# isomorphic polynomial rings

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.

## 3 Answers

No, you could have an isomorphism that agrees with $\theta$ on $F$ yet sends $x$ to $-x$ (or to $ax$ for any fixed $a\in K-\{0\}$).

• Perfect. Thanks. – nigel Jul 20 '13 at 21:45

No, that doesn't follow. For example $x \mapsto x-1$ generates an automorphism of $F[x]$. If an isomorphism between polynomial rings were determined by the isomorphism of the base fields, that would mean that a polynomial ring had only the trivial automorphism.

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.

• This is interesting. Thanks. – nigel Jul 21 '13 at 2:27