1
$\begingroup$

It seems that Noetherian normalization is easier to prove for infinite fields than for finite fields (see e.g. here, here, or here).

However, I do not see how this restriction changes anything, since usually one avoids finite fields because they all have positive characteristic -- but then the sensible restriction is to fields of characteristic zero, not to arbitrary infinite fields, some of which also have positive characteristic.

Question: If the point isn't to exclude considering fields with positive characteristic, then why does the restriction to infinite fields when proving Noetherian normalization matter, or why does it simplify anything?

Context: I do not know much commutative algebra, so the simpler the answer, the better. In other words, no schemes please, and if you can limit discussion as much as possible to ideals of polynomial rings without drifting too far afield into varieties, I would appreciate it.

This question is based on exercise 4.5.7. p.210 in Garrity et al's Algebraic Geometry: A Problem-Solving Approach, which adds as a hypothesis that the field $k$ should be infinite, a restriction that does not occur anywhere else in the section (about proving the Weak Nullstellensatz, which to the best of my knowledge also holds for finite fields). And I was able to solve the preceding problems for fields even of characteristic two, so I don't understand how the restriction makes the problem easier to solve, since neither characteristic nor cardinality seem to pose issues.

I don't want a solution for the problem, I just want to know whether the restriction to infinite fields is either (1) necessary, or (2) convenient, and in what way it is applied to make the proof easier.

4.5.7 Let $k$ be an infinite field and $g$ be a non-constant polynomial in $k[x_1, \dots, x_n]$ (with $n \ge 2$). Prove that there exist $\lambda_1, \dots, \lambda_{n-1}$ in $k$ such that the coefficient of $x_n^d$ in $$g(x_1 + \lambda_1x_n, \dots, x_{n-1}+\lambda_{n-1}x_n, x_n)$$ is nonzero, where $d$ is the total degree of $g(x_1 + \lambda_1x_n, \dots, x_{n-1}+\lambda_{n-1}x_n, x_n)$.

If it helps, I conjecture that the aforementioned coefficient is just $$g(\lambda_1, \dots, \lambda_{n-1},1) $$ I have proved this in a special case, but still need to prove it more generally. Anyway, I don't see how the cardinality of $k$ has anything to do with the existence of solutions for $$g(\lambda_1, \dots, \lambda_{n-1},1)=1 \not=0 $$ (w.l.o.g. in a field we can normalize any non-zero value to 1 -- the point is to make the polynomial monic in $x_n$ via a change of coordinates) for arbitrary polynomials $g \in k[x_1, \dots, x_n]$ -- seemingly only characteristic would be an issue, if anything.

Edit/Note: The formula given above for the coefficient is correct only if $g$ is homogeneous. In general, if $g_{hom,d}$ denotes the homogeneous part of $g$ of degree $d$, then the coefficient is $$g_{hom,d}(\lambda_1, \dots, \lambda_{n-1},1).$$

I'm pretty sure the restriction $n\ge 2$ is just so that the corresponding affine change of coordinates isn't the identity or ill-defined by the above description. The restriction to infinite fields however is baffling to me, since it is given neither context nor motivation.

$\endgroup$
  • 3
    $\begingroup$ Finite fields have polynomials which are zero at every point but are non-zero in the polynomial ring. Thus the weak nullstellensatz ($V(I) \neq \emptyset$ if $I$ is a proper ideal) does not hold here. $\endgroup$ – Paul K Oct 9 '16 at 17:41
  • 4
    $\begingroup$ As pointed out in the comment above and the answer below, the assumption of infinite field is not necessary, but convenient. Given a non-zero $g$ as above, you would like to make it monic in one of the variables. Your argument says, over an infinite field, you can make a homogeneous linear change of variables to achieve this and the example in the answer below says, it may not be possible for finite fields. So, over finite fields, you have to make non-linear change of variables, a more difficult task. $\endgroup$ – Mohan Oct 9 '16 at 17:47
  • 2
    $\begingroup$ I think in general, whenever the infiniteness hypothesis appears, it is related the fact that over an infinite field, a polynomial is determined by its values; over a finite field, this is false. Polynomials over finite fields are not simply "polynomial functions". $\endgroup$ – user144221 Oct 9 '16 at 17:50
  • 1
    $\begingroup$ @Alejo I like this way of thinking about it too. What do you mean by "not polynomial functions"? i.e. there isn't a one-to-one correspondence between polynomials and functions (since there are only $n^n$ possible functions over any finite field but infinitely many possible non-zero polynomials)? $\endgroup$ – Chill2Macht Oct 9 '16 at 17:52
  • 2
    $\begingroup$ @William yes, exactly. It is a rather trivial yet important observation :-) In the infinite case, we have a 1-1 correspondence between polynomials and polynomial functions, so many algebra students confuse the two things, probably because we all deal with polynomials starting from a very young age at school, without really knowing what a polynomial is. $\endgroup$ – user144221 Oct 9 '16 at 18:10
3
$\begingroup$

My short answer: it's not.

