6
$\begingroup$

In what follows, ring is defined to be a commutative ring with unit ($1$).

Definition: Perhaps over-generalizing from (12) in this Math.SE answer, call any ring $R$ a Prüfer ring if, for all non-zero ideals $I,J,K$ of $R$, one always has $I \cap J + I \cap K = I \cap (J+K)$. (I.e. the modular law holds with equality without extra hypotheses.) $R$ doesn't have to be a domain.

Question: Given a ring $R$, under what conditions does $R[X,Y]$ fail to be a Prüfer ring? Always? Or when $R$ is an integral domain?

Also, as a less important side question, if $S$ is not a Prüfer ring, then is it also the case that $S[Z]$ is not a Prüfer ring? E.g. if $R[X,Y]=:S$ is not a Prüfer ring, then is it the case that also $R[X,Y,Z]\cong (R[X,Y])[Z]$ is not a Prüfer ring? So by induction all multivariate polynomial rings $R[X_1, \dots, X_n]$ with coefficients in $R$ are not Prüfer rings? (Related question)

Attempt: I think I have a proof that works at least whenever $R$ is an integral domain, although maybe it works more generally. Take $I= \langle X + Y \rangle$, $J= \langle X \rangle$, $K = \langle Y \rangle$. Then the left hand side is: $$\langle X + Y \rangle \cap \langle X \rangle + \langle X + Y \rangle \cap \langle Y \rangle = \langle (X + Y)X \rangle + \langle (X + Y)Y \rangle \,.$$ (I think, I'm not even sure about this step.) Then the right hand side would be: $$\langle X + Y \rangle \cap (\langle X \rangle + \langle Y \rangle) = \langle X + Y \rangle \cap \langle X , Y \rangle = \langle X + Y \rangle \,,$$ since clearly $\langle X + Y \rangle \subseteq \langle X, Y \rangle$. And then I think, but I'm not sure, regardless of the coefficients $R$, or at least when $R$ is an integral domain, one has that $$X+Y \not\in \langle (X + Y)X \rangle + \langle (X + Y)Y \rangle \quad \text{even though obviously} \quad X+Y \in \langle X + Y \rangle \,. $$

Thus $\langle (X + Y)X \rangle + \langle (X + Y)Y \rangle \subsetneq \langle X + Y \rangle$ and $R[X,Y]$ is not a Prüfer ring?

Background: A counterexample for the modular law failing given here is the three ideals given above in $\mathbb{Z}[X,Y]$. However, I couldn't figure out how that counterexample depended on the coefficient ring being $\mathbb{Z}$ instead of $\mathbb{R}$ or $\mathbb{C}$ or anything else. I had thought that the coefficient ring being $\mathbb{Z}$ mattered somehow because $\mathbb{Z}[X]$ (univariate polynomials) is given as a "well-known" ring that fails to be a Prüfer ring on Wikipedia. However, this other answer on Math.SE gives not only $\mathbb{Z}[X]$ as a non-example of a Prüfer ring, but also $\mathbb{Q}[X,Y]$. So I feel like I don't understand the "essence" or the "big idea" behind the counterexamples at all. (Probably related question)

$\endgroup$

1 Answer 1

7
$\begingroup$

The rings you describe are known as Arithmetical Rings in the commutative algebra literature.

Definition A ring $R$ is called Arithmetical if for all $I, J, K$, it holds $I \cap (J + K) = I \cap J + I \cap K$, i.e. their ideals form a distributive lattice.

Arithmetical rings are also characterized by the property that they are locally chain rings, i.e. the ideals of $R_\mathfrak{p}$ are totally ordered for any prime $\mathfrak{p}$ (this is due to Jensen, the proof is very short and you can see it here).

The term Prüfer Ring is reserved for a class of rings that in general extend the class of Arithmetical rings (though agree in the case of domains). The usual definition of Prüfer ring extending to rings with zero divisors is due to M. Griffin, and restricts attention to regular ideals (i.e. ideals containing a non-zero divisor). I'd suggest you take a look at his seminal paper, Prüfer Rings with Zero Divisors. In particular, in Theorem 13, you will find that one possible definition is

Definition A ring $R$ is called Prüfer if for any ideals $I,J,K$, at least one of which is regular, it holds $I \cap (J + K) = I \cap J + I \cap K$

With terminology out of the way, let's look at your proof that $R[x,y]$ is never arithmetical. It's a nice proof! And it works for any ring $R$, not just domains. Let's fill in your two gaps.

Verifications In the ring $R[x,y]$ the following are true
(1) $(x+y)\cap(x) = (x+y)x$
(2) $x + y \notin (x+y)(x,y)$

Proof

(1) Let $f \in (x+y) \cap (x)$. Write $f = xg = (x+y)h$. We will show that $x + y$ divides $g$, therefore $x(x+y)$ divides $f$ as desired. Consider the equation $xg = (x+y)h$ modulo the ideal $(y)$. We get the equivalence $xg \equiv xh$ mod $y$, and since $x$ is a regular element of $R[x] = R[x,y]/(y)$, this implies $g \equiv h$ mod $y$. Thus we can write $g + yq = h$ for some $q \in R[x,y]$. Substituting this into $xg = (x+y)h = (x+y)(g + yq)$. Cancelling the $xg$ from both sides you get $yg + (x+y)yq = 0$, and since $y$ is a regular element of $R[x,y]$, cancelling the $y$s gives $g + (x+y)q = 0$, and we're done.

