# Form of maximal ideals in an algebraicaly closed polynomial ring

I have been trying to prove the following bijection which is a consequence of the nullstellansatz $$\{\text{maximal ideals of }\mathbb{C}[x_1,\dots,x_n] \} \leftrightarrow \{\text{points in }\mathbb{A}^{n}_{\mathbb{C}}\}$$

and I am struggling to prove that all maximal ideals in $\mathbb{C}[x_1,...x_n]$ have the form $\langle x_1 - a_1, .... , x_n - a_n\rangle$.

I have the following idea but it seems incorrect:

1. Take a maximal ideal $M$ in $\mathbb{C}[x_1, \dots, x_n]:= C$.
2. Then $C/ M$ is a field.
3. Consider the ideals $\langle x_i + M \rangle$ in $C/M$.
4. As these are ideals in a field they are either the zero ideal or the whole ideal.
5. If they are all the zero ideal then we are done as then $\langle x_1,....,x_n \rangle$ is in $M$.
6. Otherwise some $\langle x_i + M \rangle = C/M$ and by the correspondence theorem for rings when I pull this back to $C$ I get $\langle x_i \rangle = C$ which is a contradiction as $1 \not \in \langle x_i \rangle$.

I think this is wrong though as I don't think I have used algebraically closed anywhere. Thus I could use this to prove all maximal ideals in $\mathbb{R}[x]$ are of the forem $\langle x - a \rangle$ which is false as $\langle x^2 + 1 \rangle$ is maximal.

Is there a way to make this work?

• $\langle x_i+M\rangle$ pulls back to $\langle x_i\rangle+M$ – anon May 2 '15 at 20:18

You had a proof that you were suspicious of, and you also had what looked like a counterexample to a statement that your proof looked like it proved. Did you try going through the steps of the proof with your counterexample? Here is what would have happened:

1. Take a maximal ideal $(x^2 + 1)$ in $\mathbb{R}[x]$.
2. Then $\mathbb{R}[x]/(x^2 + 1)$ is a field (it's $\mathbb{C}$).
3. Consider the ideal $(x) + (x^2 + 1)$ in $\mathbb{C}$ (it's the ideal generated by $i$, which is all of $\mathbb{C}$).
4. As this is an ideal in a field it is either the zero ideal or the whole field (it's the whole field).
5. When I pull this back to $\mathbb{R}[x]$ I don't get $(x) = \mathbb{R}[x]$. As anon's comment says, I get $(x) + (x^2 + 1) = \mathbb{R}[x]$.

Anyway, this is not a statement that admits a completely formal proof. You really have to put in some amount of actual work. The key point is that $\mathbb{C}[x_1, x_2, \dots x_n]$ modulo a maximal ideal is always $\mathbb{C}$ (rather than some bigger field, as above). There are various ways to show this; one relatively cheap way (which depends on $\mathbb{C}$ being uncountable as well as algebraically closed) is to show that $\mathbb{C}[x_1, x_2, \dots x_n]$ modulo a maximal ideal cannot contain a transcendental element (over $\mathbb{C}$) because it's too small (at most countable-dimensional).

As pointed out above the method I posted is incorrect.

Here is a way to prove it using the weak nullstellansatz which states that: $$J \subsetneq \mathbb{C}[x_1, \dots, x_n] \implies V(J) \neq \emptyset.$$

Suppose $J$ is maximal. By definition $J$ is not the whole ring and so by the weak nullstellansatz $V(J) \neq \emptyset$.

Now notice that $J \subseteq I(V(J))$ and so $I(V(J))$ is either the entire ring or $J$. As $V(J)$ is not empty $I(V(J)) \neq \mathbb{C}[x_1, \dots, x_n]$ and so we are left with $J=I(V(J))$.

Now to see the form of $J$ is what we think it should be let $(a_1, \dots, a_n) \in V(J)$. Clearly $P:=\{(a_1,\dots,a_n)\} \subseteq V(J)$ and by the inclusion reversing of $I(\cdot)$ we see, $$J=I(V(J)) \subseteq I(P),$$

Using the maximality of $J$ again we get what we want, $$J = I(P) =\langle x_1 - a_1, \dots, x_n - a_n \rangle.$$