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For any commutative ring $R$ let $R[x]$ denote the ring of polynomials with coefficients in $R$. Any polynomial $p \in R[x]$ naturally induces a function $\hat{p} :R \rightarrow R$. In some cases, a nonzero polynomial will induce the zero function. For example, with $R=\mathbb{Z}_6$, the polynomials $x^5 +3x^2+2x$, $3x^4 + 5x^3 + 4x$, and $x^4 - x^3 - x^2 + x$ all induce the zero function. Some things that I already know:

  • If $R$ is finite, $R = \{r_1, r_2, ... r_n\}$, then $(x-r_1)(x-r_2)...(x-r_n)$ induces the zero function.
  • If $R=\mathbb{Z}/p\mathbb{Z}$ then the ideal $I$ of polynomials inducing the zero function is generated by $x^p - x$ (and in fact this coincides with the polynomial given above).
  • If $R$ is an infinite integral domain, then no nonzero polynomial induces the zero function.
  • If $R$ is infinite but not an integral domain, then there may or may not be nonzero polynomials that induce the zero function. (I have examples of each.)

It's this last point that brings me here. I would like to know if there are necessary and sufficient conditions on $R$ that ensure that $\hat{p}=0$ only for $p=0$. Such conditions would have to be stronger than "infinite" but weaker than "infinite integral domain". Does anybody know what those conditions are?

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    $\begingroup$ In rings that are Cartesian products of integral domains $\hat p=0$ implies $p=0$. (We can argue component-wise). Interesting question. $\endgroup$ – P Vanchinathan Mar 21 '14 at 4:55
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    $\begingroup$ I don't think so! I think you need them all to be infinite. For example every element in a Cartesian product of $\mathbb{Z}/2\mathbb{Z}$s satisfies $x^2-x=0$. And if you allow non-monic polynomials (whose coefficients are only non-zero in one component) then much worse things can happen. I think the question is "wild" and there will be no reasonable classification. $\endgroup$ – eric Mar 21 '14 at 7:03
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    $\begingroup$ From the case of domains we get: If $\mathfrak{p}$ is a prime ideal with $|R/\mathfrak{p}| = \infty$, then all coefficients of a "bad" polynomial are in $\mathfrak{p}$. So e.g. a reduced ring with $|R/\mathfrak{p}| = \infty$ for all minimal primes qualifies. -- A notational request: Could you please write $\mathbb{Z}/p$ instead of what I see as $p$-adics in the second point (or just remove it, as it is really just a special case of the first)? $\endgroup$ – Torsten Schoeneberg Mar 21 '14 at 10:54
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    $\begingroup$ In a Cartesian product $R=\prod_i R_i$ of rings, $\hat{p}=0$ implies $p=0$ iff all $R_i$'s have the same property. Has anybody an example of an infinite, indecomposable ring (i.e., no idempotents except $0$, $1$) such that $\hat{p}=0$ for some $p\neq 0$? $\endgroup$ – Frieder Ladisch Mar 21 '14 at 13:47
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    $\begingroup$ @FriederLadisch, great question. Consider $R = \mathbb{Z}/(2) [x_1, x_2, \dots]/(x_1, x_2, \dots)^2$. This is a local algebra over the field with two element whose maximal ideal has square zero, and which is isomorphic to $\mathbb{Z}/(2)$ modulo that maximal ideal. Thus $p(x) = x^2(1-x)$ satisfies $\hat{p} = 0$ for this ring. $\endgroup$ – Manny Reyes Mar 21 '14 at 13:54
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$R$ has a nonzero polynomial that induces the zero function if and only if there are ideals $I$, $J$ such that $I$ is nontrivial, $IJ=0$, and $R/J$ is a ring satisfying the following condition:

There exists $n$ such that, for any $n$ elements $x_1, \dots x_n \in R/J$, the discriminant $\prod_{i<j} (x_i-x_j)=0$.

