# Small step in proving Hilbert's (Weak) Nullstellensatz

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!

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 $a_i$.
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?
• 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