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I am teaching an extremely basic course in algebraic geometry and I would like to find more exercises on Hilbert's basis theorem, a result from commutative algebra. Unfortunately, I have not been able so far to find interesting exercises in all the books and lecture notes that I've seen.

Could you advise me some exercises on this topic?

agleaner
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    Show that the subring of K[x] which contains all polynomials which are missing x is noetherian. (This is k[x^2,x^3] which is isomorphic to the coordinate ring k[x,y]/(x^2-y^3).) – user26857 Feb 11 '20 at 07:01
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    Similarly, show that the subring of k[x] containing all polynomials f with the property f(0)=f(1) is noetherian. – user26857 Feb 11 '20 at 07:09
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    Btw, not all subrings of k[x] are noetherian, but those containing k are noetherian. – user26857 Feb 11 '20 at 07:10
  • I like these exercises! Thank you. Do you think there is a place that contains more of such exercises? I would solve the first exercise by saying that this ring is finitely generated over $k[x]$. Is this the simplest solution? – agleaner Feb 11 '20 at 09:34
  • You are both right. I guess I wanted to say that this is a quotient of $k[x_1,x_2]$ by an ideal... How would one quickly deduce this statement from Hilbert basis theorem? – agleaner Feb 11 '20 at 18:23
  • You may find some of the exercises in http://www.maths.usyd.edu.au/u/oyacobi/commalg/Exercises3-2016.pdf to be useful. – Gerry Myerson Feb 12 '20 at 22:27

5 Answers5

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A) Obligatory exercise, to be committed to memory.
Any finitely generated algebra over a noetherian ring is noetherian.

B) If $A$ is a PID and $f\in A$ , then the fraction ring $A_f=S^{-1}A$ with $S=\{1,f,f^2,f^3,\cdots\}$ is noetherian.
Amusing example: the ring of all decimal numbers, i.e. those that can be written with finitely many digits after the decimal point (like $-3.1415926535$), is noetherian.

C) The image of $f:\mathbb R\to \mathbb R^n: t\mapsto (t,t^2,\cdots, t^n)$ is an algebraic variety.

D) If $k$ is a field, then any $k$-subalgebra of the $k$-algebra $k[X]$ is noetherian.

E) Is a noetherian algebra over a field finitely generated over that field?

F) a) Is the $\mathbb Q$-algebra $\mathbb Q[1+\sqrt 2,1+2\sqrt 2,\cdots , 1+n\sqrt 2,\cdots ]$ finitely generated over $\mathbb Q$ ?

$\; \; \:$ b) Is the polynomial ring in infinitely many variables $\mathbb Q[X_1,X_2,\cdots , X_n,\cdots ]$ finitely generated over $\mathbb Q$?
$\; \; \:$ c) Is the ring $\mathbb Q(X)$ noetherian? Is it a finitely generated algebra over $\mathbb Q$ ?

NB These are not necessarily corollaries of Hilbert's theorem but are definitely very related to it.

Solution of D)
Let $A\subsetneq k[X]$ be a $k$-subalgebra and let $f(X)=X^n+q_1X^{n-1}+\cdots+q_n\in A \quad (n\geq 1)$.
Then the monic polynomial $T^n+q_1T^{n-1}+\cdots+q_n-f(X)\in (k[f(X)])[T]$ kills $X$, so that $k[X]$ is integral over $k[f(X)]$ and thus finite over $k[f(X)]$ ( Atiyah-Macdonald,Remark page 60).
And now for the killing: $k[f(X)]$ is a noetherian ring by A) and $k[X]$ is a noetherian module over $k[f(X)]$ by finiteness (Atiyah-Macdonald, Proposition 6.5).
Then the $k[f(X)]$-submodule $A$ of $k[X]$ is finitely generated over $k[f(X)]$ as a module and a fortiori as an algebra over the noetherian ring $k[f(X)]$.
Reapplying A), we see that $A$ is a noetherian ring. Et voilà!