So $R$ is glued together out of an arbitrary ring and a ring that satisfies a specific polynomial identity.


Proof of only if: Let $I$ be the ideal generated by the coefficients of the nonzero polynomial and let $J$ be the ideal of zero divisors of $I$. We will show that for all $n$ elements $x_1,\dots x_n \in R$, $I\prod_{i<j} (x_i-x_j)=0$, so that $R/J$ satisfies this identity. To see this, take any $n$ elements $x_1,\dots,x_n\in R$ and take the Vandermonde matrix. Multiplying this by the vector of coefficients of the nonzero polynomial gives $0$. So multiplying on the other side by the adjugate matrix we still get $0$, but a matrix times its adjugate is just the identity times the determinant, so the determinant time each coefficient of the polynomial is $0$. The determinant of the Vandermonde matrix is just the discriminant $\prod_{i<j} (x_i-x_j)$.

Proof of if: Let $a$ be a nonzero element of $I$, let $n$ be the smallest $n$ such that $a\prod_{i<j} (x_i-x_j)=0$ for all $x_1,\dots,x_n \in R$, so that we have some $y_1,\dots y_{n-1}\in R$ that satisfy $a\prod_{i<j} (y_i-y_j)\neq0$ . Then all $x$ satisfy:

$$a\prod_{i<j} (y_i-y_j) \prod_{i=1}^{n-1} (x-y_i)=0$$

and the leading coefficient of that polynomial is nonzero.


Since $R/I$ is arbitrary, I think we should study rings satisfying $\prod_{i<j} (x_i-x_j)=0$. Such rings have all residue fields bounded in size by $n$, so all prime ideals are maximal and their spectra are totally disconnected topological spaces. Since the geometry of this kind of space is not very easy to classify, we might want to restrict our attention to the local rings:

If $R$ is a local ring with maximal ideal $m$, there is some $n$ such that, for all $x_1,\dots x_n \in R$, $\prod_{i<j} (x_i-x_j)=0$ if and only if $R/m$ finite and there is some $N$ such that every element $x\in m$ satisfies $x^N=0$.


Proof of only if: If $R/m$ is infinite, take $x_1,\dots x_n$ to be lifts of distinct elements in $R/m$. Set $N=\frac{n^3-n}{6}$. For all $x \in m$ we can take $x_i= x^i$ for $1\leq i \leq n$. Then

$$0=\prod_{i<j} (x_i-x_j) = \prod_{i<j} x^i (1-x^{j-i})$$

which is $x^{\frac{n^3-n}{6}}$, times a unit, so $x^N=0$.

Proof of if: If $a_1,\dots,a_m$ are lifts of all the elements of the residue field, $\prod_{i=1}^m (x-a_i)^N$ is monic and vanishes for all $x$, so the bottom row of the Vandermonde matrix of any $x_1,\dots, x_{mN}$ is a weighted sum of the previous rows, and so the determinant of the Vandermonde vanishes.


I don't think one can improve the classification much beyond this. Consider the examples $\mathbb Z[a,b]/(a^2,ab,2a)$, which is just slightly off from a nice infinite integral domain, but every element satisfies the identity $a x(x-1)=0$, and $\mathbb F_2[a_1,a_2,\dots]/(a_1^2,a_2^2,\dots)$, which is an extension of Manny Reyes's example where every element satisfies the identity $x^2(x-1)^2=0$. But I would be excited to see any further insights.

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  • $\begingroup$ In your proof of the only if part, you cannot take arbitrary $n$: in order to be able to multiply the Vandermonde matrix with the vector of coefficients of the polynomial, you must take $n=1+k$, being $k$ the degree of the polynomial. $\endgroup$ – Matemáticos Chibchas May 15 '17 at 17:09
  • $\begingroup$ @MatemáticosChibchas That's correct. I'm pretty sure that's what I meant when I was writing this, but forgot to state it. $\endgroup$ – Will Sawin May 15 '17 at 17:14
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I can prove a cleaner criterion in the Noetherian case. This does not carry over to the general case, as shown by Will Sawin's example in the comments below.

