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Background, Notation, Definitions: Given a set $X$, I define the set $M(X)$ of monomials with $X$-indeterminates to be the set of elements of $\omega^X$ having finite support. Given $m_0,m_1\in M(X)$, I define the operation $*$ on $M(X)$ by $$(m_0*m_1)(x):=m_0(x)+_\omega m_1(x).$$ $\langle M(X),*\rangle$ is then a commutative, cancellative monoid, with the zero element of $\omega^X$ as the identity.

Given a ring $R$, it is then natural to define the set $R[X]$ of polynomials with $R$-coefficients and $X$-indeterminates to be the set of elements of $R^{M(X)}$ having finite support. We define the addition and multiplication operations $\oplus$ and $\odot$ on $R[X]$ in terms of the addition and multiplication operations $+$ and $\cdot$ on $R$ as follows: $$(p_0\oplus p_1)(m):=p_0(m)+p_1(m)$$ $$(p_0\odot p_1)(m):=\underset{m_0*m_1=m}{\sum_{m_0,m_1\in M(X)}}p_0(m_0)\cdot p_1(m_1).$$ Then $\langle R[X],\oplus,\odot\rangle$ is a ring. It will be commutative when $R$ is, with unity when $R$ has one.

I define the function $\deg:M(X)\to\omega$ by $$\deg(m):=\sum_{x\in X}m(x),$$ and the function $\sigma:\bigl(R[X]\smallsetminus\{0_{R[X]}\}\bigr)\to\omega$ by $$\sigma(p):=\max\{\deg(m):m\in M(X),p(m)\ne0_R\}.$$ It is readily seen that $\deg(m_0*m_1)=\deg(m_0)+_\omega\deg(m_1)$ and that $\sigma(p_0\odot p_1)\le\sigma(p_0)+_\omega\sigma(p_1)$ whenever $p_0,p_1,p_0\odot p_1\ne 0_{R[X]}$.

I define the set $H(R,X)$ of homogeneous polynomials with $R$-coefficients and $X$-indeterminates to be the set of all $p\in R[X]\smallsetminus\{0_{R[X]}\}$ such that $$\sigma(p)=\min\{\deg(m):m\in M(X),p(m)\ne0_R\}.$$ It is readily seen that $H(R,X)\cup\{0_{R[X]}\}$ is a sub-semigroup of $\langle R[X],\odot\rangle$ (a sub-monoid if $R$ is unital).

The Actual Question: It seems clear to me that $R$ has the zero product property ($a\cdot b=0_R$ implies $a=0_R$ or $b=0_R$) if and only if $\langle H(R,X),\odot\rangle$ is a semigroup. In that case, the restriction of $\sigma$ to $H(R,X)$ should be a semigroup homomorphism--that is, $\sigma(h_0\odot h_1)=\sigma(h_0)+_\omega\sigma(h_1)$.

Unfortunately, I have been banging my head against the wall trying to prove these for some time now. In particular, I'm having trouble showing that whenever $R$ has the zero product property, then $h_0\odot h_1\ne0_{R[X]}$ whenever $h_0,h_1\in H(R,X)$. I've tried to proceed by induction on the cardinalities of the supports of $h_0,h_1$, but I can't figure out how to make the induction step click.

Any suggestions, hints, or nice proofs of this?

