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I want to prove that every maximal ideal $m \subset \mathbb{C}[X_0,X_1]$ verifies: $m=(X_0 - \alpha, X_1 - \beta), (\alpha, \beta) \in \mathbb{C}^2$.

I've read that $m=(X_0 - \alpha, X_1 - \beta)$ , i,e, (the ideal of every polynomial $f$, with $f(\alpha, \beta)=0)$, is a maximal ideal, but I don't know to show that EVERY maximal ideal is equal to one of this.

If any of you knows a detailed proof it would help me a lot.

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  • $\begingroup$ This is called Hilbert's Nulstellensatz. Any textbook in algebraic geometry should contain this theorem somewhere in the beginning. $\endgroup$ – user96815 Oct 30 '13 at 12:42
  • $\begingroup$ Thanks, I didn't know the name of the theorem. Do you know where can I find a detailed proof of it? $\endgroup$ – Miguel Hernández Oct 30 '13 at 12:53
  • $\begingroup$ Also, you can use LaTeX on this site! $\endgroup$ – user64687 Oct 30 '13 at 12:56
  • $\begingroup$ @MiguelHernández : Just walk into your library and pick up a textbook of algebraic geometry that pleases you! I haven't to this date seen a textbook of Alg. Geom. without a proof of this theorem. $\endgroup$ – user96815 Oct 30 '13 at 13:03
  • $\begingroup$ Dear Miguel, If you google "Nullstellensatz", you will get an enormous number of hits, some of which will be proofs of the theorem. Regards, $\endgroup$ – Matt E Oct 30 '13 at 15:12
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This follows from proposition 7.9 in Atiyah-Macdonald's introduction to commutative algebra. The proposition says that if $k$ is a field and $E$ is a finitely generated $k-$algebra. Then $E$ is a finite algebraic field extension of $k$. This is proved in the text. Now as a corollary: Let $m$ be a maximal ideal of $R = \mathbb{C}[x_0,x_1]$. Then $R/m$ is a field and it's clearly a finitely generated $\mathbb{C}$-algebra. Hence by the prop it's a finite algebraic extension of $\mathbb{C}$, but $\mathbb{C}$ is algebraically closed hence this extension is isomorphic to $\mathbb{C}$ as a $\mathbb{C}$-algebra. Hence we have isomprhism $\varphi : R/m \to \mathbb{C}$. Let $a_0$,$a_1$ be the image of respectively $x_0$ and $x_1$. Then we see that $f(x_0,x_1) = x_0 - a_0$ and $g(x_o,x_1) = x_1 - a_0$ is contained in the kernel, hence we must have $(x_0 -a_0, x_1 -a_1) \subset m$ but by what you have already proved we must have equality.

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