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I want to prove that although $K$ is a field that IS NOT algebraically closed, every maximal ideal in $K[x_1, \ldots, x_n]$ can be generated by $n$ elements.

To prove this, I am following the next steps:

i) Every maximal ideal in $K[x]$ is generated exactly by ONE element.

Induction Hypothesis: Every maximal ideal in $K[x_1, \ldots, x_{n-1}]$ can be generated by $n-1$ elements.

ii) Let $\mathfrak{m}$ be a maximal ideal in $K[x_1, \ldots, x_n]$ and let $\mathfrak{p} = \mathfrak{m} \cap K[x_1, \ldots, x_{n-1}] $. I have considered the extensions:

$$K \subset {K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}} \subset K[x_1, \ldots, x_n] / \mathfrak{m}.$$

I have proved that $K[x_1, \ldots, x_n] / \mathfrak{m}$ is an algebraic extension of $K$ and $\mathfrak{p}$ is a maximal ideal that is generated by $n-1$ elements.

Now I have to consider the extension:

${K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}} \subset ({K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}})[x_n]$ and to find a generator of $\mathfrak{m} / \mathfrak{p} = \overline{\mathfrak{m}} \subset ({K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}})[x_n] $

but I don't know how to do that. Then I have to conclude that $\mathfrak{m}$ can be generated by $n$ elements. I would appreciate some help.

Thank you very much.

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  • $\begingroup$ Isn't that just a maximal ideal in the polynomial ring over a field? $\endgroup$
    – Hagen von Eitzen
    Apr 20, 2015 at 21:02
  • $\begingroup$ I have to prove this exactly following the steps :) I have edited the question, it is an algebraic extension of K $\endgroup$
    – Mathdone
    Apr 20, 2015 at 21:06

1 Answer 1

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Let $R=K[x_1, \ldots, x_{n-1}]$, and $\mathfrak p$ a maximal ideal of $R$. Then $(R/\mathfrak p)[x_n]\simeq R[x_n]/\mathfrak p[x_n]$. In $R[x_n]/\mathfrak p[x_n]$ the ideal $\mathfrak m/\mathfrak p[x_n]$ is principal. That's all.

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  • $\begingroup$ I don't get why we have that isomorphism, in the left side, we map $\mathfrak{p}$ to zero but in the right side we also map the variable $x_n$ that you have just added. $\endgroup$
    – Mathdone
    Apr 20, 2015 at 21:22
  • $\begingroup$ @user127432 $\mathfrak p[x_n]$ is the ideal of $R[x_n]$ which consists of all polynomials whose coefficients are in $\mathfrak p$. In other words, $\mathfrak p[x_n]$ is the extension of $\mathfrak p$ to $R[x_n]$. (When talk about the contraction of an ideal just take a look at the "dual" operation, the extension of an ideal. Both are clearly explained in Atiyah and Macdonald.) $\endgroup$
    – user26857
    Apr 20, 2015 at 21:27
  • $\begingroup$ Ok, thank you so much, I only have to understand why that ideal is principal $\endgroup$
    – Mathdone
    Apr 20, 2015 at 21:45
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    $\begingroup$ @user127432 For this use i) and notice that $R/\mathfrak p$ is a field. $\endgroup$
    – user26857
    Apr 20, 2015 at 21:50

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