(2) Observe that $x+y$ is a regular element of $R[x,y]$, indeed if $(x+y)f = 0$ then order the monomials of $f$ lexicographically with $x < y$ and note that if $f_{ij}$ is the coefficient of the smallest monomial of $f$, then $f_{ij}$ is also the coefficient of the smallest monomial of $(x+y)f$. Since $x+y$ is regular in $R[x,y]$, we have that $x+y \in (x+y)(x,y)$ iff $1 \in (x,y)$. $\square$

This settles your question about when multi-variate polynomial rings are arithmetical: Never! However, from the end of your post I think it worthwhile to look deeper into the univariate case, so that we can get some insight into why e.g. $\mathbb{Z}[x]$ can't be arithmetical but $K[x]$ is arithmetical for any field $K$.

Main Point If $R[x]$ is Prüfer (in the sense of Griffin), then $R$ is Von Neumann Regular.

I'm going to cheat a little bit (in the context of your question) and take for granted the characterization that finitely generated regular ideals are invertible, and I'll leave it to you to produce a proof directly from the arithmetical property if you so desire. To see the connection between the arithmetical property and invertible regular ideals, use Jensen's argument in combination with the fact that regular ideals are invertible iff they are locally principal.

Proof of Main Point: What we'll do is fix $a \in R$ and consider the ideal $(a,x) \subseteq R[x]$.

Since $(a,x)$ is regular, it is invertible, and we get $(a,x)J = R[x]$ where $J$ is an $R[x]$-submodule of $T(R[x])$ and $(a,x)J \subseteq R[x]$.
Therefore we have elements $h_1, h_2$ of $T(R[x])$ such that $ah_i \in R[x], xh_i \in R[x]$, and $ah_1 + xh_2 = 1$.

Let's say that $ah_1 = f_a, xh_1 = f_x$, and $ah_2 = g_a, xh_2 = g_x$. From the first two equations, we see that $a$ divides the lowest coefficient of $f_a$, since $af_x = xf_a$ and hence $af_{x1} = f_{a0}$. From the third and fourth, we see that $ag_x = xg_a$, so that $a$ annihilates the lowest coefficient of $g_x$.

Now after substitution, our relation $ah_1 + xh_2 = 1$ turns into $f_a + g_x = 1$. Multiplying through by $a$ and examining the lowest coefficient, we then have $a^2f_{x1} = a$. Since $a$ was arbitrary, we've thus shown that $R$ is Von Neumann Regular. $\square$

(I should mention that there's also a converse to this: Univariate polynomial rings over Von Neumann Regular rings are $1$-dimensional semi-hereditary Bezout domains. See this paper of Gilmer for the semi-hereditary part, and look for the paper of Gilmer and Shores, Semigroup Rings as Prüfer Rings, for the Bezout part).

Two immediate corollaries to this are:

If $R$ is not arithmetical, then neither is $R[x]$.

Proof It's enough to check that Von Neumann Regular rings are arithmetical. Referring to Jensen's characterization above this is obvious because VNR's are locally fields.

And though you've already settled this point, we now have another way of seeing that

A polynomial ring in more than one indeterminate is never Prüfer, and a fortiori never arithmetical.

Proof An easy argument by consideration of Krull dimension. VNR's are zero-dimensional, and adding indeterminates strictly increases dimension.

So the moral here is that it's a really big deal for the ideals of $R[x]$ to form a distributive lattice, such a big deal that $R$ has to be locally a field. In particular if $\mathbb{Z}[x]$ were to be arithmetical, then $\mathbb{Z}$ would have to be a field; and if $\mathbb{Q}[x,y]$ were to be arithmetical, then $\mathbb{Q}[x]$ would have to be a field!

$\endgroup$
9
  • 1
    $\begingroup$ They're also called, at least in the noncommutative case, (right/left) distributive rings. Although I do recognize the term "arithmetical" as very common in the commutative case. $\endgroup$
    – rschwieb
    Jun 4, 2019 at 20:08
  • 1
    $\begingroup$ @rschwieb Do you know of any results about polynomials in commuting variables over non-commutative rings analogous to the ones I mentioned in this post? I'm familiar with the fact that if $R[x]$ is left semi-hereditary then $R$ is Von Neumann Regular, and that (I think?) the converse fails, but I'm not sure what happens when we replace "semi-hereditary" by "distributive" $\endgroup$ Jun 4, 2019 at 22:39
  • 1
    $\begingroup$ You specifies noncommutative rings, I know, but have you seen this: ams.org/journals/proc/1974-045-02/S0002-9939-1974-0352165-2/… $\endgroup$
    – rschwieb
    Jun 4, 2019 at 22:59
  • 1
    $\begingroup$ @rschwieb Yes, I'm familiar. See my "Main Point" above and the parenthetical below it's proof, all of which are stronger claims than Camillo makes. $\endgroup$ Jun 4, 2019 at 23:07
  • 1
    $\begingroup$ Not to say that there's anything novel in this post, just that it was pretty extensive in addressing the commutative case and I am wondering specifically about the noncommutative case. All I'm aware of is this note of Pillay which addresses the implication $R[x]$ left-semihereditary then $R$ is VNR ams.org/journals/proc/1980-078-04/S0002-9939-1980-0556614-9/…, and apparently the converse holds when $R[x]$ is assumed to be coherent (according to Faith) but not in general (I haven't seen the counterexample) $\endgroup$ Jun 4, 2019 at 23:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .