Relative big polynomial rings

See [ESS, Proposition 3.2] ∎

Suppose $I$ has finite height. Then, by definition, $I$ is contained in a prime of finite height, which is finitely generated by Proposition 2.1.

Now suppose $I$ is contained in a finitely generated non-unital ideal, say $(f_{1},\ldots,f_{r})$. Each $f_{i}$ uses only finitely many variables, and so $(f_{1},\ldots,f_{r})$ is extended from a finite variable subring. Performing such an extension does not change height [ESS, Proposition 3.3]. ∎

The natural homomorphism $E\otimes_{F}R/\mathfrak{p}\to S/\mathfrak{q}$ is surjective. We thus find

where $\dim$ denotes Krull dimension. Since $\dim(R/\mathfrak{p})=n-\operatorname{ht}_{R}(\mathfrak{p})$ and $\dim(S/\mathfrak{q})=n-\operatorname{ht}_{S}(\mathfrak{q})$, the result follows. ∎

For any $J\subset I$ we have $\operatorname{ht}_{R}(J)\leq\operatorname{ht}_{R}(I)$, and so $\sup_{\alpha\in\mathcal{I}}\operatorname{ht}_{R}(J_{\alpha})\leq\operatorname{ht}_{R}(I)$. We now prove the reverse inequality. First, suppose that the $J_{\alpha}$ are finitely generated. If $\sup_{\alpha}\operatorname{ht}_{R}(J_{\alpha})$ is infinite then there is nothing to prove, so suppose it is a finite number $c$. Passing to a cofinal subset, we may as well suppose that $\operatorname{ht}_{R}(J_{\alpha})=c$ for all $\alpha$. For each $\alpha$, let $\mathcal{P}_{\alpha}$ be the set of prime ideals of height $c$ containing $J_{\alpha}$; this set is finite by hypothesis. Of course, if $\alpha\leq\beta$ then $\mathcal{P}_{\beta}\subset\mathcal{P}_{\alpha}$. By a standard compactness result, we have $\bigcap_{\alpha\in\mathcal{I}}\mathcal{P}_{\alpha}\neq\varnothing$. We can thus find a prime $\mathfrak{p}$ of height $c$ such that $J_{\alpha}\subset\mathfrak{p}$ for all $\alpha$. Since $I=\bigcup_{\alpha\in\mathcal{I}}J_{\alpha}$, we thus find $I\subset\mathfrak{p}$, and so $\operatorname{ht}_{R}(I)\leq c$.

We now treat the general case. For each $\alpha$, let $\{J_{\alpha,\beta}\}_{\beta\in\mathcal{K}_{\alpha}}$ be the finitely generated ideals contained in $J_{\alpha}$. Then

where in the first and last step we used the previous case. ∎

Let $I$ be a finitely generated ideal of $R$ of height $c$. Then $I$ is extended from an ideal $I^{\prime}$ of some finite variable subring $R_{0}$ of $R$. Since such extensions do not affect height [ESS, Proposition 3.3], we see that $I^{\prime}$ has height $c$ also. Let $\mathfrak{q}_{1},\ldots,\mathfrak{q}_{r}$ be the minimal primes above $I^{\prime}$ in $R_{0}$. Now, let $\mathfrak{p}$ be height $c$ prime of $R$ containing $I$. Then $\mathfrak{p}$ is finitely generated by Proposition 2.1, and thus extended from a prime $\mathfrak{p}^{\prime}$ of a finite variable subring $R_{1}$ of $R$, which we can assume contains $R_{0}$. The ideal $I^{\prime}R_{1}$ has height $c$ and $\mathfrak{q}_{1}R_{1},\ldots,\mathfrak{q}_{r}R_{1}$ are the minimal primes over it. Since $\mathfrak{p}^{\prime}$ contains $I^{\prime}R_{1}$ and has the same height, it follows that $\mathfrak{p}^{\prime}=\mathfrak{q}_{i}R_{1}$ for some $i$, and thus $\mathfrak{p}=\mathfrak{q}_{i}R$. We thus see that there are only $r$ choices for $\mathfrak{p}$. ∎

