Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working through my notes and I'm stuck in the middle of the proof of Hilbert's Nullstellensatz.

(Hilbert's Nullstellensatz) Let $k$ be an algebraically closed field. Let $J$ be an ideal in $k[x_1, \dots , x_n]$. Let $V(J)$ denote the set of $x$ in $k$ such that all $f$ in $J$ vanish on them. Let $U \subset k^n$ and let $I(U)$ denote the set of $f$ in $k[x_1, \dots , x_n]$ that vanish on $U$. Then $$ r(J) = I(V(J))$$ where $r$ denotes the radical of $J$.


Let me go through the proof as far as I understand it:

$\subset$: Easy. Let $p \in r(J)$. Then $p^k \in J$ which means $p(x)^k = 0$ for all $x$ in $V(J)$. Hence $p(x) = 0$ for $x$ in $V(J)$ hence $p \in I(V(J))$.

$\supset$: Assume $f \notin r(J)$. Then for all $k>0$, $f^k \notin J$. We know that there exists a prime ideal $p$ such that $J \subset p$ and $f^k \notin p$ for all $k>0$. To see this we use the same argument used in the proof of proposition 1.8. on page 5 in Atiyah-MacDonald: Let $\Sigma$ be the set of all ideals that do not contain any power of $f$. We order $\Sigma$ by inclusion and use Zorn's lemma to get a maximal element $p$. We claim $p$ is prime. Assume neither $x \notin p$ nor $y \notin p$ (then we want to show $xy \notin p$). Then $p + (x), p + (y)$ are ideals properly containing $p$ hence neither of them is in $\Sigma$ hence $f^n \in p + (x)$ and $f^m \in p + (y)$. Now $f^{n+m} \in (p + (x)) (p + (y)) = p^2 + (x)\cdot p + (y)\cdot p + (xy) \subset p + (xy)$ so $p + (xy) \notin \Sigma$. Hence $p$ is properly contained in $p + (xy)$ hence $xy$ cannot lie in $p$.

So we have $p$ is a prime ideal containing $J$. Now consider the map $$ k[x_1, \dots, x_n] \xrightarrow{\pi_1} k[x_1, \dots, x_n]/p \xrightarrow{i} (k[x_1, \dots, x_n]/p) [\overline{f}^{-1}] =: B([\overline{f}^{-1}]) \xrightarrow{\pi_2} B[\overline{f}^{-1}] /m$$

where $\overline{f}$ denotes $\pi_1 (f)$ and $m$ is some maximal ideal in $B[\overline{f}^{-1}]$. We may assume $f \neq 0$ so that $\overline{f} \neq \overline{0}$. $\overline{f}^{-1}$ is an element of the field of fractions of $B$ so we may adjoin it to $B$ to get a new ring.

Since we only adjoined one element and otherwise only took quotients, the thing coming out on the RHS is a finitely generated $k$-algebra (because $ k[x_1, \dots, x_n]$ is).

Now by theorem 5.24 in Atiyah-MacDonald we know that $k \cong B[\overline{f}^{-1}] /m$.

The proof now finishes as follows: "Let $t_1, \dots, t_n$ denote the images of $x_1 , \dots, x_n$ under this composite ring homomorphism.

(*)By construction, $g \in J \implies g \in p \implies g(t_1, \dots, t_n) = 0 \implies (t_1 , \dots, t_n ) \in V(J)$.

(**)On the other hand, $f(t_1 , \dots, t_n )$ is precisely the image of $f$ in $B[\overline{f}^{-1}] /m$, which is a unit. $\implies f(t_1 , \dots, t_n ) \neq 0 \implies f \notin I(V(J))$."

Question 1: What is the line (*) showing? I think we want to show $f \notin I(V(J))$, where does this come in here?

Question 2: Why is $f(t_1 , \dots, t_n )$ a unit?

