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Let $R$ be an infinite, characteristic zero, commutative ring. I can furthermore suppose it is reduced and indecomposable (no nontrivial nilpotents or idempotents).

My question is whether there is a nonzero polynomial $f\in R[x]$ which is identically zero on $R$.

Note: it is easy to show that there are polynomials with infinitely many roots: let $R=\mathbb Z[s]/(2s)$ and consider $f\in R[x]$ given by $f(x)=\overline{s}x^2+\overline{s}x$. All integers are roots of $f$.

But my question is whether we can have $f$ vanishing on all of $R$, not just on an infinite subset.

On the other hand, if we further mod out by $s^2$, turning $f$ (I believe) identically vanishing, we create a nilpotent element.

A technique that I tried is trying and produce a Vandermonde matrix $V$ associated to the elements $a_1,...,a_k$ of $R$ that be a nonzero element, so that $V$ multiplied by the matrix of coefficients of the canonical basis $e_i$ of polynomials of degree up to $k$, with its $i$-th element replaced by the coefficients of $f$, $f_i$'s, would have two proportional columns and yield $\det(V)\,f_i=0$ and therefore, if I can manage to make $\det(V)$ regular, I will get $f_i=0$.

But I believe we may have reduced rings where all elements are zerodivisors, so I am currently trying to modify this Vandermonde argument, by using the very coefficients of $f$ as $a_i$'s and create a nice Vandermonde lattice.

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  • $\begingroup$ What does "characteristic zero" mean for a ring? $\endgroup$ May 1, 2017 at 18:42
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    $\begingroup$ All "rational integers" n_R are nonzero, where n_R=n*1_R=1_R+...+1_R (n times) $\endgroup$ May 1, 2017 at 19:23
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    $\begingroup$ @darijgrinberg The characteristic of a ring $R$ is the kernel of the unique ring homomorphism $\mathbb Z \to R$. $\endgroup$ May 1, 2017 at 20:44
  • $\begingroup$ The MO question mathoverflow.net/questions/160986/… and answer by Will Sawin seems relevant. $\endgroup$ May 1, 2017 at 21:45
  • $\begingroup$ Yes, Joe, thanks, I read this (and left a comment). However, I was interested in knowing whether reduced indecomposable would be a sufficient condition, which is not contained in the (though exact) characterization by Will Sawin. R. Van Dobben de Bruyn answered perfectly my question. My only doubt left (which is not included in my question, but I was somehow looking for a proof speculating about this) is on any possible restriction on the amount of zerodivisors or of integers zerodivisors that a reduced indecomposable ring may have. $\endgroup$ May 1, 2017 at 23:06

1 Answer 1

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This is impossible. We deal with the Noetherian case first, and then deduce the general case.

Lemma. Let $R$ be a reduced infinite Noetherian ring without idempotents, and let $\mathfrak p \subseteq R$ be a prime. If $R/\mathfrak p$ is finite, then there exists a prime $\mathfrak q \subsetneq \mathfrak p$. For every such $\mathfrak q$, the ring $R/\mathfrak q$ is infinite.

Proof. Since $R/\mathfrak p$ is a finite domain, it is a finite field, hence $\mathfrak p$ is maximal. If there does not exist such $\mathfrak q$, then $\mathfrak p$ is minimal as well. This means that the constructible set $V(\mathfrak p) = \{\mathfrak p\}$ is stable under generisation and specialisation, hence it is a clopen subset of $\operatorname{Spec}(R)$; see e.g. [Tag 0542]. Since $R$ has no nontrivial idempotents, $\operatorname{Spec} R$ is connected, so this forces $\mathfrak p$ to be the unique prime ideal. But since $R$ is reduced, we conclude that $\mathfrak p = (0)$, so $R$ itself is finite, contradiction.

Hence, $\mathfrak p$ cannot be minimal. Thus, there exists $\mathfrak q \subsetneq \mathfrak p$. But $R/\mathfrak q$ cannot be finite, because a finite domain is a field, which would say that $\mathfrak q$ is maximal. $\square$

Remark. We used the Noetherian hypothesis in the statement that $V(\mathfrak p)$ is constructible. In a general ring, there is a retrocompactness assumption on the complement $D(\mathfrak p)$, which is satisfied for example if $\mathfrak p$ is (the radical of) a finitely generated ideal. See [Tag 04ZC] for details.

