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 have solutions to Exercises 5.16–5.19 in Atiyah–Macdonald's Introduction to Commutative Algebra, but not in the order desired; I find myself needing later exercises to do earlier ones, and it's been frustrating me. Online solution sets (I count five, in various stages of completion) seem to either not notice this problem or treat it as something too obvious to merit consideration.

For context, the first part of 5.16 is Noether's normalization lemma, which is fine, and 5.18 is Zariski's Lemma, which is proved in the text two times, and is accessible to (at least) two natural proofs at this point.

What I can't do is to obtain the second part of 5.16 without using 5.17, or prove 5.17 without using 5.18. Moreover, 5.19 specifically asks one to prove 5.17 using 5.18 (which I can do), so the strong implication is that the original solutions should not require use of Zariski's Lemma.

$k$ is an infinite field, assumed algebraically closed in 5.17 but not 5.16.

The first part of 5.17 is that if an affine variety $X$ in $k^n$ has ideal $I(X)$ a proper subset of $k[t_1,\ldots,t_n]$, then $X$ is not empty. This follows immediately from the second part of 5.16.

The second part of 5.16 is that for any subvariety $X$ of $k^n$ there are an $r \leq n$ and a linear map $k^n \to k^r$ taking $X$ onto all of $k^r$. The natural candidate map follows from Noether normalization, but my approach to surjectivity seems to require the second part of 5.17.

The second part of 5.17 is that every maximal ideal of $k[t_1,\ldots,t_n]$ is of the form $(t_1 - a_1,\ldots,t_n - a_n)$ for some $a_i \in k$. From this one can show a similar result for the coordinate ring of a subvariety of $k^n$.

So this mess will be fixed if I can prove the second parts of 5.16 and 5.17 without reference to later material.

In more detail, my current approach to the second part of 5.16 is as follows. Noether normalization shows $A$ is integral over some polynomial ring $A' := k[y_1,\ldots,y_r]$ in $r \leq n$ indeterminates. The proof of Noether normalization, at least in case $k$ is infinite, gives the $y_i$ as $k$-linear combinations of the natural generators $x_1,\ldots,x_n$ of $A$, the coordinate functions on $X$. If we say $y_i = \sum a_{ij} x_j$, then the projection $k^n \to k^r$ should be given by $(z_1,\ldots,z_n) \mapsto \big(\sum_{j=1}^n a_{1j}z_j,\ldots,\sum_{j=1}^n a_{rj}z_j,\big)$. It's not immediately obvious to me that it is surjective. However, letting $\textrm{Max}(A) \subseteq \textrm{Spec}(A)$ denote the set of maximal ideals of $A$, results of Chapter 5 show that the map $\textrm{Max}(A) \to \textrm{Max}(A')$ induced by $k^n \to k^r$ is surjective. If we know each maximal ideal of $A$ is of the form $(x_1-c_1,\ldots,x_n-c_n)$ for some $(c_1,\ldots,c_n) \in X$, then we can identify $X$ and $\textrm{Max}(A)$, and that will show the map $X \to k^r$ is surjective. That seems, however, not to be what they want, and also like too much work.

My current approach to the second part of 5.17 is to use the weak Nullstellensatz (if $k$ is algebraically closed and $\mathfrak a$ is a proper ideal of $k[t_1,\ldots,t_n]$, then there is at least one point of $k^n$ at which $\mathfrak a$ vanishes). The weak Nullstellensatz implies the first part of 5.17, but the reverse implication is not obvious to me. This implication is certainly proved by, e.g., the strong Nullstellensatz, but that would be completely missing the point.

Update: To clarify, the text of 5.16, excluding a long hint about the first part, is as follows.

Let $k$ be a field and let $A \neq 0$ be a finitely generated $k$-algebra. Then there exist elements $y_1,\ldots,y_r \in A$ which are algebraically independent over $k$ and such that $A$ is integral over $k[y_1,\ldots,y_r]$. We shall assume that $k$ is infinite. (The result is still true if $k$ is finite, but a different proof is needed.)

...

From the proof it follows that $y_1,\ldots, y_r$ may be chosen to be linear combinations of $x_1,\ldots,x_n$. This has the following geometrical interpretation: if $k$ is algebraically closed and $X\!$ is an affine algebraic variety in $k^n$ with coordinate ring $A \neq 0$, then there exists a linear subspace $L$ of dimension $r$ in $k^n$ and a linear mapping of $k^n$ onto $L$ which maps $X\!$ onto $L$ [Use Exercise 2].