Cameron Buie
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  • "Natural"... this is certainly not the natural definition. It suggests the wrong direction for functoriality (it looks contravariant in $X$ but it should be covariant). – Qiaochu Yuan May 03 '13 at 19:36
  • @Qiaochu: Perhaps a poor choice of definitions is why I've been having such trouble proving it. What definition(s) would you suggest, instead? – Cameron Buie May 03 '13 at 19:52
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    The definition should be in terms of the universal property (the construction can be whatever you want as long as you can show that it satisfies the universal property). Also, in general you're using way too much rigor and way too much notation. It's just going to confuse you (it's confusing me, anyway). I can't even parse your definition of homogeneous polynomial to make sure it's what I think it is. What is $\rho$? – Qiaochu Yuan May 03 '13 at 20:04
  • Are you just trying to say the following? Let $X$ be a set. Let $R$ be a ring, possibly noncommutative and nonunital. Let $R[X]$ be the ring of polynomials $\sum r_i x^i$ where $i = (i_1, ... i_n)$ is a multi-index and $x^i = x_1^{i_1} ... x_n^{i_n}$ (and $x_i \in X$). The degree of a monomial in $R[X]$ is $\sum i_j$, and an element of $R[X]$ is homogeneous if it's a sum of monomials of the same degree. Yes? And the question is under what conditions a product of nonzero homogeneous polynomials is nonzero homogeneous? (The correct definition of homogeneous includes $0$.) – Qiaochu Yuan May 03 '13 at 20:10
  • I am trying to show that a product of (nonzero) homogeneous polynomials in $R[X]$ is a nonzero polynomial, so long as $R$ has the zero product property. – Cameron Buie May 03 '13 at 20:21
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    @QiaochuYuan, there is no basis for the claim that «the definition should be in terms of the universalproperty». That only reflects your education. – Mariano Suárez-Álvarez May 03 '13 at 21:56
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    @Mariano: with all due respect, I think there is. If a mathematical object has more than one possible construction, I don't think it's a good idea to define the object in terms of one of the possible constructions because the question remains: before we perform one of these constructions, what is it, exactly, that we're constructing? By analogy, suppose I tell you I'm trying to construct a table. If you ask me "what's a table?" I'm not going to finish constructing the table and then say "there. That's a table." I'm going to say "a table is something you can place in a room and place other..." – Qiaochu Yuan May 04 '13 at 01:25
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    "...things on top of" or something like that. In other words, I'll define a table for you in terms of the function I want it to serve, not in terms of a construction of a particular table. The construction is secondary, since ultimately all tables serve the function of being a table and that's why we care about tables; it's the function that is primary. Analogously, the function of a universal object is to be the thing that satisfies its universal property; when we construct it, the reason we are trying to construct it and not some other thing is because of that property. – Qiaochu Yuan May 04 '13 at 01:26
  • More specifically, suppose you are trying to explain to a student what the polynomial ring $\mathbb{F}_p[x]$ over a finite field is. The student protests that he thinks it should be the ring of polynomial functions $\mathbb{F}_p \to \mathbb{F}_p$. You explain, no, that turns out to be the wrong thing. Why is it the wrong thing? Because it doesn't satisfy the correct universal property with respect to $\mathbb{F}_p$-algebras; in particular, you can't evaluate such a thing on an element of an extension of $\mathbb{F}_p$. Before you construct $\mathbb{F}_p[x]$ you need to explain... – Qiaochu Yuan May 04 '13 at 01:30
  • ...why you are constructing $\mathbb{F}_p[x]$ instead of $\mathbb{F}_p[x]/(x^p - x)$, and in particular why it is the former rather than the latter that deserves the name "the ring of polynomials over $\mathbb{F}_p$," and this isn't a question that can be answered by referring to the construction of $\mathbb{F}_p[x]$. – Qiaochu Yuan May 04 '13 at 01:30
  • Rota makes a similar point in an essay I can't find about how the multitude of possible constructions of the real numbers makes it clear that mathematicians have some Platonic object in mind that they are trying to grasp in various ways rather than any particular one of those constructions being "the" real numbers (he put it more elegantly than this but it's been awhile since I've read it). – Qiaochu Yuan May 04 '13 at 01:39
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    @QiaochuYuan, for the most part, all that is true but irrelevant really :-) In this specific context, we have someone trying to prove a basic fact about polynomials: the universal property of polynomial rings is at best an interesting distraction! I certainly agree that the universal property of polynomial rings is quite important (it tends to be much less so, though, when the coefficient rings are non-commutative, for you have to litter what are pretty statements which extra conditions —in fact, the universal property quite rarely, if at all, invoked in non-commutative contexts...) – Mariano Suárez-Álvarez May 04 '13 at 07:13
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    (cont.) but imposing it as a deontological necessity at a point where this particular statement arises is —if nothing else— pedagogically disingenuous. It is clear that the categorical outlook helps you reason (and, as it turns out, it helps me quite a lot), but it is the most difficult part of being helpful to others to come to terms with the fact that others may probably be different to ourselves in many ways, and in this aspect in particular. – Mariano Suárez-Álvarez May 04 '13 at 07:14
  • @CameronBuie 1. Let me see if I've understood your notation: $\omega$ denotes the set of non-negative integers, and $+_{\omega}$ is the sum of such numbers. 2. I think your question reduces to the following: if $f,g\in R[X_1,\dots,X_n]$ are non-zero homogeneous polynomials, and $R$ has the zero product property, then $fg\ne0$. Doesn't this follow from "$R$ has the zero product property $\implies$ $R[X_1,\dots,X_n]$ has the zero product property"? (I suppose the indeterminates commute with the elements of $R$.) – user26857 Nov 27 '15 at 17:40
  • @user26857: If I recall correctly (it's been a while), I was trying to prove that $R[X]$ has the zero product property whenever $R$ does, using the result on $H(R,X)$ as an intermediate result, so I couldn't use it. – Cameron Buie Nov 27 '15 at 23:13
  • @CameronBuie Isn't easier to prove directly that $R[X]$ has the zero product property whenever $R$ does? If the set $X$ has only one element this is standard; if $|X|<\infty$ apply induction, and the general case reduces to the finite case. (Am I missed something?) – user26857 Nov 27 '15 at 23:16
  • @user26857: Not that I know of, but I'll have to track down my old work on the problem. My thanks for the recommendation! – Cameron Buie Nov 28 '15 at 03:25