Thank you for your help.

share|improve this question
1  
$f(t_1,\dots,t_n)$ is a unit because $f$ is invertible in $B[f^{-1}]$ and remains so in the homomorphic image $B[f^{-1}]/m$, but the image of $f$ is precisely $f(t_1,\dots,t_n)$. The point of ($*$) is that there is one distinguished point $(t_1,\dots,t_n)$ in $B[f^{-1}]/m\times\cdots\times B[f^{-1}]/m\simeq k^n$ that kills all the polynomials in $J$, and thus all the polynomials in $I(V(J))$, so it should also kill $f$ (by definition). Since it doesn't by ($**$), this is the desired contradiction. –  Olivier Bégassat Jul 19 '12 at 15:46
    
@OlivierBégassat Oh, of course! Thank you. I confused myself. It's $\overline{f}(t_1, \dots , t_n)$ really, and it's a unit because we made it a unit by adjoining $\bar{f}^{-1}$. Then $\pi_2$ is a ring homo hence maps units to units, et voilà. –  Rudy the Reindeer Jul 19 '12 at 16:33
    
@OlivierBégassat As for question 2: I'm not sure I understand. We don't want a contradiction. We assumed $f \notin r(J)$ and we want to show $f \notin I(V(J))$. –  Rudy the Reindeer Jul 19 '12 at 16:35
    
I don't think so: you assume $f\in I(V(J))$, and want to show that this implies $f\in\sqrt{J}=r(J)$. To do so, you reason by contradiction: suppose there was $f\in I(V(J))$ not in $r(J)$, then... The conclusion is that $f$ doesn't kill $(t_1,\dots,t_n)\in V(J)$ (by ($*$)) which contradicts the assumption $f\in I(V(J))$. –  Olivier Bégassat Jul 19 '12 at 16:43
    
@OlivierBégassat But the lecturer started the proof by writing "Assume $f \notin r(J)$...." Then he goes on to showing that if we assume that, we get a prime ideal which we can use to construct a composite quotient map. And at the end he concludes the proof with $\implies f \notin I(V(J))$. –  Rudy the Reindeer Jul 19 '12 at 16:46

1 Answer 1

up vote 2 down vote accepted

This is a comment made into an answer. Suppose $f$ is any polynomial not in $r(J)$.

Question 2 $f(t_1,\dots,t_n)$ is a unit because $f$ is (by construction) invertible in $B[f^{-1}]$ and remains so in the homomorphic image $B[f^{-1}]/m$. But the image of $f$ is precisely $f(t_1,\dots,t_n)$.

Question 1 The point of $(*)$ is that there is one distinguished point $(t_1,\dots,t_n)$ in $B[f^{-1}]/m×⋯×B[f^{-1}]/m≃k^n$ that kills all the polynomials in $J$. Put another way, $$\mathrm{the~point~ of~}(*)\mathrm{~ is~ that~}(t_1,\dots,t_n)\in V(J)$$ and thus all the polynomials in $I(V(J))$ must be killed by it. By $(**)$, $f$ does not take this point to $0$ and so $$\mathrm{the~point~ of~}(**)\mathrm{~ is~ that~}f\notin I(V(J)).$$ By now the authors have shown that $f\notin r(J)\Rightarrow f\notin I(V(J))$ i.e. $k[x_1,\dots,x_n]\setminus r(J)\subset k[x_1,\dots,x_n]\setminus I(V(J))$ i.e. $$I(V(J))\subset r(J).$$

share|improve this answer
    
But one can prove it directly (without contradiction) and then one does not need (*). Right? (I'd written this in a comment to the question but I made some unfortunate typos) –  Rudy the Reindeer Jul 19 '12 at 17:37
1  
I haven't used contradiction this time, I think both points are essential. Without the first point $(*)$, I wouldn't know how t deduce the second one... –  Olivier Bégassat Jul 19 '12 at 17:38
    
I think I was confused because I thought that $x_i$ are variables (namely, they take values in $k$). But they are elements of the ring $k[x_1, \dots, x_n]$ and hence the image of $(x_1, \dots , x_n)$ is one point in $k^n$. Does that make sense? –  Rudy the Reindeer Jul 19 '12 at 17:45
1  
Absolutely, the $x_i$ are polynomials! And the fact that $f$ is sent to $f(t_1,\dots,t_n)$ depends on this! Indeed, $f$ is an algebraic expression in the elements $x_1,\dots,x_n$, maybe $x_1^4(x_2-5x_3^{17})+1-x_7^3$, and thus its image under the ring homomorphism $B[f^{-1}]\rightarrow B[f^{-1}]/m$ is the corresponding algebraic expression in the images of $x_1,\dots,x_n$ i.e. the $t_1,\dots,t_n\in (B[f^{-1}]/m\simeq)k$, so with our example $t_1^4(t_2-5t_3^{17})+1-t_7^3$. –  Olivier Bégassat Jul 19 '12 at 17:51
1  
It's always a pleasure :D –  Olivier Bégassat Jul 19 '12 at 17:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.