Lemma. Let $(R,\mathfrak p)$ be an Artinian local ring, and let $f \in R[x]$ of degree $n$. If there exists a set $S \subseteq R$ with $|\operatorname{im}(S \to R/\mathfrak p)| > n$ such that $f(r) = 0$ for all $r \in S$, then $f = 0$.

Proof. We proceed by induction on the length of $R$. The case $\ell(R) = 1$ is the case where $R$ is a field, which is trivial (this is where the hypothesis $|S| > n$ is used). In general, let $(R,\mathfrak p)$ be local Artinian. Then we have a filtration $$0 = I_0 \subsetneq \ldots \subsetneq I_s = R$$ of $R$-submodules (i.e. ideals) such that each of the subquotients $I_i/I_{i-1}$ has length $1$. Let $R' = R/I_1$, and note that $\ell(R') = \ell(R) - 1$. Therefore, by the induction hypothesis, we have $f \in I_1 R[x]$. Since $\ell(I_1) = 1$, we have $I_1 \cong R/\mathfrak p$. Thus, $I_1$ is principal, say $I_1 = (r_1)$, and $\operatorname{Ann}(r_1) = \mathfrak p$.

Hence $f = r_1 g$ for some $g \in R[x]$, which we may assume has degree $n$. Then $r_1g(r) = 0$ for all $r \in S$, so $g(r) \in \operatorname{Ann}(r_1) = \mathfrak p$ for all $r \in S$. Applying the induction hypothesis this time to $R/\mathfrak p$, we see that $g \in \mathfrak p R[x]$. Hence, $r_1 g = 0$, since $\mathfrak p \cdot r_1 = 0$. $\square$

Theorem. If $R$ is Noetherian, then there exists $f \in R[x]\setminus\{0\}$ of degree $\leq n$ with $f(r) = 0$ for all $r \in R$ if and only if $R$ has an associated prime $\mathfrak p$ such that $|R/\mathfrak p| \leq n$.

Proof. If $\mathfrak p = \operatorname{Ann}(r_0)$ is such that $|R/\mathfrak p| \leq n$, then let $\{r_1,\ldots,r_m\}$ be a set of representatives for $R/\mathfrak p$, and consider $f = r_0 \prod_{i=1}^m (x-r_i)$. Then for each $r \in R$, we have $\prod (r-r_i) \in \mathfrak p$, hence $f(r) = 0$ since $\mathfrak p \cdot r_0 = 0$. Moreover, the degree of $f$ is $m \leq n$.

Conversely, assume $|R/\mathfrak p| > n$ for every associated prime $\mathfrak p \subseteq R$, and let $f \in R[x]$ of degree $\leq n$ be such that $f(r) = 0$ for all $r \in R$. The natural map $R \to \prod_{\mathfrak p \in \operatorname{Ass}(R)} R_{\mathfrak p}$ is injective [Tag 0331]. Let $S_\mathfrak p \subseteq R_\mathfrak p$ be the image of $R \to R_\mathfrak p$ and let $f_\mathfrak p$ be the image of $f$ in $R_\mathfrak p$, for each $\mathfrak p \in \operatorname{Ass}(R)$.

Since $|R/\mathfrak p| > n$, the image of $S_\mathfrak p$ in $R_\mathfrak p/\mathfrak p = \operatorname{Frac}(R/\mathfrak p)$ has more than $n$ elements. Applying the lemma above to $R_\mathfrak p/\mathfrak p^m$ with the set $S_\mathfrak p/\mathfrak p^m$, we get that $f_\mathfrak p \in \mathfrak p^mR_\mathfrak p[x]$. By Krull's intersection theorem [Tag 00IP], this implies $f_\mathfrak p = 0$. Since this holds for all associated primes $\mathfrak p$, we conclude that $f = 0$. $\square$

Corollary. If $R$ is Noetherian, then there exists $f \in R[x]\setminus\{0\}$ with $f(r) = 0$ for all $r \in R$ if and only if $R$ has an associated prime $\mathfrak p$ such that $R/\mathfrak p$ is finite. $\square$

Remark. The first part of the proof of the does not use the Noetherian hypothesis. Moreover, the argument actually works when there exists any ideal $I$ (not necessarily prime) occurring as the annihilator of some element such that $R/I$ is finite.

