Let $R$ be a finitely generated algebra over a commutative ring $R_0$. My question is why $R$ is a homomorphic image of $R_0[x_1,x_2,…,x_r]$ for some $r$? Could one explicitly define the homomorphism?
[Math] Finitely generated algebra over a ring
abstract-algebracommutative-algebra
Best Answer
I'm assuming $R$ is commutative, given your tag. However, note that if you don't assume $R$ is commutative the claim is false: the real quaternions are finitely generated over $\mathbb{R}$ by $i$, $j$, and $k$; but $\mathbb{R}[x_0,\ldots,x_r]$ is commutative for every $r$, and a noncommutative ring cannot be a homomorphic image of a commutative ring.
If $R$ is commutative, since $R$ is finitely generated over $R_0$, it follows that there exist $k_0,\ldots,k_r\in R$ such that $\langle R_0,k_0,\ldots,k_r\rangle=R$ (where $R_0$ is identified with the subring $R_0\cdot 1$ of $R$).
By the universal property of $R_0[x_1,\ldots,x_r]$, identity map on $R_0$ extends to a unique ring homomorphism $\varphi\colon R_0[x_1,\ldots,x_r]\to R$ such that $\varphi(r) = r$ for every $r\in R_0$, and $\varphi(x_i)=k_i$ for $i=0,\ldots,r$. Since the image contains a generating set for $R$, we conclude that $\varphi$ is onto, and the desired result follows from the Isomorphism Theorems.