# Prime Ideals of Extension Fields

• Apr 14th 2009, 07:16 PM
Coda202
Prime Ideals of Extension Fields
Let E/F be an extension of fields. For a in E let phi_a be the evaluation homomorphism defined by: phi_a: F[x] --> E; f --> f(a)
Show that for all a in E, ker(phi_a) is a prime ideal in F[x].
I know that ker(phi_a) is an ideal of F[x], I'm having trouble showing that it is prime.
• Apr 14th 2009, 07:32 PM
NonCommAlg
Quote:

Originally Posted by Coda202
Let E/F be an extension of fields. For a in E let phi_a be the evaluation homomorphism defined by: phi_a: F[x] --> E; f --> f(a)
Show that for all a in E, ker(phi_a) is a prime ideal in F[x].
I know that ker(phi_a) is an ideal of F[x], I'm having trouble showing that it is prime.

$\displaystyle \frac{F[x]}{\ker \phi}$ is isomorphic to a subring of $\displaystyle E$ and hence it has to be an integral domain because $\displaystyle E$ is a field and thus an integral domain. so $\displaystyle \ker \phi$ is a prime ideal of $\displaystyle F[x].$