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This problem is from a qualifying exam.

I am to show that if $k$ is a field and if $n > 1$, then no maximal ideal of $k[x_1, \dots, x_n]$ is principal.

I understand why this is true if $k$ is algebraically closed, since every maximal ideal is of the form $(x_1 - a_1, \dots, x_n - a_n)$ for some $a_i \in k$.

My attempt for $k$ an arbitrary field. Let $m \subset k[x_1, \dots, x_n]$ be a maximal ideal and let $K$ be the algebraic closure of $k$. Let $M \subset K[x_1, \dots, x_n]$ be the push-forward ideal of $m$, namely, $M = K[x_1, \dots, x_n]m$. Assume $M$ is not the unit ideal, and so is contained in the maximal ideal $(x_1-a_1, \dots, x_n - a_n)$ of $K[x_1, \dots, x_n]$ for some $a_i \in K$.

Suppose $m$ is principal and generated by some irreducible $f \in k[x_1, \dots, x_n]$. Observe that the subset $A$ of $(x_1 - a_1, \dots, x_n - a_n)$ consisting of all polynomials whose coefficients are in $k$ is an ideal of $k[x_1, \dots, x_n]$ containing $m$, and therefore $A = m$ by maximality. For each $i$, let $f_i \in k[x_1, \dots, x_n]$ be a polynomial of minimal degree such that $x_i - a_i$ divides $f_i$ in $K[x_1, \dots, x_n]$ -- such a polynomial exists since each $a_i$ is algebraic over $k$. In particular, $f_i \in A = m$. Since $f$ divides $f_i$ in $k[x_1, \dots, x_n]$, it follows by the minimality of $f_i$ that $f_i = f$ for all $i$. We gather $x_i - a_i$ divides $f$ for all $i$.

Note that if any of the $a_i$ are in $k$ then we are done because $f = f_i = x_i - a_i$, again by the minimality of $f_i$, which is a contradiction since $x_j - a_j \nmid x_i - a_i$ for $i \neq j$.

However, I cannot find the contradiction in the case all the $a_i$ are in $K - k$. Does anyone see how to proceed along this particular route? If not, I welcome other suggestions. I understand there are arguments using the Krull dimension, but given that Krull dimension is not covered in our algebra course, I'm hesitant to use such an argument on an exam. I think it is intended we use Nullstellensatz, but I'm not entirely sure.

EDIT: As discussed in the comments, one also must account for the case $M$ is unital, if this strategy is to be kept.

Elías Guisado Villalgordo
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Doug
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    Why $M$ is not the unit ideal? thanks. – Matemáticos Chibchas Sep 05 '16 at 22:33
  • Because $m$ is not, and we're just multiplying $m$ by polynomials to get $M$ – Doug Sep 05 '16 at 22:34
  • This is not sufficient; when you talk about "polynomials" you are being ambiguous (are they with coefficients in $K$ or in $k$?). – Matemáticos Chibchas Sep 05 '16 at 22:59
  • Multiplying $m$ by polynomials with coefficients in $K$. This cannot decrease the degree, so $1$ remains outside of $M$. – Doug Sep 05 '16 at 23:02
  • Now I see... but note that you are already assuming that $m$ is principal at this point. In the general case (that is, $m$ not principal) your degree argument doesn't work. Only on the next paragraph you are assuming that $m$ is principal; you must correct that. – Matemáticos Chibchas Sep 05 '16 at 23:06
  • Are you saying I need $m$ principal to say $M$ is not unital? This is not true. Every element of $m$ is a polynomial, so when I multiply $m$ by $K[x_1, ..., x_n]$ the same degree argument shows $1$ is not in $M$. – Doug Sep 05 '16 at 23:13
  • Yes, I am saying that. What do you mean by "degree argument"? – Matemáticos Chibchas Sep 05 '16 at 23:34
  • $m \subset K[x_1, ..., x_n]$. The latter space has the property that $deg_i (fg) = deg_i(f) + deg_i(g)$ for all $i$ where $deg_i$ denotes the degree in the $i^{th}$ variable (see below). Therefore the $deg_i$ of an element of $m$ cannot decrease upon multiplying by any element in $K[x_1, ..., x_n]$. So $1$ is not in $M$ as at least one of $deg_i$ is nonzero for any element of $m$. – Doug Sep 05 '16 at 23:40
  • But you are calculating the $i$-th degree of a sum $\sum_j p_j g_j$, with $p_j\in K[x_1,\ldots,x_n$ and $g_j\in m$, and you keep talking about the degree of a single element of $m$. – Matemáticos Chibchas Sep 05 '16 at 23:48
  • I agree with you now. One needs to address the possibility the push forward is unital. – Doug Sep 05 '16 at 23:54
  • No. Just assume in advance that $m$ is principal, so your degree argument works. On the other hand, I strongly believe that the result remains true for arbitrary $m$, but this is another story. – Matemáticos Chibchas Sep 06 '16 at 00:42
  • Elementary proofs (that is, without Nullstellensatz) in this thread – user26857 Apr 05 '19 at 20:04

