I'm addressing people familiar with the vernacular of logicians.(The tag commutative algebra is due anyway.)

TL;DR The following atypical proof of Hilbert's Basis Theorem puzzle me. The reason, at the end.

Satz 1. Let $A\subseteq M\models T_{\rm id}$ and let $p(x)\subseteq L_{\rm at^\pm}(A)$, where $x$ is a finite tuple. Then there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p_+(x)$, where $p_+(x)=p(x)\cap L_{\rm at}(A)$.


$T_{\rm id}\ $ theory of integral domains in the language of rings

$L_{\rm at}(A)\ $ atomic formulas (or, up to equivalence, polynomial equations)

$L_{\rm at^\pm}(A)\ $ atomic and negated atomic formulas

$\vdash\ $ logic consequence over $T_{\rm id}\cup{\rm Diag}\langle A\rangle_M$


Only to provide some context. Hilbert's Basis Theorem translated in a model theoretical language claims the following

Satz 2. Let $A\subseteq M\models T_{\rm id}$ and let $p(x)\subseteq L_{\rm at}(A)$, where $x$ is a finite tuple. Then there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p(x)$.

Edit 1: when $A=M$ is a field this is the translation of $M[x]$ is Noetherian, a weaker form of Hilbert's Basis Theorem, enough for my purposes.

Edit 2: Edit 1 is not completely correct. As remarked by Alex Kruckman, the definition of $\vdash$ above, entails that types closed under $\vdash$ correspond to radical ideal. Hence Satz 2 only claims that in $M[x]$ radical ideal are finitely generated.

Obviously Satz 2 $\Rightarrow$ Satz 1. The converse implication follows from a standard compactness argument. Parenthetically, Hilbert's Nullstellensatz easily follows from Satz 1, by quantifier elimination.

Proof of Satz 1

By induction on the length of $x$.

If $x$ is the empty tuple, the claim is trivial.

Now, let $x$ be finite tuple and let $y$ be a single variable. Let $p(x,y)\subseteq L_{\rm at^\pm}(A)$. We write $p(x)$ for the set of formulas $\varphi(x)\in L_{\rm at^\pm}(A)$ such that $p(x,y)\vdash\varphi(x)$.

Assume as induction hypothesis that there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p_+(x)$. We prove Sats 1 with $x,y$ for $x$.

Let $\psi(x,y)\in L_{\rm at}(A)$ have minimal degree (in $y$) among the formulas such that $p(x,y)\vdash\psi(x,y)$ and $\psi(x)\nvdash\psi(x,y)$. Let $n>0$ be the degree of $\psi(x,y)$, so we may assume $\psi(x,y)$ has the form $t(x,y)=0$ where

$$t(x,y)=t_n(x)\cdot y^n + t'(x,y),$$

$t'(x,y)$ is a polynomial of degree $<n$, and $p(x)\nvdash t_n(x)=0$. We claim that

$$(1)\quad p_+(x,y)\dashv t_n(x)\neq0\; \wedge\; \psi(x)\;\wedge\; \psi(x,y)$$

Suppose not and let $s(x,y){=}0\in p_+(x,y)$ be a counter example to 1 of minimal degree. By the minimaly of $n$ we may assume that $s(x,y)$ has the form

$$s(x,y)=s_{n+i}(x)\cdot y^{n+i}+s'(x,y)$$

for some polynomial $s'(x,y)$ of degree $<n+i$ and some $i\ge0$. Clearly $$(2)\quad p(x,y)\vdash s_{n+i}(x)\cdot t(x,y)\cdot y^{i} - t_n(x)\cdot s(x,y) = 0.$$

It is also clear that the polynomial in (2) has degree $<n+i$. This contradicts the choice of $s(x,y)$. A contradiction that proves the proposition.


Note that an elementary argument proves Satz 1 for tuples of length 1. Therefore I would expect the induction to go as follows. Let $a$ realize $p(x)$ in some large model. Apply to $p(a,y)$ the elementary for tuples of length 1 and obtain that $\psi(a,y)\vdash p_+(a,y)$ for some conjunction of formulas in $p(a,y)$. Conclude that Satz 1 holds for $p(x,y)$.

The strategy above does not work. Hence I suspect that a stronger claim than Satz 1 holds (that would make the strategy work).

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    $\begingroup$ dear Primo, apologies for being dense, but I'm not entirely sure what your question is – are you asking for feedback about the proof strategy you outline in the "Question" section, or about whether you can strengthen the statement of Satz 1, or about the first proof of Satz 1 that you give? $\endgroup$ Jun 14 at 16:55
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    $\begingroup$ You are right. I kind of expected such a remark.... I feel that the proof of Satz 1 is unnatural hence I suspect that Satz 1 can be strengthened. Feedback is appreciated. (E.g. has anybody encountered other atypical proofs of Hilbert's Basis Theorem?) $\endgroup$ Jun 14 at 17:05
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    $\begingroup$ dear Primo, that makes sense, thank you :) ... also, forgive me if this is a silly question, but how do you recover Hilbert's basis theorem from Satz 2? if I'm not mistaken, what Satz 2 says gives us in algebraic language is that, for any integral domain $R$, if $I$ is an ideal of $R[t_1,\dots,t_n]$, then there exist $f_1,\dots,f_m\in I$ such that $\mathcal{V}(I)=\mathcal{V}(\langle f_1,\dots,f_m\rangle)$. it's not obvious to me how to recover Hilbert's basis theorem from this... $\endgroup$ Jun 14 at 17:57
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    $\begingroup$ There are a few things about the motivation section that puzzle me. Maybe you can clarify. (1) Don't Satz 1 and Satz 2 say exactly the same thing? Satz 2 is a special case of Satz 1 (where we don't allow any negated atomic formulas in $p$), and Satz 1 follows by applying Satz 2 to $p_+$. No compactness necessary. (2) If $M$ is a non-Noetherian integral domain, then $M[x]$ is not Noetherian. Did you mean to require $M$ to be a field? (3) I don't think Satz 2 really expresses Notherianity of $M[x]$. Given a set of polynomials, their logical consequences via $\vdash$ as you defined it is not ... $\endgroup$ Jun 16 at 2:37
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    $\begingroup$ ... the ideal $J$ they generate. Instead, it is the radical $\sqrt{J}$, or it is the unit ideal if $\sqrt{J}$ contains any non-zero element of $M$ (since in this case the formula is inconsistent). So if I have that right, Satz 2 says that for every ideal $I$ in $M[x]$ such that $I\cap M=0$, there are finitely many elements of $I$ such that $I$ is contained in the radical of the ideal they generate. $\endgroup$ Jun 16 at 3:09

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