If $I$ is prime then it is finitely generated (Proposition 2.1) and the result follows from the corresponding result for finite variable polynomial rings and the fact that extending to larger polynomial rings does not change height [ESS, Proposition 3.3]. Now suppose that $I$ is a general ideal of finite height $c$. Let $\mathfrak{p}$ be a height $c$ prime containing $I$. Then $I\cap A\subset\mathfrak{p}\cap A$, and the latter has height $\leq c$. Thus the result follows. ∎

Let $\mathfrak{p}_{0}\subset\cdots\subset\mathfrak{p}_{r}=\mathfrak{p}$ be a strict chain of primes in $S^{-1}R$. Contracting gives a strict chain of primes in $R$. Thus $\operatorname{ht}_{R}(\mathfrak{q})\geq\operatorname{ht}_{S^{-1}R}(\mathfrak{p})$.

Now let $\mathfrak{q}_{0}\subset\cdots\subset\mathfrak{q}_{r}=\mathfrak{q}$ be a strict chain of primes in $R$. Since $\mathfrak{q}$ is the contraction of an ideal from $S^{-1}R$, it is disjoint from $S$, and so all the ideals $\mathfrak{q}_{i}$ are as well. Thus extending gives a strict chain, and so $\operatorname{ht}_{S^{-1}R}(\mathfrak{p})\geq\operatorname{ht}_{R}(\mathfrak{q})$. ∎

Let $c=\operatorname{ht}_{S^{-1}R}(S^{-1}I)$. If $c=\infty$ there is nothing to prove, so suppose $c$ is finite. Let $\mathfrak{p}$ be a height $c$ prime of $S^{-1}R$ containing $S^{-1}I$. Let $\mathfrak{q}$ be the contraction of $\mathfrak{p}$. Then $\operatorname{ht}_{R}(\mathfrak{q})=c$ by the proposition. Since $I\subset\mathfrak{q}$, we thus have $\operatorname{ht}_{R}(I)\leq c$. ∎

Let $f$ be a non-zero element of $I$. As in [ESS, Lemma 4.9], we can find a linear change of variables $\gamma$ in finitely many of the $x$’s so that $\gamma(f)$ is monic in $x_{1}$. We then have $\operatorname{ht}_{\mathcal{R}_{>1}}(\gamma(I)\cap\mathcal{R}_{>1})=\operatorname{ht}_{\mathcal{R}}(I)-1$. Indeed, if $I$ is fintiely generated, this is [ESS, Corollary 3.8] (which easily reduces to the finite variable case), while Propositions 2.5 and 2.6 allow us to reduce to this case. It follows from Proposition 2.7 that $\operatorname{ht}_{\mathcal{R}_{>n}}(\gamma(I)\cap\mathcal{R}_{>n})<\operatorname{ht}_{\mathcal{R}}(I)$ for all $n\geq 1$. Taking $n$ larger than the variables used in $\gamma$, we thus have $\operatorname{ht}_{\mathcal{R}_{>n}}(I\cap\mathcal{R}_{>n})<\operatorname{ht}_{\mathcal{R}}(I)$. Thus, by induction on height, the result follows. ∎

Let $c=\operatorname{ht}_{\mathcal{R}}(I\mathcal{R})$. If $c=\infty$ there is nothing to prove, so suppose $c$ is finite. Let $\mathfrak{p}$ be a height $c$ prime of $\mathcal{R}$ containing $I\mathcal{R}$. Let $n$ be such that $\mathfrak{p}$ is contracted from a prime $\mathfrak{p}^{\prime}$ of $\operatorname{Frac}(\mathcal{R}_{>n})[x_{1},\ldots,x_{n}]$, necessarily of height $c$ (by Proposition 2.8). Let $\mathfrak{q}^{\prime}$ be the contraction of $\mathfrak{p}^{\prime}$ to $\operatorname{Frac}(\mathcal{R}^{\flat}_{>n})[x_{1},\ldots,x_{n}]$. Then $\mathfrak{q}^{\prime}$ has height $\leq c$ by Proposition 2.3. Let $\mathfrak{q}$ be the contraction of $\mathfrak{q}^{\prime}$ to $\mathcal{R}^{\flat}$. Then $\mathfrak{q}$ has the same height as $\mathfrak{q}^{\prime}$ by Proposition 2.8, which is $\leq c$. Since $I$ is contained in $\mathfrak{q}$, it too has height $\leq c$. ∎

Let $\delta(n)$ be the dimension of $\Sigma_{n}(M)/M$. Let $x_{1},x_{2},\ldots$ be elements of $\Sigma(M)$ such that $x_{1},x_{2},\ldots,x_{\delta(n)}$ is a basis of $\Sigma_{n}(M)/M$ for each $n$. Note that $x_{1},x_{2},\ldots$ is a basis for $\Sigma(M)/M$.

