How do ring morphisms work when dealing with polynomial rings?

Question

I've been trying to understand how ring morphisms work when dealing with polynomial rings, but I'm having a little bit of trouble understanding how they work.

Let $R,S$ be rings (they need not be commutative nor have a multiplicative identity) and $f:R\to S$ a ring morphism. We know that this induces a ring morphism $f^*:R[x]\to S[x]$ given by $f^*(a_0+a_1x+\cdots+a_nx^n)=f(a_0)+f(a_1)x+\cdots+f(a_n)x^n$. My problem now is: is every ring morphism $F:R[x]\to S[x]$ of the form $F=g^*$ for some $g:R\to S$?

I don't know whether it is or not the case, but if it is I'm having trouble proving it.

My first approach would be that any polynomial can be thought as a sum and product of constant polynomials and $x^k$ for some $k$ and in that sense, then $F(a_0+\cdots a_nx^n)=F(a_0)+\cdots F(a_n)F(x)^n$, so if I knew that $F$ sends constant polynomials to constant polynomials and polynomials of the form $x^n$ to themselves, then I would have the desired result, but this is something I haven't been able to prove yet.

Only thing I know is that for $c\in R$, $F(c)=b_0+\cdots+b_mx^m$ with $b_i\in S$ for all $i$, but I can't deduce from there that $b_i=c$ if $i=0$ and $b_i=0$ otherwise. Same case with $x^n$ .

If this is not the case, however, and I'm trying to prove something false, then how do these morphisms work? Is there a way to more or less understand their behaviour?

Any help is thanked for :)

Answer 1

The result is not true:

Take $R = S = \mathbb{Z}[y]$. Then $R[x] = S[x] = \mathbb{Z}[x,y]$ has an automorphism given by sending $x$ to $y$ and $y$ to $x$. This cannot be of your desired form, as it sends (for example) the polynomial $x$ to something in $S$, but any map of your desired form must send it to something of the form $sx$, where $s \in S$, which does not lie in $S$.

What your approach gives you is a grading on $R[x]$ (indeed, the very prototypical example of a grading). That is: a collection of subsets $R_i$ with $R = \bigoplus R_i$ as additive groups, such that if $a \in R_i$ and $b \in R_j$, then $ab \in R_{i+j}$: the grading is precisely by $x$-degree, so that $R_i = \{ax^i | a \in R\}$. Then each element of $R[x]$ is a sum of one element from each $R_i$, as you've written.

With that, then, the morphisms that do what you ask for are precisely the graded morphisms (with respect to this grading): that is, the morphisms $\varphi: R[x] \to S[x]$ such that $\varphi(R_i) \subseteq S_i$ for all $i$. Your problems, though, are twofold:

1. Gradings are very much not unique, and there's nothing that particularly privileges your choice of grading over any other.
2. It is not true, in general, that all morphisms are graded morphisms with respect to your particular grading. Indeed, if $R$ has a non-trivial grading, then it has non-zero morphisms $R \to R_0$, where $R_0$ is the graded component containing the identity (the simplest example is $E_0$, which is just the projection onto the $R_0$ summand). This cannot be graded. In the case of polynomial rings, things are actually worse for you: in $R[x]$, the above map is precisely evaluation at $0$. There is a similar evaluation morphism $E_s$ for any other $s \in R$, which has the same problem. Indeed, the only ring for which your result does hold is the trivial ring $0$ (just because then $0[x] = 0$, so the only homomorphism $0[x] \to 0[x]$ is the zero homomorphism $0 = 0^*$.

Answer 2

A ring homomorphism $F : R[x] \to S[x]$ can send $x$ to any element of $S[x]$; for a particular value of $F(x)$, the image of the coefficient ring $F(R)$ must land in the centralizer $Z_{S[x]}(F(x))$ of $F(x)$. This is a complete description of all possible ring homomorphisms. Equivalently, $F$ can restrict to an arbitrary ring homomorphism $R \to S[x]$, and then $F(x)$ must commute with the image.

This construction simplifies a lot if $R$ and $S$ are commutative; then $F(x)$ is any element of $S[x]$ and $F(R)$ can be arbitrary. So your morphisms are less than fully general for two reasons:

• $F(R)$ need not be contained in $S$, and may contain non-constant polynomials, and
• $F(x)$ need not be $x$, but can in fact be any element of $S[x]$ in general.