You're right that something like this can be used to prove Noether's normalization lemma for infinite fields (see this SE post). I just want to note (as Mohan hinted to in a comment) that it doesn't require any extra technology or work to prove Noether normalization for any field $k$, i.e. that any finitely-generated $k$-algebra is finitely generated as a module over a subring isomorphic to a polynomial ring over $k$. All it requires is a slightly more clever change of variables, which is much more enlightening, anyway (in my opinion). The point of this problem you're working on is essentially to show that the nonzero polynomial $g$ can be made monic in one of the variables after possibly changing the variables first. So, while this particular way of doing it may be easier when $k$ is infinite, you can accomplish the same goal without that assumption. In other words, characteristic is actually not an issue, unless you make it one. Here is a way to do this without ever having to think about $\mbox{char}(k)$ and without any need for extra machinery:

Let $N \in \mathbb N$ be larger than the total degree of $g$. Consider the following change of variables: $x_m \mapsto x_m + x_n^{N^m}$ for $m < n$ and $x_n \mapsto x_n$. You will see that, when you apply the change of variables to $g$, it will be monic in $x_n$: if you look at each nonzero term of $g$ under the change of variables, it will expand to have a unique largest degree term which is a nonzero constant times $x_n$ raised to some power; moreover, these power-of-$x_n$ terms coming from expanding the different nonzero terms of $g$ will all be distinct, by the uniqueness of base-$N$ decomposition of integers; therefore, there is a largest one, and it is the largest degree term of $g$.

$\endgroup$
  • $\begingroup$ I will have to try writing out this change of variables later to see if I understand it -- the exponent of $x_m$ is $N^m$, right, i.e. N^m? I guess my question is why the power of the largest degree term isn't divided by the characteristic of the field? Or is that why we choose $N^m$ rather than just $N$? Sorry for the dumb question -- I'm not really good at non-zero characteristic $\endgroup$ – Chill2Macht Oct 18 '16 at 20:08
  • 2
    $\begingroup$ Consider a nonzero term $cx_1^{a_1}\cdots x_n^{a_n}$, i.e. $c\ne 0$. After changing variables it becomes $c(x_1+x_n^N)^{a_1}\cdots (x_{n-1}+x_n^{N^{m-1}})^{a_{n-1}}x_n$. It's clear that there is a unique largest-degree term in the expansion of this, namely $cx_n^{a_n+a_1N+\dots+a_{n-1}N^{n-1}}$. This sort of thing happens for each term in $g$, and all of these kinds of terms are distinct, because $N > a_i$ for all $i$ by definition, so this exponent is uniquely determined by $(a_1,\dots,a_n)$. It follows that there is a largest-degree one, and $g$ is now monic in $x_n$. $\endgroup$ – user379719 Oct 18 '16 at 20:15
  • 1
    $\begingroup$ Whether the characteristic of the field divides the exponent is irrelevant. $\endgroup$ – user379719 Oct 18 '16 at 20:16
  • $\begingroup$ Oh OK yes this makes sense actually -- sorry I thought for some reason $x_i^p = 0$ if the characteristic is $p>0$; it would be $p\cdot x_i=0$ and that doesn't matter since the leading term already doesn't vanish, so the leading term under this new change of coordinates won't vanish either (i.e. deliberately because of the construction), which is all we want. $\endgroup$ – Chill2Macht Oct 18 '16 at 20:23
  • 1
    $\begingroup$ Yep! I hope seeing a proof for general $k$ that isn't really any harder than the proof for infinite $k$ is a satisfying answer to your original question. $\endgroup$ – user379719 Oct 18 '16 at 20:27
2
$\begingroup$

Generally, in the proof of Noether normalization, there is a step in the proof where you have a nonzero polynomial $f(x_1, x_2, \ldots, x_n) \in k[x_1, x_2, \ldots, x_n]$ and you need to choose $a_i \in k$ such that $f(a_1, a_2, \ldots, a_n) \ne 0$. If $k$ is finite, this may not be possible, e.g., $x^p-x$ is a finite field of order $p$.

$\endgroup$
  • $\begingroup$ This is related to menag's comment about non-zero polynomials which are zero at every point, right? And that can only happen for finite fields (I guess because of a generalized version of the fundamental theorem of algebra to arbitrary algebraically closed fields)? I just want to make sure I understand before accepting. $\endgroup$ – Chill2Macht Oct 9 '16 at 17:43
  • 1
    $\begingroup$ Correct. Actually, you don't even need the field to be algebraically closed. In the one-variable case, we're using the theorem that says a nonzero polynomial has only finitely many roots (which holds in any field). In larger number of variables, choose any value for one of the variables, then proceed by induction. $\endgroup$ – Ted Oct 9 '16 at 17:47
  • 1
    $\begingroup$ For my future reference -- proof that any nonzero polynomial in one variable (over any integral domain, which includes all fields as special cases) has only finitely many roots: math.stackexchange.com/a/1137202/327486 $\endgroup$ – Chill2Macht Oct 9 '16 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.