2 Answers2

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Krull’s principal ideal theorem. If $A$ is Noetherian and $f$ is a non-unit, then any prime ideal $P$ that is minimal with respect to containing $(f)$ is necessarily of height at most one.

In particular, if $(f)$ is prime, it's of height at most one. Now make a Krull dimension argument to show that $(f)$ can't be maximal. Also, don't forget to justify why you're in a Noetherian ring.

user26857
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syusim
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  • I am looking for an answer that doesn't use the Krull dimension, possibly using Nullstellensatz as I have attempted to do. I edited my question to state this explicitly. – Doug Aug 26 '16 at 02:40
  • @user26857 The justification of why the ring is Noetherian is to verify that we can actually apply the theorem. I just mentioned that because it seemed like the asker wanted to be as thorough as possible.

    Also, a possible argument for the Krull dimension part is to say that since we're in an f.g. $k$-algebra, $\dim k[x_1, \dots, x_n]/(f) = \dim k[x_1, \dots, x_n] - height((f))$. In particular it's $n-1$. Now use the correspondence theorem to get that this is the size of a chain of primes in $k[x_1, \dots, x_n]$ containing $(f)$. Since $n \geq 2$, $(f)$ can't be maximal.

     – syusim Aug 27 '16 at 05:04
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You're almost done: what you showed is that if $f_i(x)$ is the minimal polynomial of $a_i$ over $k$, then $f_i(x_i) \in \mathfrak{m}$ is divisible by $f(x_1, ..., x_n)$ in $k[x_1, ..., x_n]$, hence they are equal. Since $n \geq 2$ this contradicts that $f$ is irreducible. (In other words, it's not possible for all $f_i$ to be equal to $f$, because the $f_i$ involve different variables from each other.)

user26857
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anon
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  • What I can say for sure is that $f = (x_1 - a_1)...(x_n - a_n)g$ for some polynomial $g$ with coefficients in $K$. I don't see that this immediately contradicts the irreducibility of $f$ over $k$. I don't think we can say, a priori, that the $f_i$ involve different variables from each other. – Doug Aug 26 '16 at 04:29
  • In characteristic zero you can say it like this: if $x_i - a_i$ divides $f$, a polynomial in $k[x_1, ..., x_n]$, as polynomials in $K[x_1, ..., x_n]$, then also all the galois conjugates $x_i- \sigma(a_i)$ of $x_i-a_i$ divide $f$, so the minimum polynomial of $a_i$ divides $f$. In other words, if $\mathbf{M} = (x_1-a_1, ..., x_n-a_n)$ contains $\mathbf{m} K[x_1, ..., x_n]$, then so do all the galois conjugates of $\mathbf{M}$. (If you talk about minimal polynomials directly then what you say makes sense in positive characteristic too.) – anon Aug 26 '16 at 04:56
  • Ah, I think I understand what you are saying. If $m_i(X) \in k[X]$ is the minimal polynomial of $a_i$ over $k$, then $x_i-a_i$ divides $m_i(x_i)$ in $K[x_1, \dots, x_n]$, hence $m_i(x_i) \in m$ and $f$ divides $m_i(x_i)$ in $k[x_1, \dots, x_n]$. We want to conclude $f$ and $m_i(x_i)$ are associates for all $i$, a contradiction. Possibly a silly question: By assumption $m_i(x_i)$ is irreducible in $k[x_i]$; do we know for sure it remains irreducible in $k[x_1, \dots, x_n]$? – Doug Aug 26 '16 at 05:37
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    Yes, that's it. For the irreducibility, $k[x_1, ..., x_n]$ is graded by degree in each of the variables $x_1,..., x_n$; that is, the degree of a product $fg$ in a given variable $x_j$ is the sum of the degrees of $f$ and $g$ in $x_j$. This implies that a polynomial in which only a proper subset of variables appears does not factor into a product where the other variables appear. – anon Aug 26 '16 at 16:07