Let $R=\mathbb{Q}[x_1,x_2,\ldots,x_n]$ be a ring of multivariate polynomials over $\mathbb{Q}$. Let $V=\mathop{\rm span}\{x_1,x_2,\ldots,x_n\}$ be the linear subspace spanned by the indeterminates. Let ${\frak p}\lhd R$ be a prime ideal of $R$ which trivially intersects $V$.

Question: Do we always have a maximal ideal ${\frak m}\lhd R$ such that ${\frak p\subseteq m}$ and ${\frak m}\cap V=\{0\}$?


I think the answer is yes when $n=2$.

Let $R=\mathbb{Q}[x,y]$; since $R$ is a two-dimensional the only non-trivial case is for height-1 primes, and since $R$ is a UFD we can suppose that the prime ideal $P$ is generated by an irreducible polynomial $f(x,y)$. We distingush two cases.

Case 1: $f$ is linear. Since $P\cap V=(0)$, there are $\alpha,\beta,\gamma\in\mathbb{Q}$, with $\gamma\neq 0$ (since $P\cap V=(0)$), such that $f(x,y)=\alpha x+\beta y+\gamma$. By applying a linear substitution (which does not change $V$), we can suppose that $f(x,y)=y+\gamma$. Let $g\in\mathbb{Q}[z]$ be an irreducible polynomial of degree $>1$, and define $M:=(g(x),y+\gamma)R$.

Case 2: $f$ is not linear. Without loss of generality, suppose that the $x$-degree of $f$ is $>1$. By Hilbert's irreducibility theorem, there are infinitely many $y_0\in\mathbb{Q}$ such that $g(x):=f(x,y_0)$ is irreducible; in particular, we can choose $y_0$ different from $0$ and such that the degree of $g$ is $>1$. Define $M:=(g(x),y-y_0)R$, and note that $P\subseteq M$ (write every $y^k$ of $f(x,y)$ as $(y_0+(y-y_0))^k$: the $y_0^k$ goes to build $g$, while all the rest is a multiple of $y-y_0$).

In both cases, we have $M:=(g(x),y-y_0)R$ for some $y_0\neq 0$; I claim that $M\cap V=(0)$. Clearly $x\notin M$ (otherwise $g(0)\in M$, against the fact that $g$ is irreducible and not linear). Suppose $M$ contains $\alpha x+y$: then, $\alpha x+y-(y-y_0)=\alpha x+y_0\in M$. Hence, $M$ should contain both $g(x)$ and $\alpha x+y_0$, which is impossible since (again) $g$ is irreducible of degree $>1$.

Therefore, $M$ is your maximal ideal.

For $n>2$, I think a similar proof should work, with $M$ being in the form $(f_1(x_1),f_2(x_2),\ldots,f_{n-1}(x_{n-1}),f_n(x_n))$, with each $f_i$ irreducible and at most one of them linear; however, you probably need some better control on degrees of the generators of $P$ (for example, a generator of $P$ may be linear in two variables but in the other ones).


This is a consequence of Zorn's Lemma. Let $P$ be the set of all proper ideals of $R$ that contain $\mathfrak p$ and intersect $V$ trivially. This set $P$ is partially ordered by set inclusion. If we can show that every totally ordered subset has an upper bound in $P$, then $P$ has at least one maximal element $\mathfrak m$. This ideal $\mathfrak m$ is then a (proper) maximal ideal that contains $\mathfrak p$ and satisfies $\mathfrak m\cap V=\{0\}$.

At this point, the remainder of the proof proceeds along standard arguments about unions of nested ideals. If you haven't seen this argument before, you should check out the proof here and try making the necessary alterations yourself.

If you are still interested in the details, I'll write them out below:

Let $T\subseteq P$ be a totally ordered subset. Either $T=\emptyset$ or not. If $T=\emptyset$, then $\mathfrak p\in P$ is an upper bound for $T$. Otherwise, define

$$ \mathfrak m:=\bigcup\limits_{\mathfrak q\in T}\mathfrak q.$$

By construction, $\mathfrak m\supseteq\mathfrak q\supseteq\mathfrak p$ for each $\mathfrak q\in T$, and it is a routine check to see that $\mathfrak m$ is an ideal of $R$. Suppose for the sake of contradiction that $\mathfrak m=R$. Then we would have

$$1\in\mathfrak m = \bigcup\limits_{\mathfrak q\in T}\mathfrak q,$$

which implies there is at least one $\mathfrak q_0\in T\subseteq P$ with $1\in\mathfrak q_0$... but then this implies that $\mathfrak q_0=R$, which contradicts the fact that $P$ only contains proper ideals. Thus we conclude that $\mathfrak m$ is indeed proper.

Finally to verify that $\mathfrak m$ is actually an element of $P$, we need to check that $\mathfrak m\cap V=\{0\}$:

\begin{align*} \mathfrak m\cap V &= \left(\bigcup\limits_{\mathfrak q\in T}\mathfrak q\right)\cap V\\ &= \bigcup\limits_{\mathfrak q\in T}\left(\mathfrak q\cap V\right)\\ &= \bigcup\limits_{\mathfrak q\in T}\{0\}\;=\;\{0\}. \end{align*}

Thus every totally ordered subset $T\subseteq P$ has an upper bound in $P$, so Zorn's Lemma is satisfied.$\hspace{1cm}\blacksquare$

  • 1
    $\begingroup$ How do you know you ideal $\frak m$ is maximal in $R$? $\endgroup$ May 30 '18 at 19:52
  • 1
    $\begingroup$ @Sean Sanford: Observe that the statement is false if $\mathbb{Q}$ is replaced by $\mathbb{C}$: the prime ideal $(0)\subseteq \mathbb{C}[X_1,\dots,X_n]$ has trivial intersection with $V$ but every maximal ideal has non-trivial intersection with $V$ by the Nullstellensatz. On the other hand, your argument "seems" to work in general, and as Adam Przezdziecki points out, I don't see why $\mathfrak{m}$ should be maximal. $\endgroup$ May 30 '18 at 20:04
  • $\begingroup$ @FilippoAlbertoEdoardo Actually when $n=1$ almost every maximal ideal intersects $V$ trivially. But your statement is true when $n\geq 2$. $\endgroup$ May 31 '18 at 1:28
  • $\begingroup$ @AdamPrzeździecki You are right, and I don't see a way of proving it. $\endgroup$ May 31 '18 at 1:34
  • $\begingroup$ @FilippoAlbertoEdoardo You make a good point. Since the validity of the claim depends on the base ring, and my argument doesn't reference this at all, it cannot possibly be salvaged without explaining why this works for $\mathbb Q$ and not $\mathbb C$. $\endgroup$ May 31 '18 at 1:37

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