
Lemma 10.114.4. Let $k$ be a field. Let $S = k[x_1, \ldots , x_ n]/I$ for some ideal $I$. If $I \neq (1)$, there exist $r\geq 0$, and $y_1, \ldots , y_ r \in k[x_1, \ldots , x_ n]$ such that (a) the map $k[y_1, \ldots , y_ r] \to S$ is injective, and (b) the map $k[y_1, \ldots , y_ r] \to S$ is finite. In this case the integer $r$ is the dimension of $S$. Moreover we may choose $y_ i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots , x_ n]$ generated by $x_1, \ldots , x_ n$.

Proof. By induction on $n$, with $n = 0$ being trivial. If $I = 0$, then take $r = n$ and $y_ i = x_ i$. If $I \not= 0$, then choose $y_1, \ldots , y_{n-1}$ as in Lemma 10.114.3. Let $S' \subset S$ be the subring generated by the images of the $y_ i$. By induction we can choose $r$ and $z_1, \ldots , z_ r \in k[y_1, \ldots , y_{n-1}]$ such that (a), (b) hold for $k[z_1, \ldots , z_ r] \to S'$. Since $S' \to S$ is injective and finite we see (a), (b) hold for $k[z_1, \ldots , z_ r] \to S$. The last assertion follows from Lemma 10.111.4. $\square$

Comment #2947 by Dario Weißmann on

Nitpick: $I$ should obviously be a proper ideal. It's hard to fit a field into the zero-ring.

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