Atiyah-Macdonald Exercises 5.16-5.19

Question

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).]

Answer 1

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.

Answer 2

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 $\varphi\circ v=z$.