Edit: Warning!
The assertion D) becomes false if the field $k$ is replaced by a ring, even a nice noetherian one like $\mathbb Z$:
For example the $\mathbb Z$-subalgebra (which just means the subring!) $A\subset \mathbb Z[X]$ defined by $$A=\mathbb Z\oplus 2\mathbb Z \cdot X\oplus 2\mathbb Z \cdot X^2\oplus 2\mathbb Z \cdot X^3\oplus\cdots =\mathbb Z\oplus (\bigoplus_{i\geq 1} 2\mathbb Z \cdot X^i)$$ is not noetherian since its ideal $$ \langle 2X,2X^2,2X^3,\cdots\rangle \subset A$$ is not finitely generated .

Georges Elencwajg
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  • C is obvious: just write down the defining ideal. You probably want to generalise to arbitrary exponents. – darij grinberg Feb 11 '20 at 21:28
  • Another one, with a more olympiad puzzle character: https://math.stackexchange.com/questions/2189658/a-subring-of-mathbb-q-which-is-not-noetherian – darij grinberg Feb 11 '20 at 21:33
  • @darij: What's a simple proof that the for arbitrary exponents the image is algebraic, i.e. coincides with the zero locus of some ideal? The result is true for arbitrary polynomials too (not necessary monomials) but I don't know an elementary proof. – Georges Elencwajg Feb 11 '20 at 22:02
  • Sorry, I was wrong about arbitrary exponents: The image of $\left(t^2,t^2\right)$ is not an algebraic variety... I assumed the field is algebraically closed. – darij grinberg Feb 11 '20 at 22:04
  • Claim: Let $K$ be a field that has $n$-th roots of all elements for all $n \geq 1$. (This is weaker than requiring $K$ to be algebraically closed.) Let $a_1, a_2, \ldots, a_m$ be nonnegative integers. Then, the image of the map $f : K \to K^m, \ t \mapsto \left(t^{a_1}, t^{a_2}, \ldots, t^{a_m}\right)$ is an algebraic variety. Proof sketch. WLOG assume that all $a_i$ are positive (since otherwise, we can throw away the constantly-$1$ coordinates). For any $i$ and $j$, set $g_{i,j} = \gcd\left(a_i, a_j\right)$. In the polynomial ... – darij grinberg Feb 11 '20 at 22:23
  • ... ring $K\left[x_1, x_2, \ldots, x_m\right]$, let $I$ be the ideal generated by all differences of the form $x_i^{a_j/g_{i,j}} - x_j^{a_i/g_{i,j}}$. Then, it suffices to show that the image of $f$ is the vanishing set of $I$. So let us show this. This means showing that if $\left(z_1, z_2, \ldots, z_m\right) \in K^m$ is a point in the vanishing set of $I$, then there exists some $t \in K$ such that all $i$ satisfy $z_i = t^{a_i}$ (because the converse of this statement holds for obvious reasons). So let us show this. ... – darij grinberg Feb 11 '20 at 22:25
  • ... Let $\left(z_1, z_2, \ldots, z_m\right) \in K^m$ be a point in the vanishing set of $I$. We must find a $t \in K$ such that all $i$ satisfy $z_i = t^{a_i}$. If some $z_i$ is $0$, then so are all other $z_i$ (here we are using the positivity of the $a_i$), and thus we can just take $t = 0$. Hence, we can WLOG assume that all $z_i$ are nonzero. Hence, they belong to the abelian group $K^\times$. Now, we are left with a purely group-theoretical problem, namely showing that if $z_1, z_2, \ldots, z_m$ are $m$ elements of an abelian group $G$ (written multiplicatively) ... – darij grinberg Feb 11 '20 at 22:26
  • ... satisfying $z_i^{a_j / g_{i,j}} = z_j^{a_i / g_{i,j}}$ for all $i$ and $j$ (with $a_i$ and $g_{i,j}$ as before), then there exists some $t \in G$ such that all $i$ satisfy $z_i = t^{a_i}$. How do we show this? We WLOG assume that $G$ is finitely generated (since we can replace $G$ by the subgroup generated by $z_1,z_2,\ldots,z_m$). Thus, by the Structure Theorem for Finitely Generated Abelian Groups, $G$ is isomorphic to a direct sum ... – darij grinberg Feb 11 '20 at 22:28
  • ... of $\mathbb{Z}$'s and $\mathbb{Z}/p^i\mathbb{Z}$ (with $p$ prime and $i$ a positive integer). It thus suffices to solve our problem for $G = \mathbb{Z}$ and for $G = \mathbb{Z}/p^i\mathbb{Z}$. This is straightforward. – darij grinberg Feb 11 '20 at 22:28
  • Wait, oops, this is not actually straightfroward. Sorry. It's probably wrong as stated. – darij grinberg Feb 11 '20 at 22:33
  • Dear Georges, many thanks for the answers. How would you prove D? – agleaner Feb 11 '20 at 23:28
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    Dear aglearner: I have added a solution of D in my answer. – Georges Elencwajg Feb 12 '20 at 10:45
  • D is already proved on this site: https://math.stackexchange.com/questions/349328/any-subring-of-kx-that-contains-k-is-noetherian-not-all-of-such-subrings – user26857 Feb 12 '20 at 12:41
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Here's an exercise from Ravi Vakil. The game of Chomp starts with a half-infinite checkboard with squares labeled $(n, m)$ for integers $n, m\geq 0$, with a cookie on each square. Two players take turns choosing a cookie (i.e., not an empty square) and eating all the cookies on that square and the ones that are not to the left or below it; that is, if a player chooses the cookie on $(n, m)$, then they eat all the cookies on $(n', m')$ for $n' \geq n$ and $m'\geq m$. The cookie on $(0, 0)$ is poisoned, and the player who eats it immediately loses. Prove that any game of Chomp ends after finitely many moves.

