# Why should this ideal be maximal? [duplicate]

This question already has an answer here:

My textbook, Algebraic Geometry: A Problem-Solving Approach by Garrity et al gives two versions of the Weak Nullstellensatz. A proof for the first version is given; a proof for the second version is not. However, no proof that the two are equivalent is given.

Question: why is $\langle x_1-a_1, \dots, x_n - a_n \rangle$ "clearly [a] maximal" ideal in $k[x_1,\dots,x_n]$?

This question is the part of the accepted answer which answers my question about equivalence of the forms of the Weak Nullstellensatz which I don't understand: Equivalence of weak forms of Hilbert's Nullstellensatz

Context (which is unnecessary to understand the question and unnecessarily long):

Does anyone know where I can find a proof showing that the two versions of the Weak Nullstellensatz (which are given explicitly below) are equivalent?

I am confused because the first version of the Weak Nullstellensatz seems to deal with proper ideals, while the second version of the Weak Nullstellensatz deals with maximal ideals, something which has never been mentioned before in the textbook. Besides principal ideals, the only other type of ideal which the authors had mentioned yet is radical ideals (with reference to the Strong Nullstellensatz). I know what maximal ideals are from my course on abstract algebra, but it is confusing to me that the authors don't explain how they are related to the other types of ideals already mentioned. Are all radical ideals maximal? A reference for that would also be nice.

Note: this question is very similar to this one, in fact, arguably the same. However, the accepted answer to that question uses a different version of the Hilbert Basis theorem then the one I am familiar with, so using that answer would only lead me to ask how the two different versions of the Hilbert Basis theorem are equivalent, but first I would have to clarify exactly which version the person who answered that question had in mind. (It seems to have something to do with maximal ideals, which again were not mentioned at all in my textbook until the second version of the Weak Nullstellensatz was given, with no context or proof of its equivalence. Maximal ideals didn't even come up during their proof of the Hilbert Basis Theorem.)

Versions of Theorems Used in My Textbook

Weak Nullstellensatz -- Version 1 Let $k$ be an algebraically closed field and let $I$ be a proper ideal of the polynomial ring $k[x_1, \dots, x_n]$. Then $V(I)\not=\emptyset$, i.e., there exists $(a_1, \dots, a_n) \in k^n$ such that $f(a_1, \dots, a_n)=0$ for all $f\in I$.

This implies that $V(I)=\emptyset \iff I=k[x_1,\dots,x_n]$.

Weak Nullstellensatz -- Version 2 Let $k$ be an algebraically closed field. An ideal $i$ in $k[x_1, \dots, x_n]$ is maximal if and only if there are elements $a_i \in k$ such that $I$ is the ideal generated by the elements $x_i - a_i$; that is, $I = \langle x_1 - a_1, \dots, x_n - a_n \rangle$.

Hilbert Basis Theorem If $R$ is a Noetherian ring, then $R[X]$ is also a Noetherian ring.

Strong Nullstellensatz Let $k$ be an algebraically closed field and let $I$ be an ideal of the polynomial ring $k[x_1, \dots,x_n]$. Then $I(V(I))=Rad(I)$.

What I tried: Because $k[x_1,\dots,x_n]$ is Noetherian, by the Hilbert Basis Theorem because $k$ is Noetherian because $k$ is a field, all of its ideals are finitely generated, and every non-empty set of proper ideals has a maximal element, which is thus a maximal ideal.

Version 2 implies Version 1: If $I$ is a proper ideal, then it is a subset of some maximal ideal $\tilde{I}$ (which is then also proper), possibly with $I=\tilde{I}$. Then $V(I) \supset V(\tilde{I})$ (see this) because $I \subset \tilde{I}$. By version 2 of the weak Nullstellensatz, $\tilde{I}=\langle x_1 - a_1, \dots, x_n - a_n \rangle$, so $(a_1, \dots, a_n) \in V(\tilde{I}) \subset V(I) \not=\emptyset$.

Version 1 implies Version 2: Let $I$ be a maxmial ideal. Then it is proper, so by Version 1 of the Weak Nullstellensatz, $V(I)\not=\emptyset$. By the Hilbert Basis theorem, $I$ must be finitely generated, say $I=\langle f_1, \dots, f_r \rangle$. There exists $(a_1, \dots, a_n)$ such that $f(a_1, \dots, a_n)=0$ for all $f \in I$. Now we know that $I \subset I(V(I)) \subset I(\{a_1, \dots, a_n \})$ (because $\{a_1, \dots, a_n\} \subset V(I)$, see here). Now $I(\{a_1, \dots, a_n \})\not=\emptyset$, because $x_1 -a_1, \dots, x_n - a_n \in \langle x_1 - a_1, \dots, x_n - a_n \rangle \subset I(\{a_1, \dots, a_n \})$, and because $V(I)\not=\emptyset \iff I \subsetneq k[x_1,\dots, x_n]$ by version 1 of the Weak Nullstellensatz, $I(\{a_1, \dots, a_n \})$ is a proper ideal. Since $I \subset I(\{a_1, \dots, a_n \})$ and $I$ is maximal, $I=I(\{a_1, \dots, a_n \})$. So $\langle x_1 - a_1, \dots, x_n - a_n \rangle \subset \langle f_1, \dots, f_r \rangle = I$. So $r \ge n$. I am not sure what to do from here.