Let $y$ be an element of the $t$-adic closure of $M$. We can thus write $y=\sum_{j=0}^{\infty}z_{j}$, where $z_{j}\in M\cap t^{j}\Pi$. Since $t^{-j}z_{j}\in\Sigma_{j}(M)$, we can write $t^{-j}z_{j}=b_{j}+c_{1,j}x_{1}+\cdots+c_{a(j),j}x_{a(j)}$ with $b_{j}\in M$ and $c_{i,j}\in\mathbf{C}$. We thus find

Let $\epsilon(i)$ be the minimal value of $j$ such that $\delta(j)\geq i$, with the convention $\epsilon(i)=\infty$ if $\delta(j)<i$ for all $j$. Thus in the inner sum above, we can write $j\geq\epsilon(i)$. Note that $\epsilon(i)\geq 1$ for all $i$ and $\epsilon(i)\to\infty$ as $i\to\infty$. Now, the first sum above belongs to $M$ by (a). And the inner sum in the second sum converges to an element of $A$ that is dividible by $\epsilon(i)$. We conclude that for any $y$ in the $t$-adic closure of $M$, we can write

where $b\in M$ and $f_{i}(t)\in t^{\epsilon(i)}A$.

We now apply this to $x_{i}$, which belongs to the $t$-adic closure of $M$ by (c). We can thus write

where $b_{i}\in M$ and $f_{i,j}\in t^{\epsilon(j)}A$. Let $\underline{x}$ and $\underline{b}$ be the infinite column vectors $(x_{1},x_{2},\ldots)$ and $(b_{1},b_{2},\ldots)$, and let $A$ be the infinite square matrix given by $A_{i,j}=f_{i,j}$. We then get the linear equation

The entries of $A$ all belong to the maximal ideal of $A$. In fact, for any $n$ there exists an $m$ such that all columns to the right of the $m$th column of $A$ are divisible by $t^{n}$. It follows that $1-A$ is invertible, and so the entries of $\underline{x}=(1-A)^{-1}\underline{b}$ belongs to $M$. Thus $x_{i}\in M$ for all $i$, which completes the proof. ∎

Let $A$ be the set of variables appearing in $f_{1}(0),\ldots,f_{r}(0)$, which is finite. Let $S^{\prime}\subset S$ be the polynomial ring in the $A$ variables, and define $E^{\prime}$ like $E$, but using $S^{\prime}$ instead of $S$. Since $S^{\prime}$ is noetherian, it follows that $E^{\prime}$ is a finitely generated module, and so $\overline{E}^{\prime}=E^{\prime}_{d}/(E^{\prime}_{d}\cap S_{+}^{\prime}E^{\prime})$ is finite dimensional.

We have a natural inclusion $E^{\prime}\to E$. We claim that the induced map $\overline{E}^{\prime}\to\overline{E}$ is surjective, which will complete the proof. Thus suppose $(g_{1},\ldots,g_{r})$ is a degree $d$ element of $E$. Write $g_{i}=\sum_{e}g_{i,e}x^{e}$, where the sum is over all monomials in variables not in $A$, and $g_{i,e}\in S^{\prime}$. Each tuple $(g_{1,e},\ldots,g_{r,e})$ belongs to $E$ and so if $e\neq 0$ then $(x^{e}g_{1,e},\ldots,x^{e}g_{r,e})$ belongs to $S_{+}E$. Hence $(g_{1},\ldots,g_{r})$ and $(g_{1,0},\ldots,g_{r,0})$ are equal in $\overline{E}$. Since the latter belongs to $E^{\prime}$, the result follows. ∎