For the connection with the Hilbert basis theorem, consider $(n, m)$ as representing the monomial $X^n Y^m\in R[X, Y]$. More generally, consider the analogous game on $(\mathbb{Z}^{\geq 0})^d$.

anomaly
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  • That's cool! How have you learned about this exercise? Are there more like this? – agleaner Feb 11 '20 at 23:23
  • @agleaner: I took an introductory algebraic geometry class from him. (He's a great teacher, in addition to obviously being a great mathematician.) – anomaly Feb 12 '20 at 00:02
  • Anomaly, that's cool. Do you think there are some notes of the class that you took, or is this the following one: http://virtualmath1.stanford.edu/~vakil/17-145/ ? – agleaner Feb 12 '20 at 09:32
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    @agleaner: That's not the particular class I took from him (it was at a different university and a first- or second-year grad class, so in particular assuming a bit more commutative algebra than that one), but there are plenty of notes of his online. Try this, for example: https://math.stanford.edu/~vakil/725/funprobs.pdf – anomaly Feb 12 '20 at 14:41
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Not an application but a variation on the proof technique: show that if $A$ is a noetherian commutative ring, then so is the ring of formal power series $A[[t]]$.

If you are teaching algebraic geometry, then one should mention that every zero set (in $n$-dimensional affine space over a field) is the union of finitely many irreducible zero sets, i.e. zero sets coming from prime ideals. This is essentially the geometric translation of the statement that in a noetherian ring, every radical ideal is the intersection of finitely many prime ideals.

Bib-lost
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The following results, although easy, are used in "absolute Noetherian reduction" and can be proved by the Hilbert basis theorem:

  • Every ring is a filtered colimit of Noetherian rings.

  • Every $k$-algebra is a filtered colimit of Noetherian $k$-algebras for a field $k$.

If you have not introduced the language "filtered colimit", you can just say "union of subrings" instead.

You can have them adapt the proof of the Hilbert basis theorem to show the "formal Hilbert basis theorem" that if $A$ is Noetherian, then so is $A[[x]]$.

Lukas Heger
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  • Thanks Lukas. This doesn't quite look like an exercise that I can give to students in the course in algebraic geometry... – agleaner Feb 09 '20 at 10:33
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Exercise: Show that a system of infinitely many polynomial equations can be reduced to a system of finitely many polynomial equations.

Indeed, let $S$ be a set of polynomials. Then $(S)$ is an ideal and hence finitely generated, say with generators $P_1, \dots, P_n$. We have

$$V(S) = V((S)) = V(P_1, \dots, P_n)$$

and we have reduced any system of polynomials to a finite set of polynomials.

J. De Ro
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    A system of infinitely many polynomial equations defined over a noetherian ring is equivalent to a system of finitely many polynomial equations. – Bib-lost Feb 11 '20 at 14:11
  • Yes, this was implicit in my answer when I used that an ideal was finitely generated. – J. De Ro Feb 11 '20 at 14:15