I have been looking over multiple proof regarding Hilbert's basis theorem. Namely, if $R$ is a commutative noetherian ring, then $R[x]$ is also notherian.
I understand, for the most part, the construction of this proof. My question is, can Hilbert's basis theorem be written as an if and only if statement? In other words, is it true that if $R[x]$ is noetherian then $R$ is also noetherian? If it can be, then I have an idea on how to prove it, but I would like some feedback.
Proof: Let $R[x]$ be noetherian. Let $$\phi: R[x] \rightarrow R$$ be the homomorphism of rings where $$\phi(f) = f(0).$$ It follows then that $\phi(R) = R$. In other words, $\phi$ is onto. Now, it is not hard to check that the image of a noetherian ring is noetherian. Thus $R$ is noetherian. $\square$
My question is, is my proof is correct? Furthermore, if it is correct, how come a lot of standard algebra texts don't state this theorem as an if and only if?
Your proof is correct! I think the reason algebra texts don't state Hilbert's Basis Theorem as a biconditional is that this direction is much more obvious (and less often useful). I also have one nitpick: you should say "the homomorphism" instead of "a homomorphism" because you are clearly singling out a unique map.