Since $(g_{1}(0),\ldots,g_{r}(0))$ maps to 0 in $\overline{E}$, we can find elements $(a_{1,i},\ldots,a_{r,i})\in E$ for $1\leq i\leq n$ of degree $<d$ such that

for some $b_{i}\in S$. Now, we have

Since $a_{1,i}f_{1}+\cdots+a_{r,i}f_{r}$ is divisible by $t$, its quotient by $t$ belongs to $\operatorname{Sat}(I)$. It also has degree $<d$, and so it belongs to $I$ by our initial setup. Thus we can write

for elements $c_{i,j}\in R$. Letting $c_{i}=c_{i,1}+\cdots+c_{i,r}$, we thus have

Now, write $g_{i}=g_{i}(0)+tg_{i}^{\prime}$. Then

This completes the proof. ∎

The key lemma gives a surjection $Z_{k}\cong X_{k}/\ker(\rho|_{X_{k}})\to Y_{k}/Y_{k-1}$, and $Z_{k}$ is finite dimensional. ∎

Let $(h_{1},\ldots,h_{r})\in X_{k+m}$ be such that $\rho(g_{1},\ldots,g_{r})=\rho(h_{1},\ldots,h_{r})$. We have

We have

The key lemma shows that the first term belongs to $Y_{k-1}$, while we have just seen that the second belongs to $t^{m}Y_{k+m}$. ∎

The sets $Z_{k}$ form a descending chain in the finite dimensional space $\overline{E}$, and so they stabilize. Let $\ell$ be the point at which they stabilize, so that $Z_{\ell+n}=Z_{\ell}$ for all $n$. We claim that $Y_{k}=Y_{\ell}$ for all $k\geq\ell$, which will establish the result: indeed, $\operatorname{Sat}_{\leq d}(I)$ will then be generated by $I$ and $Y_{k}$, and since $Y_{k}/Y_{0}=Y_{k}/I_{d}$ is finite dimensional, we are only adding finitely many generators to $I$. We prove this by applying Proposition 4.1 with $M=Y_{\ell}$. We check the three axioms:

(a) Suppose $x_{0},x_{1},\ldots\in Y_{\ell}$. We can thus write $x_{i}=t^{-\ell}(g_{1,i}f_{1}+\cdots+g_{r,i}f_{r})$, and so $\sum_{i\geq 0}x_{i}t^{i}=t^{-\ell}(g_{1}f_{1}+\cdots+g_{r}f_{r})$ where $g_{j}=\sum_{i\geq 0}g_{j,i}t^{i}$. This clearly belongs to $Y_{\ell}$.

(b) We have $\Sigma_{k}(Y_{\ell})=Y_{k+\ell}$, and we have already seen that $Y_{k+\ell}/Y_{\ell}$ is finite dimensional for all $k$.

(c) Let $(g_{1},\ldots,g_{r})\in X_{k}$ with $k>\ell$. Then $\rho(g_{1},\ldots,g_{r})\in Z_{k+m}$ for all $m$. Thus, by the previous lemma, we have $\pi_{k}(g_{1},\ldots,g_{r})\in Y_{k-1}+t^{m}Y_{k+m}$. We thus see that $Y_{k}\subset Y_{k-1}+t^{m}Y_{k+m}\subset Y_{k-1}+t^{m}R$. Applying this inductively, we find $Y_{k}\subset Y_{\ell}+t^{m}R$, for any $m\geq 0$. Thus $Y_{k}$ is contained in the $t$-adic closure of $Y_{\ell}$, for all $k\geq\ell$.

Proposition 4.1 now applies, and shows that $Y_{\ell}$ is saturated. Thus $Y_{k}=Y_{\ell}$ for all $k\geq\ell$, which establishes the result. ∎

Suppose (a) holds, and write $f=\sum_{j=1}^{n}g_{j}h_{j}$, where each $g_{j}$ and $h_{j}$ is homogeneous of positive degree. Letting $\partial_{i}=\frac{\partial}{\partial x_{i}}$, we have

Thus (b) holds.

Now suppose (b) holds. Let $g_{1},\ldots,g_{n}$ be positive degree homogeneous elements such that $\partial_{i}f\in(g_{1},\ldots,g_{n})$ for all $i$. Write $\partial_{i}f=\sum_{j=1}^{n}h_{i,j}g_{j}$. Then by Euler’s identity, we have