TITLE: isomorphic polynomial rings
QUESTION [5 upvotes]: I'm certain that this is a dumb question, but I'll ask anyway.
I know that if $\theta : F \to K$ is a field isomorphism then we get an induced isomorphism $\varphi:F[x] \to K[x]$ such that $\varphi|F = \theta$. We construct such a $\varphi$ by setting $\varphi|F = \theta$ and $\varphi(x)=x$. My question is the following: do we need to set $\varphi(x)=x$, or does this just follow? In other words, if we have an isomorphism $\varphi:F[x] \to K[x]$ such that $\varphi|F = \theta$ then do we necessarily have $\varphi(x)=x$?
Thanks.

REPLY [4 votes]: This is a good question. In addition to the other good answers, perhaps we should think of the polynomial ring $k[x]$ as the universal $k$-algebra on one generator, in the sense that given a $k$-algebra $A$ and element $a\in A$, there is a unique $k$-algebra hom $k[x]\rightarrow A$ such that $x\rightarrow a$. 
This points out that the collection of isomorphisms $k[x]\rightarrow k[x]$ is indexed by $k$-algebra generators $x'$ of $k[x]$, and sending $x\rightarrow x'$.
Arguing on degree of polynomials, the generators are exactly $ax+b$ with $a\not=0$. Thus, $x\rightarrow ax+b$ (and extend by multiplicativity and $k$-linearity) is the most general self-isomorphism.