When and how $\phi : R \to S$ induces $\varphi : R[x_1,\cdots,x_n] \to S$?

Question

Suppose we are given $R, S$ - rings with identity, and $\phi : R \to S$ a ring homomorphism such that $\phi(1_R) = 1_S$, and for $s_1,s_2,\cdots,s_n \in S$ we have $s_is_j = s_js_i$ for all $i,j$ and $\phi(r)s_i = s_i\phi(r)$ for all $r \in R$ for all $i$.

The text tells me that then there exists a unique homomorphism $\varphi : R[x_1,\cdots,x_n] \to S$ such that $\varphi|R = \phi$ and $\varphi(x_i) = s_i$.

How do I show go about proving it? Does this completely determine $R[x_1,\cdots,x_n]$? And how?

Answer 1

well, the morphism is defined on a generating set, i.e. $\varphi(\sum \alpha_{i,j} x_i^j):=\sum \phi(\alpha)_{i,j} s_i^j$ this is now clearly well defined by the definition of the polynomial algebra, respectively free algebra (if you allow multiindexes as $j$) and is precisely the morphism you want.

i.e. we just proved existence.

for uniqueness: assume that there are 2 morphisms $\varphi,\varphi'$ with that property and let $a \in R[x_1,...,x_n]$, then $a=\sum \alpha_{i,j} x_i^j)$ in particular by linearity: $$\varphi(a)=\varphi(\sum \alpha_{i,j} x_i^j)=\sum \varphi(\alpha_{i,j} x_i^j)=\sum \varphi(\alpha_{i,j})\varphi( x_i^j)=\\ \sum \phi(\alpha_{i,j}) s_i^j\\= \sum \varphi'(\alpha_{i,j})\varphi'( x_i^j)=\sum \varphi'(\alpha_{i,j} x_i^j)=\varphi'(\sum \alpha_{i,j} x_i^j)=\varphi'(a).$$ so since a was arbitrary $\varphi=\varphi'$ which proves uniqueness.

Answer 2

Any element of $R[x_1,\ldots,x_n]$ is a polynomial in the $x_i$, so is of the form $$\sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} a_I\prod\limits_{j=1}^nx_j^{i_j},$$ with $a_I \in R$.

Thus, we can define $$\varphi\left(\sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} a_I\prod\limits_{j=1}^nx_j^{i_j}\right) = \sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} \phi(a_I)\prod\limits_{j=1}^ns_j^{i_j}$$ and you can easily show that this is a well-defined homomorphism, and has the desired properties, which gives us existence.

Now, if $\psi$ is another such homomorphism, then, since it is a homomorphism, $$\psi\left(\sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} a_I\prod\limits_{j=1}^nx_j^{i_j}\right) = \sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} \psi(a_I)\prod\limits_{j=1}^n\psi(x_j)^{i_j}$$ but since $\psi(a_I) = \phi(a_I) = \varphi(a_I)$, and $\psi(x_j) = s_j = \varphi(x_j)$, this is equal to

$$\sum\limits_{\substack{I\in \mathbb{N}^n\\I = \{i_1,\ldots,i_n\}}} \phi(a_I)\prod\limits_{j=1}^ns_j^{i_j},$$

so $\varphi(x) = \psi(x)$ for any $x \in R[x_1,\ldots,x_n]$, hence $\varphi = \psi$, so we have uniqueness.

I have no idea what you mean by "completely determine $R[x_1,\ldots,x_n]$", because it was already completely determined by its definition.