In Miles Reid, undergraduate commutative algebra, I read the following:

"Suppose that $k$ is an algebraically closed field and that $A=k[x_1,...,x_n]$ is a finitely generated $k$-algebra of form $A=k[X_1,...,X_n]/J$ where $J$ is an ideal of $k[X_1,...,X_n]$. (Here I'm using the notation to mean that $x_i=X_i$ mod $J$). Then every maximal ideal of $A$ is of form $(x_1-a_1,...,x_n-a_n)$ for some point $(a_1,...,a_n) \in V(J)$"

  1. Have I understood it right if I say $x_i \in k[X_1,...,X_n]/J$?

  2. Can someone explain what he means by "Here I'm using the notation....".

  3. $k$ is algebraically closed, what does this imply? Why do we need this?


First of all, "Here I'm using the notation to mean that $x_i=X_i \pmod J$" is a way to say that the author indicate with $x_i$ the CLASS of the single variable $X_i$ in the quotient $k[X_1, \dots, X_n]/J$.

Now, we need the assumption of $k$ algebraically closed, why? To understand this let me show you a counterexample:

Think to $\mathbb{Q}[x,y]$ and one of its quotients, say $$\mathbb{Q}(\sqrt{2})[x]=\mathbb{Q}[x,y]/(x^2-2)$$ The ideal $(y^2+1)$ is maximal here but non in the form as above!

The result of which Reid is talking for is strictly related to one of the most important theorem of commutative algebra and algebraic geometry: the Hilbert's Nullstellensatz.

In fact, in one of its many equivalent form, this theorem says:

If $k$ is an algebraically closed field, the maximal ideal of the polynomial ring $k[X_1, \dots,X_n]$ are all of the form $(X_1-\alpha_1, \dots, X_n-\alpha_n)$

Your assert comes directly from this an by the correspondence of maximal ideals of a ring $A$ with maximal ideals of its quotient $A/J$.

  • $\begingroup$ Thank you for your answer. So for the first question, this simply means that $x_i = X_i + J$? $\endgroup$ – user117449 Nov 23 '14 at 15:50
  • $\begingroup$ Yes, you're right $\endgroup$ – Sabino Di Trani Nov 23 '14 at 16:25
  • $\begingroup$ The proof for the theorem states that the ideals of $A$ are given by ideals of $k[X_1,, $\endgroup$ – user117449 Nov 23 '14 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy