 (x-b)b(x-g)c +(x-)a b(x-b)b-1(x-g)c + (x-)a(x-b)bc(x-g)c-1 • In general: if , then

  7. Distinct Roots Theorem TheoremA nonconstant polynomial that splits completely over a field has n distinct roots if and only if f(x) and its formal derivative f(x) are relatively prime proof Let 1, . . . , m be the distinct roots of f(x) and let ki be the number of times i appears in the linear factorization of f(x). Then and hence If ki >1 for some i, then (x-i) is a common factor of f(x) and f(x) and hence they are not relatively prime. If all ki equal 1, then f(x) and f(x) have no common factors and hence are relatively prime.

  8. Existence and Uniqueness • By Theorem 2, there is an extension field K of GFp such that the polynomial splits completely over K • That is, there are elements of K such that, as a polynomial over K, we have • We claim that the set of roots of form a subfield of K • To see this we need to show that the set of roots is closed under addition, multiplication, and taking additive and multiplicative inverses. • Let i, j be roots of • Then • Thus the set of roots is closed under addition

  9. Existence and Uniqueness • We are trying to show that the set of roots of thepolynomial in an extension field K form a subfield • We have already shown the set of roots is closed under addition • Again, let i, j be roots of in K • Then

  10. Existence and Uniqueness • Next we show that the set of roots of in K are closed under taking additive inverses • Assuming p > 2, then pn is odd and thus • Thus • For the case p = 2, the same argument may be used, except that -1 raised to the power is now 1. Since -1 = 1 mod 2, the argument still. holds. • Finally, we show that the set of roots is closed under taking multiplicative inverses of nonzero elements. • Since we have shown that the set forms a field, we know that for each prime p and positive integer n, there is a field with exactly pn elements.

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