<p>Critical Step in the Proof that all Splitting Fields of a Polynomial are Isomorphic.</p><p>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. </p><p>E</p><p> F F¢</p><p> F[x] F¢[x]</p><p>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) >.</p><p>E</p><p> F F¢</p><p> F(a) F[x] F¢[x] @ a F[x} < p(x) ></p><p>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¢</p><p> F F¢</p><p> F(a) F[x] F¢[x] F¢(b) @ @ a b F[x} F¢[x] < p(x) > < p(x) ></p><p>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) >.</p><p>E E¢</p><p> F F¢</p><p> F(a) F[x] F¢[x] F¢(b) @ @ a  b F[x} F¢[x] < p(x) > < p(x) ></p><p>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).</p><p>E E¢</p><p> F F¢</p><p> F(a) F[x] F¢[x] F¢(b) @ @ a  b F[x} F¢[x] < p(x) > < p(x) ></p><p> 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:</p><p>() q( f(a) ) = b( (a (f(a))) = b ( (f(x) + < p(x) >) ) = b ( (f(x)) + < p(x) >) = ( f) (b).</p><p>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. </p><p>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 =  .</p>