1 Answers1

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Let $f = \sum r_i x^i$ and $g = \sum s_j x^j$ be two polynomials. I claim that in the product $fg$ there is a term whose coefficient has the form $r_i s_j$ for some $i, j$. This suffices to prove the claim under the assumption that $R$ has no zero divisors, and reduces the claim to the following straightforward geometric argument:

Since $f, g$ both have finitely many terms, we may assume WLOG that $X$ is finite, say $|X| = n$. Let

$$\text{supp}(f) = \{ i \in \mathbb{Z}^n : r_i \neq 0 \}$$

denote the support. Then $\text{supp}(f)$ and $\text{supp}(g)$ are two finite sets of points in $\mathbb{Z}^n \subset \mathbb{R}^n$. Let $H$ be a hyperplane in $\mathbb{R}^n$ such that none of its translates passes through two or more points of either $\text{supp}(f)$ or $\text{supp}(g)$ (a generic hyperplane will have this property). Let $v \in \mathbb{R}^n$ be a vector orthogonal to $H$, and say that a point in a subset of $\mathbb{R}^n$ is extremal if $\langle v, - \rangle$ attains a maximum there. By construction, $\text{supp}(f)$ and $\text{supp}(g)$ have unique extremal points $i_0, j_0$ (if there is more than one extremal point then some translate of $H$ passes through all of them).

Now, $\text{supp}(fg)$ is contained in the Minkowski sum $\{ i + j : i \in \text{supp}(f), j \in \text{supp}(g) \}$. Furthermore, $\langle v, i + j \rangle = \langle v, i \rangle + \langle v, j \rangle$, from which it follows that

$$\langle v, i + j \rangle \le \langle v, i_0 + j_0 \rangle$$

with equality iff $i = i_0, j = j_0$. In particular, $i + j \neq i_0 + j_0$ unless $i = i_0, j = j_0$. Hence the coefficient of $x^{i_0 + j_0}$ in $fg$ is $r_{i_0} s_{j_0}$ and the conclusion follows.

The geometric picture is visualizable when $n = 2$. Here imagine two collections of points in the plane, and take e.g. "leftmost points" (although you may have to tilt the plane slightly if there is more than one leftmost point in each collection).

Qiaochu Yuan
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  • (+1) Interesting...never would have thought to approach it this way. I'll sit down and explore this soon, and try to digest it all. Thanks! – Cameron Buie May 03 '13 at 20:43
  • @Cameron: one motivation for this argument is to try to generalize the obvious argument in the case of polynomials in one variable, where in a product of polynomials $f, g$ of degree $n$ and degree $m$ a unique pair of terms contributes to the coefficient of $x^{n+m}$ because $n, m$ are the unique largest degrees of terms in $f, g$. In more than one variable there's a pesky problem with multiple terms of the same degree, so it's necessary to break this symmetry somehow. See also http://en.wikipedia.org/wiki/Monomial_order ; above I used more or less the weight order, but I think the above... – Qiaochu Yuan May 03 '13 at 20:48
  • ...argument works with any monomial order. – Qiaochu Yuan May 03 '13 at 20:50