Although in general rings not every annihilator is contained in an associated prime, this is true for annihilators $I$ such that $R/I$ is finite. Indeed, there are only finitely many primes in $R/I$, and a maximal one occurring as an annihilator is associated (see Eisenbud's Commutative Algebra, Prop. 3.4). Thus, allowing such $I$ does not give a more general class of examples.

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    $\begingroup$ Nice, this characterization is indeed cleaner than mine. A counterexample in the non-Noetherian case is the ring of continuous $\mathbb F_2$-valued functions on the Cantor set (i.e. $\mathbb F_2[x_1,x_2,\dots]/(x_i^2-x_i)$) which satisfies the identity $x^2-x=0$ but doesn't have any associated primes because no element has a prime annihilator. $\endgroup$ – Will Sawin May 2 '17 at 15:12
  • $\begingroup$ @WillSawin: that's a very nice example. So my description does not do much beyond the Noetherian case. $\endgroup$ – R. van Dobben de Bruyn May 4 '17 at 2:57
  • $\begingroup$ By $im (S \to R/\mathfrak p)$, do you mean the image of $S$ in $R/\mathfrak p$ ? Where is it used that this has cardinality more than $n$ ? $\endgroup$ – user111524 May 21 '18 at 19:18
  • $\begingroup$ Do you mean to say the image of $S_{\mathfrak p}$ in $R_{\mathfrak p}/\mathfrak p$ has " more than $n$ elements " rather than "at least $n$ elements " ? $\endgroup$ – user111524 May 21 '18 at 19:46
  • $\begingroup$ @users: yes, I mean the image of $S$ in $R/\mathfrak p$. The assumption on the cardinality is needed for the length $1$ case. You are right about at least $n$ elements; thanks. $\endgroup$ – R. van Dobben de Bruyn May 22 '18 at 3:26
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As I am not allowed to write comments, I write this as an answer, although it is only a partial answer.

  1. If $(R,\mathfrak{m})$ is local with $R/\mathfrak{m}$ finite and $\mathfrak{m}^n=0$ for some $n$, then there is a bad polynomial for $R$. It is obtained as in Manny Reyes comment. In this case one can choose a bad monic polynomial $P$ for $R/\mathfrak{m}$ (as in the question) and hence $P^n$ is not $0$.
  2. The assumption that $\mathfrak{m}^n=0$ for some $n$ is necessary here. As the following example shows: $$R=\mathbb{F}_p[t,t^{1/p},t^{1/p^2},t^{1/p^3},\dots]/(t^p)$$ is a local ring with maximal ideal $\mathfrak{m}=(t,t^{1/p},t^{1/p^2},t^{1/p^3},\dots)$ and every element in $\mathfrak{m}$ is nilpotent. But for every polynomial $P(x)\in R[x]$ there is some $t^q$ such that $P(t^q)\ne 0$.
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My guess is that either R is nonreduced or R is a cartesian product of rings (a decomposable ring), which is equivalent to say that either the polynomial 1 or x are reducible.

Surely, the existence of arbitrarily large regular(nonzerodivisor) Vandermonde determinants implies the nonexistence of such polynomials.

For other rings (ex. (Z[s]/(2s)) you have polynomials with infinitely many roots (sx^2+sx, vanishing on all integers) but you cannot have them vanish on all of R unless you create a nilpotent element (ex. Z[s]/(2s,s^2))

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