I had forgotten about that hint... Here is what Ex. 5.2 says.

Let $A$ be a subring of a ring $B$ such that $B$ is integral over $A$, and let $f\colon A \to \Omega$ be a homomorphism of $A$ into an algebraically closed field $\Omega$. Show that $f\!$ can be extended to a homomorphism of $B$ into $\Omega$. [Use (5.10).]

share|improve this question
2  
can't you ask questin in short? –  A.G Mar 3 '11 at 11:03
9  
I thought the context would be helpful. Evidently I was wrong. –  jdc Mar 3 '11 at 11:32
1  
@jdc "I would say that generally speaking the average user provides too little rather than too much information." - Moderator Yuan, meta.math.stackexchange.com/questions/1711/… I think it's safe to say you're in fair straits. –  Uticensis Mar 3 '11 at 12:32
    
I don't have the book handy: can you explain what it means to take $X$ onto all of $k^r$? (Certainly this won't necessarily be true if one works with $k$-valued points and $k$ is not algebraically closed, so do you really mean $\overline{k}$-valued points, or perhaps just maximal spectra, or even prime spectra?) –  Matt E Mar 3 '11 at 13:21
1  
Oh dear. According to the book (Exercise 1.27), a variety is a subset $X$ of a linear space $k^n$ such that for some collection of polynomials $f_\alpha \in k[t_1,\ldots,t_n]$, we have $X = \{(a_1,\ldots,a_n) \in k^n : \forall \alpha\ \big(f_\alpha(a_1,\ldots,a_n) = 0\big)\}$. This is, I think, the classical definition. –  jdc Mar 4 '11 at 8:37

2 Answers 2

up vote 9 down vote accepted

You haven't said what it means in this context for $X$ to be "nonempty", but I presume it means that there is a $\overline{k}$-valued point of $X$. (Certainly $X$ need not admit a $k$-valued point if $k$ is not algebraically closed, so considering $\overline{k}$-valued points seems to be the most sensible interpretation.)

With this interpretation, the non-emptiness of $X$ follows directly from Noether normalization together with Exercise 5.2:

If $A$ is the coordiante ring of $X$, assumed to be non-zero, then Noether normalization gives an integral injective map $k[y_1,\ldots,y_r] \hookrightarrow A,$ for some $0 \leq r \leq n$. Exercise 5.2 then guarantees that any map $k[y_1,\ldots,y_r] \to \overline{k}$ can be extended to a map $A \to \overline{k}$; this is the desired surjectivity.

[Added:] Having read the comments below, let me try again. Rereading the second part of 5.16, I see that in fact $k$ is to be assumed algebraically closed, so that $k = \overline{k}$; this simplifies the above discussion, by removing the need to talk about $\overline{k}$-valued points.

So we have $X \subset k^n$ cut out by some non-unit ideal $I \subset k[x_1,\ldots,x_n]$, an integral injection $k[y_1,\ldots,y_r] \hookrightarrow A := k[x_1,\ldots,x_n]/I,$ and our goal is to prove that the induced map $X \to k^r$ is surjective. The argument written above proves this: any point of $k^r$ corresponds to a map $k[y_1,\ldots,y_r] \to k.$ Exercise 5.2 extends this to a map $A \to k.$ This gives a point of $X$ mapping to the given point of $k^r$. QED.

share|improve this answer
1  
I don't think I'm understanding. In Exercise 5.16, as reproduced above (which I should have one initially, sorry), nothing is said about $X$ being non-empty, although for the map $X \to L$ to be surjective, we do need each fiber to be nonempty. –  jdc Mar 4 '11 at 8:15
    
I am (embarrassingly) not familiar with the terminology of $k$-valued points, which I should learn. Since the book doesn't use (or indeed, mention) this language, it seems an answer should be possible without utilizing it; and unfortunately, that's all I would understand at the moment. Apologies for my obtuseness. –  jdc Mar 4 '11 at 8:25
    
