Suppose F and F Are Fields and Is an Isomorphism Between Them
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Critical Step in the Proof that all Splitting Fields of a Polynomial are Isomorphic.
Suppose F and F are fields and is an isomorphism between them:. As is a ring homomorphism it induces a ring homomorphism : F[x] F[x], defined i i by ( ai x ) (ai )x . Since is an isomorphism, so is . Let p(x) F[x] be an irreducible polynomial over F. Let p(x) φ(p(x)) F[x] . Then p(x) is also irreducible over F¢. Suppose E is an extension of F and a is a root of p in E, i.e. p(a) = 0.
E
F F¢
F[x] F¢[x]
We know that we can define an epimorphism from F[x] to F(a) by f(x) f(a). The kernel of this epimorphism is < p(x) >, and so by the First Isomorphism Theorem for Rings, F[x] / < p(x) > F(a). Let be the isomorphism : F(a). F[x] / < p(x) >, defined by a(f(a)) = f(x) + < p(x) >.
E
F F¢
F(a) F[x] F¢[x] @ a F[x} < p(x) >
Similarly, suppose E is an extension of F¢ and b is a root of p(x) in E, i.e. p(b) 0 . We can define an epimorphism from F¢ [x] to F¢ (b) by f(x) f(b). The kernel of this epimorphism is < p(x) >, and so by the First Isomorphism Theorem for Rings, F¢ [x] / < p(x) > F¢ (b). Let be the isomorphism b : F¢ [x] / < p(x) > F¢(b), defined by b( g(x) + < p(x) > ) = g(b). E E¢
F F¢
F(a) F[x] F¢[x] F¢(b) @ @ a b F[x} F¢[x] < p(x) > < p(x) >
Just as the isomorphism induces the isomorphism , induces an isomorphism between F[x] / < p(x) > and F¢[x] / < p(x) >, defined by ( f(x) + < p(x) > ) = (f(x)) + < p(x) >.
E E¢
F F¢
F(a) F[x] F¢[x] F¢(b) @ @ a b F[x} F¢[x] < p(x) > < p(x) >
Let be the composition of the three isomorphisms a, , and b: q = b a. Thus q is an isomorphism from F(a) to F¢(b).
E E¢
F F¢
F(a) F[x] F¢[x] F¢(b) @ @ a b F[x} F¢[x] < p(x) > < p(x) >
q Suppose f(x) is any polynomial in F[x] . Let’s compute q( f(a) ). From the definitions of a, , and b, we get:
() q( f(a) ) = b( (a (f(a))) = b ( (f(x) + < p(x) >) ) = b ( (f(x)) + < p(x) >) = ( f) (b).
Suppose f(x) = x. Then f(a) = a, and since (f(x) ) = x, ( f) (b) = b. Therefore q(a) = q(f(a)) = (by () ) ( f) (b) = b. So q maps a into b.
Next let c be an arbitrary element of F, and suppose g(x) F[x] is the constant polynomial g(x) = c. Then g(a) = c and (g(x) ) = (c), so ( g) (b) = (c) also. Therefore q(c) = q(g(a)) = (by () ) ( g) (b) = (c). Hence q acts the same as on all elements of F; q F = .