A homomorphism between two A-algebras.

Question

A homomorphism between two algebras is described here. I want to describe a homomorphism $f:A[x_1,x_2,\dots,x_n]\to R$, where $R$ is an A-algebra. $A$ is a ring.

Obviously, $A[x_1,x_2,\dots,x_n]$ is an A-algebra.

The article says that if $A$ and $B$ are two algebras over $K$, and $k\in K$ and $x,y\in A$ then $$f(kx)=kf(x)....(1)$$ $$f(x+y)=f(x)+f(y).....(2)$$ $$f(xy)=f(x)f(y).....(3)$$

Here, let us assume $s_{1},s_{2}\in A[x_1,x_2,\dots,x_n]$. $$f(s_{1}+s_{2})=f(s_{1})+f(s_{2}).....(3)$$ seems fine. What about (1) and (3) though? Is the following the correct interpretaton of the rule:

If $m,n\in A[x_1,x_2,\dots,x_n]\setminus A$, then $$f(mn)=f(m)f(n)$$

If $x\in A,y\in A[x_1,x_2,\dots,x_n]$, then $$f(xy)=xf(y)$$

If $p,q\in A\subset A[x_1,x_2,\dots,x_n]$, then $f(pq)$ is not defined.

I'm really getting confused here. Thanks in advance for your time!

Answer 1

Edit: Looks like I assumed the ring $A$ has an identity, but I believe that is standard when talking about an algebra over a ring.

I assume you want this homomorphism to be an $A$-algebra homomorphism. The first two rules you ask about are correct, and furthermore, the first rule is true for any $m$ and $n$ in $A[x_1, \dots, x_n]$. As for the third rule, we have for $p \in A$: $$f(p) = f(p \cdot 1) = p \cdot f(1).$$ If the homomorphism is unital, then $f(1) = 1$ so that $f(p) = p$.