The only way, in my current limited viewpoint, I can think to use Exercise 5.2 is the following. A maximal ideal $\mathfrak m$ is the kernel of a map $k[y_1,\ldots,y_r] \to \bar k$, and lies the kernel $\mathfrak p$ of an extended map $A \to \bar k$. Results of Chapter 5 show that $\mathfrak p$ is maximal. Then if we know the form of maximal ideals of $A$, for example using a later exercise, we are done. I am not sure if that has anything to do with your $\bar k$-valued points, but it is my best attempt to interpret your response in terms I already know. Thank you for your patience. –  jdc Mar 4 '11 at 8:31
1  
@jdc: Dear jdc, Firstly, I had missed the fact that in the second part of 5.16 (which is what we are discussing) $k$ is assumed algebraically closed, so $k = \overline{k}$; this simplifies the above discusion. Secondly, there is no need to talk about maximal ideals. Rather, one has a point of $k^r$, and wants to lift it to a point of $X$. This follows directly from Exercise 5.2 directly as it is stated, taking $\Omega$ to be $k$. This is written out in an edit to my answer. Regards, –  Matt E Mar 4 '11 at 13:10

This is just another wording of Matt E's very nice argument. That's why I'm making it a community wiki. (I voted for the question and for Matt E's answer.)

I'll freely use Exercises 1.27 and 1.28. For any regular map $\varphi:V\to W$ between affine varieties, write $\varphi^*$ for the induced $k$-algebra morphism between coordinate rings (going into the reverse direction).

Let $V\subset k^n$ be an affine variety, let $A$ be its coordinate ring, let $B$ be the coordinate ring of $k^r$, and let $\varphi:V\to k^r$ be a regular map induced by a linear map from $k^n$ to $k^r$. Assume that $\varphi^*:B\to A$ is injective and that $A$ is integral over $\varphi^*(B)$.

Let $z$ be in $k^r$. We must find a $v$ in $V$ such that $\varphi(v)=z$.

View $z$ as a regular map from the zero dimensional affine space $\{0\}$ to $k^r$. By Exercise 5.2 there is a $k$-algebra morphism $v^*:A\to k$, coming from a $v$ in $V$ viewed as a regular map from $\{0\}$ to $V$, such that $v^*\circ\varphi^*=z^*:B\to k$, and we get $\varphi\circ v=z$.

share|improve this answer
    
Okay, that seems to work for me. I believed in the $k$-algebra homomorphism $v^*\colon A \to k$, but I couldn't see why that gave a point. I was bogged down with thinking of points as (a priori special kinds of) maximal ideals, wanting to see that that was in fact the general case, and was all set to ask the generous man about that once again. But what you are saying is that a map $A \to k$ comes from a point just using the correspondence of Ex. 1.28 between $k$-algebra homomorphisms and regular maps the other way, which somehow had never occurred to me. Thank you. –  jdc Mar 5 '11 at 11:07
1  
@jdc - Yes, exactly! --- Thanks again for such a nice question! –  Pierre-Yves Gaillard Mar 5 '11 at 11:48
    
Thank you for your kindness and patience with my glacial mental processing. –  jdc Mar 5 '11 at 20:44
    
If $X\!$ be an affine algebraic variety in $k^n$, where $k$ is an algebraically closed field, and $I(X)$ be the ideal of $X$ in the polynomial ring $k[t_1,\ldots,t_n]$, then $I(X) \neq (1)$ then $X\!$ is not empty. The book seems to say this implies every maximal ideal in the ring $k[t_1,\ldots,t_n]$ is of the form $(t_1 - a_1,\ldots,t_n - a_n)$ for some $a_i \in k$ Why? I only see how to get this using Zariski's Lemma or possibly that if $B = k[t_1,\ldots,t_n,t_{n+1}]$ is a polynomial ring, then a maximal ideal of $B$ meets $A = k[t_1,\ldots,t_n]$ in a maximal ideal of $A$. –  jdc Mar 7 '11 at 16:50
    
@jdc - You're probably referring to Exercise 5.17. Here is what I think. Let m be a maximal ideal of A:=k[t_1,...,t_n]. By the previous exercise, there are algebraically independent y_i in A/m such that A/m is integral over k[y_1,...,y_r]. Proposition 5.7 yields r=0, which implies that the composition of the natural maps k -> A -> A/m is an isomorphism, and thus that m is of the desired form. –  Pierre-Yves Gaillard Mar 7 '11 at 18:07

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.