# All maximal ideals in the ring of polynomials of are of the kind $N_p=\langle x_i-p_i:i=\overline {1,n}\rangle$ for some point p in the affine space

I am reading a proof on the coincidence of the functional field of a variety (defined by equivalence classes of regular functions) and the field of quotients of its coordinate ring. It turns out I have gaps in my understanding of introductory commutative algebra.

Lemma: All maximal ideals in the ring of polynomials $k[x_1,\dots,x_n]$ of $n$ variables over an algebraically closed field $k$ are of the kind $N_p=\langle x_i-p_i:i=\overline {1,n}\rangle$ for some point $p=(p_1,\dots,p_n)$ in the affine space $k^n$.

Proof: If $N\trianglelefteq k[x_1,\dots,x_n]$ is a maximal ideal then the factor ring $$k[x_1,\dots,x_n]/N = ( k+N)/N[x_1+N,\dots,x_n+N] \approx ( k/( k \cap N)) [x_1+N,\dots,x_n+N] = k[x_1+N,\dots,x_n+N]$$ is a finitely generated $k$-algebra.

(I am not convinced any of these equalities hold. I only found a lemma according to which the rings $( k+N)/N$ and $k/( k \cap N)$ are isomorphic but it is left as a preliminary remark and without proof.)

Then by a previous lemma the elements $x_i+N,i=\overline {1,n}$ are algebraic over $k$ so $x_i+N\in k\approx k+N/N$, since $k$ is algebraically closed. Therefore, for each $0\leq i\leq n$ exists $p_i\in k$ such that $x_i+N=p_i+N$.

(It is true that $p_i$ may be the image of $x_i$ under the said isomorphism, but how can we justify $x_i+N=p_i+N$?)

p.s. After carefully reading the proof, I resolved all further issues. Thank you for helping.

• Isn't this the (weak) Nullstellensatz? – egreg Dec 31 '13 at 17:34
• Very well then. Yet I'd like you to point out where I mixed up $k$ and $\overline k$, @YACP. – superAnnoyingUser Jan 1 '14 at 9:48
• I see which instances you mean. I did mix up. – superAnnoyingUser Jan 1 '14 at 10:01

In order to understand why $$k[x_1,\dots,x_n]/N=((k+N)/N)[x_1+N,\dots,x_n+N] \simeq (k/(k \cap N)) [x_1+N,\dots,x_n+N] = k[x_1+N,\dots,x_n+N]$$ just take a polynomial in $k[x_1,\dots,x_n]$ and see what happens with this modulo $N$: it is a "polynomial" in the residue classes $x_i+N$ of the variables $x_i$ and its coefficients are residue classes of elements in $k$, so they are in $(k+N)/N$. Now use the isomorphism theorem you've mentioned and note that $k\cap N$ is an ideal of $k$, so $k\cap N=(0)$ or $k\cap N=k$, but the last case is not possible, otherwise $N$ would contain an invertible element and thus $N=k[x_1,\dots,x_n]$, a contradiction.
Now you said that by Zariski's Lemma get $x_i+N\in k$. If you look back at the way the elements in $k[x_1,\dots,x_n]/N$ look like, you found that $x_i+N$ is in fact the residue class of a constant modulo $N$, so there exists $p_i\in k$ such that $x_i+N=p_i+N$.
• Thank you, user, your answer is helpful and convincing. I am still not entirely convinced the first equality is true, however, for your argument may be too close to restating the result. How can I see that the residues of polynomials are polynomials in the residues $x_i+N$ with coefficients residues of constants? – superAnnoyingUser Jan 1 '14 at 10:37
• Dear @Ivan, taking the residue class modulo $N$ of a polynomial $f$ means to apply a homomorphism (the canonical projection) $\pi:k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]/N$ to $f$. The homomorphism commutes with the sums and products, so $\pi$ applies on each $x_i$ that appears in $f$ and on each coefficient of $f$. But $\pi(x_i)=x_i+N$ and $\pi(a)=a+N$, for $a\in k$. (Actually you replace in $f$ every $x_i$ by $x_i+N$ and every coefficient $a$ by $a+N$.) I hope this time I was more convincing. – user89712 Jan 1 '14 at 10:54