Although I guess at this point, the answer to the aformentioned question comes into play: Equivalence of weak forms of Hilbert's Nullstellensatz

If $\langle x_1 - a_1, \dots, x_n - a_n\rangle$ is a priori maximal, then $\langle x_1 - a_1, \dots, x_n - a_n \rangle \subset I \implies \langle x_1 -a_1, \dots, x_n - a_n \rangle$, giving one direction of the if and only if.

And then for the other direction, if $I=\langle x_1 - a_1, \dots, x_n - a_n\rangle$, and $\langle x_1 - a_1, \dots, x_n - a_n \rangle$ is a priori maximal, then $I$ is maximal, thus finishing the proof of the equivalence.

Thus I have completed the proof of the equivalence of the two versions of the Weak Nullstellensatz provided I can argue why $\langle x_1 - a_1, \dots, x_n - a_n \rangle$ should be a priori maximal.

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 11 '16 at 7:08

• The quotient by that ideal is $K$, ie a field so it's maximal. – quid Oct 10 '16 at 19:08
• @quid Why is the quotient by that ideal $K$? I know that and understand why an ideal is maximal if and only if its quotient is a field. EDIT: never mind, the answer to my question is literally exercise 4.6.1 (three exercises after the theorem) and they give a good hint, so I should figure it out by myself. – Chill2Macht Oct 10 '16 at 19:13
• Can you see it for $n=1$ (or $a_1 = \dots a_n=0$). If the former, you could use induction. But a neater way should be to recognize it as the kernel of the (surjective) map $p \mapsto p(a_1, \dots, a_n)$ – quid Oct 10 '16 at 19:19

Consider the evaluation homomorphism $\varphi_{(a_1, \dots, a_n)} \colon k[x_1, \dots, x_n] \rightarrow k$ given by $$\varphi_{(a_1, \dots, a_n)} = p(a_1, \dots, a_n).$$ This homomorphism is clearly onto (consider an element of $k$ as a zero-degree polynomial in $k[x_1, \dots, x_n]$ and look at its image) and so $k[x_1, \dots, x_n] / \ker \varphi_{(a_1, \dots, a_n)} \approx k$ and $\ker \varphi_{(a_1, \dots, a_n)}$ is a maximal ideal.

Let us show that $\ker \varphi_{(a_1, \dots, a_n)} = \left< x_1 - a_1, \dots, x_n - a_n \right>$. Assume first that $a_1 = \dots = a_n = 0$ so we need to show that $\ker \varphi_{(0,\dots,0)} = \left< x_1, \dots, x_n \right>$. The inclusion $\left< x_1, \dots, x_n \right> \subseteq \ker \varphi_{(0,\dots,0)}$ is straightforward. For the other inclusion, note that if $p = \sum_{I} c_I x^I \in \ker \varphi_{(0,\dots,0)}$ we have $\varphi_{(0,\dots,0)}(p) = p(0, \dots, 0) = c_{(0,\dots,0)} = 0$ and so $p$ can be written as

$$p = \sum_{i=1}^n x_i \left( \sum_{J} c_{(j_1, \dots, j_{i-1}, j_i + 1, j_{i+1}, \dots, j_n)} c_J x^J \right) \in \left< x_1, \dots, x_n \right>.$$

For the general case, note that the map $f_{(a_1,\dots,a_n)} \colon k[x_1, \dots, x_n] \rightarrow k[x_1, \dots, x_n]$ given by $$f_{(a_1,\dots,a_n)}(p) = p(x_1 + a_1, \dots, x_n + a_n)$$ is an automorphism of $k[x_1, \dots, x_n]$ (with $f_{(a_1,\dots,a_n)}^{-1} = f_{(-a_1,\dots,-a_n)}$) and we have

$$\varphi_{(0,\dots,0)} \circ f_{(a_1,\dots,a_n)} = \varphi_{(a_1,\dots,a_n)}$$

and so

$$\ker \varphi_{(a_1,\dots,a_n)} = f_{(a_1,\dots,a_n)}^{-1} \left( \ker \varphi_{(0,\dots,0)} \right) = f_{(a_1,\dots,a_n)}^{-1} (\left< x_1, \dots, x_n \right>) = \\ f_{(-a_1,\dots,-a_n)}(\left< x_1, \dots, x_n \right>) = \left< f_{(-a_1,\dots,-a_n)}(x_1), \dots, f_{(-a_1,\dots,-a_n)}(x_n) \right> = \\ \left< x_1 - a_1, \dots, x_n - a_n \right>.$$

• This is a nice answer, which also provides good practice with the isomorphism theorems -- I appreciate it! – Chill2Macht Oct 11 '16 at 12:14