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Let $R$ be the ring $$\mathbb{R}[t_1,t_2\ldots]/(t_1^2,t_2^2,\ldots)$$

Let $p_1,\ldots p_\ell$ be polynomials in $R[x_1,\ldots,x_n]$ whose constant terms are 0.

Let $f$ be a polynomial which is zero on the zero-set of the $p$’s.

Conjecture: $f$ must be in the ideal generated by the $p$’s.

Does this conjecture hold? If so it would be a Nullstellensatz of the form $I=J(V(I))$ for these ideals.

I find this easier to reason about with a contrapositive: Let $p_1,\ldots p_\ell$ be polynomials in $R[x_1,\ldots,x_n]$ whose constant terms are 0. Then for any polynomial $f$ which is not in their ideal, there is some $s=(s_1,\ldots,s_n)\in R^n$ with all $p_i(s)=0$ and $f(s)\neq 0$.

A first example is $p=x^{k+1}$. Then $f=x^k$ is not in the ideal of $p$, and $s=t_1+\cdots +t_k$ satisfies $p(s)=0$ (because each $t_i^2$ is zero) and $f(s)=k!\prod t_i \neq 0$.

As another example, with $p_1=x^3$, $p_2=y^3$, $p_3=x^2+y^2$, and $f=x^2-y^2$, we can take $s=(t_1+t_2,t_1-t_2)$.

I once wrote down a proof of this conjecture for a large class of examples at this level of difficulty. For more complicated polynomials like $x^2+xy+y^3$ I have neither a proof nor a counterexample.

I make the conjecture because of an analogy with Smooth Infinitesimal Analysis, which I’ve explained in another answer. Specifically, this conjecture is analogous to a corrected version of the uniqueness in the generalized Kock-Lawvere axiom from Models For Smooth Infinitesimal Analysis by Ieke Moerdijk and Gonzalo Reyes; the correction, which I found by exploring this analogy, is that the ideal in the book’s statement should include the polynomials $x_i^{k+1}$.

There might be a slick topos-theoretic proof, or there might be a helpful combinatorial construction; I’d welcome any results.

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  • $\begingroup$ If the intention is that all infinitesimals square to $0$, shouldn’t the ideal in the definition of $R$ also include $t_it_j$ (on account of $(t_i+t_j)^2-(t_i-t_j)^2=4t_it_j$)? $\endgroup$ Commented Oct 10, 2022 at 17:26
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    $\begingroup$ The intention is that all infinitesimals are nilpotent, with some squaring to zero, others (like $t_i+t_j$) cubing to zero, etc., but no finite power annihilating all of them. E.g. synthetic differential geometry treats the element of area as the product of two infinitesimals, so that shouldn’t vanish. $\endgroup$
    – user44143
    Commented Oct 10, 2022 at 17:40
  • $\begingroup$ I see. That’s peculiar. $\endgroup$ Commented Oct 10, 2022 at 18:31

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