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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?

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You should map $x_i$ into the $i$th generator of $R$ over $R_0$. –  Andrea Mar 9 '12 at 21:00
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up vote 3 down vote accepted

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.

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"By the universal property of $R_0[x1,…,xr]$", what property? –  Jr. Mar 9 '12 at 23:41
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@Jr.: If $R$ is a ring with unity, $f\colon R\to S$ is a ring homomorphism, and $s_1,\ldots,s_n$ are elements of $S$ that commute with $f(r)$ for every $r\in R$, then there is a unique homomorphism $\overline{f}\colon R[x_1,\ldots,x_n]\to S$ such that $\overline{f}(r) = f(r)$ for every $r\in R$ and $\overline{f}(x_i) = s_i$ for $i=1,\ldots,n$. –  Arturo Magidin Mar 10 '12 at 4:00
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