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The Stacks project

Noether normalization

Lemma 10.115.4. 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 10.115.3. 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 10.112.4. \square


Comments (4)

Comment #2947 by Dario Weißmann on

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

Comment #6597 by suggestion_bot on

Suggested tag: Noether normalization

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