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 reading up on a proof of Hilbert's Nullstellensatz which uses the Artin-Tate lemma. I followed all of it except for one step, which is probably quite elementary, but my brain may be too fried from the rest of the proof to find the logic.

Let $k$ be an algebraically closed field, and $I$ be a maximal ideal of the polynomial ring $k[x_1,...,x_n]$. After much proof, we have that $k[x_1,...,x_n]/I=k$. Then for each $x_i$, there exists $a_i\in k$ such that $x_i-a_i\in I$.

(Finishing the proof from here, $I$ must contain the ideal $(x_1-a_1,...,x_n-a_n)$, and that ideal is maximal, so $I$ is exactly that.) What's the reason for why such an $a_i$ exists? I appreciate your help!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Consider the quotient map $f: k[x_1,\cdots,x_n]\to k$. The image of each $x_i$ is some element in $k$, call it $b_i$. Then since $f$ is a $k$-algebra homomorphism, we have that $f(x_i-b_i)=f(x_i)-f(b_i)=b_i-b_i=0$, or that $x_i-b_i\in I$, and thus $b_i$ is the desired $a_i$.

share|improve this answer

Write $A = k[x_1, \ldots, x_n]$ to save space. Here is the argument in (too much) detail: there is a canonical injection $f\colon k \to A$, and after much effort we show that the composition of this embedding and the quotient map $g\colon A \to A/I$ is an isomorphism. So for each $i$ there is a unique $a_i \in k$ such that $g(f(a_i)) = g(x_i)$. Is the result clear now?

share|improve this answer
1  
Maybe the important thing to note is that everything is being treated as a $k$-algebra, and so $A/I$ is not just isomorphic to $k$ via some relatively arbitrary ring homomorphism: there's a canonical copy of $k$ inside of it and the proof shows that this is in fact all of $A/I$. –  Dylan Moreland Aug 31 '12 at 5:19
    
Ah, I got it. If I know that $A/I=k$ , then I can just consider the commutative diagram $f:k\to A, g:A\to A/I, h: A/I\to k$ where $f$ is injective, $g$ surjective, and $h$ isomorphic. Then $g\circ f$ is an isomorphism. Hence, the result follows. Thanks! –  Dustin Tran Aug 31 '12 at 5:24

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.