There is a more elementary argument if you don't want to use constructible sets. Indeed, assume $R$ has minimal primes $\mathfrak p_i$ different from $\mathfrak p$. Since $R$ is Noetherian, there are finitely many such, and since $\operatorname{Spec}(R)$ is connected, there must be an $i$ such that $V(\mathfrak p) \cap V(\mathfrak p_i) \neq \varnothing$. But that means that $\mathfrak p \in V(\mathfrak p_i)$, i.e. $\mathfrak p_i \subseteq \mathfrak p$, contradicting minimality of $\mathfrak p$.

Lemma. Let $R$ be a reduced infinite Noetherian ring without idempotents, and let $f \in R[x]$. If $f(r) = 0$ for all $r \in R$, then $f = 0$.

Proof. Let $\mathfrak p$ be a prime for which $R/\mathfrak p$ is infinite. Denote by $\bar{r}$ the reduction of $r$ in $R/\mathfrak p$. We clearly have $\bar f(\bar r) = \overline{f(r)}$ for all $r \in R$, so we conclude that $\bar f \in (R/\mathfrak p)[x]$ vanishes at all elements of $R/\mathfrak p$. Since $R/\mathfrak p$ is an infinite domain, this forces $\bar f = 0$, i.e. $f \in \mathfrak pR[x]$. Hence, $$f \in \bigcap_{\substack{\mathfrak p\\ |R/\mathfrak p| = \infty}} \mathfrak p R[x].\label{Eq 1}\tag{1}$$ But by the lemma above, every prime for which $R/\mathfrak p$ is finite contains a prime $\mathfrak q$ for which $R/\mathfrak q$ is infinite, so the intersection in (\ref{Eq 1}) is equal to $$\bigcap_{\mathfrak p} \mathfrak p R[x].$$ That is, every coefficient of $f$ is in $\bigcap_{\mathfrak p} \mathfrak p = \mathfrak{nil}(R)$, which is $0$ by assumption. $\square$

Corollary. Let $R$ be any reduced infinite ring without idempotents, and let $f \in R[x]$. If $f(r) = 0$ for all $r \in R$, then $f = 0$.

Proof. If $R$ is a field, then it is an infinite field, so we are done. Otherwise, let $r_0 \in R\setminus\{0\}$ be an element that does not have an inverse, and let $R'\subseteq R$ be a finitely generated subring containing $r_0$ such that $f \in R'[x]$. For example, we can take the subring generated by the coefficients of $f$ and $r_0$.

Then $R'$ is reduced and has no nontrivial idempotents, and $R'$ is Noetherian since it is finitely generated. If $R'$ is finite, then it is Artinian, hence local since it has no idempotents, hence a field since it has no nilpotents. This contradicts the assumption that $r_0$ does not have an inverse.

Thus, $R'$ is infinite, and $f(r') = 0$ for all $r' \in R'$. Therefore, $f = 0$ by the lemma above applied to the ring $R'$. $\square$

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    $\begingroup$ Wow! Thank you, so much R. van Dobben, your answer looks thourough and amazing. I just need some time to process it.... $\endgroup$ May 1, 2017 at 19:28
  • $\begingroup$ Dear R. Van Dobben, I wonder whether I (understand correctly and) may simplify: If $\mathcal{p}$ is a minimal prime, then for any prime $\mathcal{q}$ take a minimal prime $\mathcal{m_q}$ contained in it $(\mathcal{m_p}=\mathcal{p})$. Now, $\cup_{q\neq p} V(\mathcal{m_q})=Spec(R)\backslash \{\mathcal{p}\}$ and the full union $\cup V(\mathcal{m_q})$=Spec(R). As $R$ is Noeth, its Spec is q-compact and we can take a finite sub-union, excluding $\mathcal{p}$, we have that $\cup_{q\neq p} V(\mathcal{m_q})$ is actually a finite union of closed sets, hence closed, so that $V(\mathcal{p})$ is open. $\endgroup$ May 2, 2017 at 0:39
  • $\begingroup$ @MarcusBarãoCamarão The $V(\mathfrak m_\mathfrak q)$ are closed sets, not open. Compactness only says something about open covers. $\endgroup$ May 2, 2017 at 0:47
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    $\begingroup$ @Vincent: ah sorry, that's what I meant. What I wrote is obviously wrong and very confusing. $\endgroup$ May 3, 2017 at 15:29
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    $\begingroup$ @users: no, it only works if the reduction of $R$ is still infinite. A counterexample is $\mathbb F_p[x_1,\ldots]/(x_1^2,\ldots)$. $\endgroup$ May 20, 2018 at 2:09

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