# Prove that an algebra is finitely generated iff it is isomorphic to the quotient ring $R[x_1,\dots,x_n]/I$

This is a basic question, but I haven't done abstract algebra in a while and not certain of the answer.

We say that an $R$-algebra $A$ is finitely generated if there exist $a_1, \dots, a_n$ such that $A=R[a_1,\dots, a_n]$. I want to prove that an $R$-algebra $A$ is finitely generated iff it is isomorphic to the quotient ring $R[x_1,\dots,x_n]/I$.

Assume $A$ is finitely generated by $a_1,\dots,a_n$. We can define a unique algebra homomorphism $f\colon R[x_1,\dots,x_n] \to A$ in the usual way (apply the homomorphism $h$ from $R$ to $A$ to the coefficients, and change $x_i$ to $a_i$). The image of $f$ is $R[a_1,\dots,a_n]$. By one of the isomorphism theorems, we conclude that $A$ is isomorphic to $R[x_1,\dots,x_n]/I$ where $I$ is the kernel of $f$.

I now struggle with the other direction. I found a proof online that says given $A$ is isomorphic to $R[x_1,\dots,x_n]/I$ by a map $g$, we can conclude that there exists a surjective ring homomorphism $f\colon R[x_1,\dots,x_n]\to A$. Am I right to understand that this homomorphism is given by applying $g$ to elements of $R[x_1,\dots,x_n]$ which are not in $I$ and mapping elements in $I$ to zero?

Now since $R[x_1,\dots,x_n]$ is generated by $x_1,\dots,x_n$, $A$ will be generated by $f(x_1),\dots,f(x_n)$. I don't have an intuitive understanding of why this is true either.

• I edited a typo in the title, but the whole question would be more readable if you learned a bit about how to format mathematics on this website. There are links from the "help" page. Commented Jan 7, 2014 at 21:30
• You mean that $A$ is commutative, don't you? The direction you're struggling with is the easiest one. Since $R[x_1,\dots,x_n]$ is finitely generated such is every epimorphic image of it, by the images of $x_1,\dots,x_n$. Commented Jan 7, 2014 at 21:34

There is a canonical surjective $R$-algebra homomorphism $$\pi:R[x_1,\ldots,x_n]\to R[x_1,\ldots,x_n]/I$$ with $\ker\pi = I$, sending each polynomial $p(x_1,\ldots,x_n)$ to the coset $p(x_1,\ldots,x_n)+I$.

Suppose we have an $R$-algebra isomorphism $$g:R[x_1,\ldots,x_n]/I\stackrel{\sim}{\to} A.$$ Then the composition $f = g\circ\pi:R[x_1,\ldots,x_n]\to A$ is a surjective $R$-algebra homomorphism. Since $f$ is an $R$-algebra homomorphism,

$$f(p(x_1,\ldots,x_n)) = p(f(x_1),\ldots,f(x_n))$$ for each $p\in R[x_1,\ldots,x_n]$. It follows that since $f$ is surjective, every element of $A$ is a polynomial in $f(x_1),\ldots,f(x_n)$ with coefficients in $R$ -- that is, $A$ is generated by these $n$ elements as an $R$-algebra.

• Thank you, this was very clear. Commented Jan 8, 2014 at 1:44

Perhaps it is simpler, notation-wise, to argue this way:

1. Every quotient of a finitely generated algebra is a finitely generated algebra.
2. An algebra that is isomorphic to a finitely generated algebra is finitely generated.
3. Therefore, an algebra isomorphic to a quotient of a polynomial ring is finitely generated.

You seem to be having trouble with 1). Carefully think of the definition of "generates". Whatever definition you are using, you can argue that a surjective map $f:A\to B$, where $A$ is generated by a set $S\subset A$, forces $B$ to be generated by the set $f(S)\subset B$.

This argument should have the form: Take $b\in B$. Then there is some $a\in A$ with $f(a)=b$. $a$ is a polynomial in elements of $S$, therefore...