Let $\Theta$ be a domain. We shall choose $d$ elements $\theta_1,\ldots,\theta_d \in \Theta$ such that any chosen $j$ elements $\theta_{i_1},\ldots,\theta_{i_j}$ form a prime ideal $(\theta_{i_1},\ldots,\theta_{i_j})$ satisfying ${\mathrm{ht}}((\theta_{i_1},\ldots,\theta_{i_j})) = j$. Especially for the ideal $\Theta_d \colon= (\theta_1,\ldots,\theta_d)$ in $\Theta$, we have ${\mathrm{ht}}(\Theta_d) = d$.

Let us consider the power ring $\Lambda \colon= \Theta[[X_1,\ldots,X_d]]$ in $d$ variables over $\Theta$.

Suppose that we have arbitrary $d$-elements $f_1,\ldots,f_d \in (X_1,\ldots,X_d)\Lambda$. Let us set $\lambda_1,\ldots,\lambda_d \in \Theta[[X_1,\ldots,X_d]]$ as $\lambda_1 \colon= \theta_1 + f_1,\ldots,\lambda_d \colon= \theta_d + f_d$.

Q. Does the ideal $\Lambda_d \colon= (\lambda_1,\ldots,\lambda_d)$ form a prime of $\Lambda$ such that ${\mathrm{ht}}(\Lambda_d) = d$?


1 Answer 1


The answer is ``yes'' if $\Theta$ is noetherian.

Indeed, order the monomials $X_1^{\alpha_1}X_2^{\alpha_2}\dots X_d^{\alpha_d}$ by the lexicographical ordering of the $(d+1)$-tuples $\left(|\alpha|,\alpha_1,\alpha_2,\dots,\alpha_d\right)$, where $\alpha=(\alpha_1,\dots,\alpha_d)$ and $|\alpha|=\sum\limits_{i=1}^d\alpha_i$.

For $g\in\Lambda$, let $\theta(g)$ denote the smallest monomial appearing in $g$ (I will call it the leading monomial of $g$). For an ideal $I\subset\Lambda$, I will denote by $\theta(I)$ the ideal generated by the leading monomials of elements of $I$. An element $g\in\Lambda$ does not belong to $I$ if and only if there exists $h\in I$ such that $\theta(g-h)\notin\theta(I)$. In particular, if $\theta(I)$ is prime then so is $I$.

We have $\theta(\lambda_i)=\theta_i$, $i\in\{1,\dots,d\}$.

Fix an $\ell\in\{1,\dots,d\}$. Let $\Lambda_\ell=(\lambda_1,\dots,\lambda_\ell)$. We say that a non-zero $\ell$-tuple $a=(a_1,\dots,a_\ell)\in\Theta^\ell$ is an $\ell$-syzygy of $\theta_1,\dots,\theta_\ell$ if $$ \sum\limits_{i=1}^\ell a_i\theta_i=0.\qquad\qquad(1) $$ I claim that all the syzygies are generated by the ``trivial'' ones of the form $s_{ij}=(0,0,\dots,0,\theta_j,0,\dots,0,-\theta_i,0,\dots,0)$ where $i<j$, $\theta_j$ is the $i$-th entry and $-\theta_i$ is the $j$-th entry, all the other entries being 0. This can be proved by induction on $\ell$. If $\ell=1$, since $\Theta$ is a domain, there are no syzygies at all. Assume that $\ell>1$ and all the $(\ell-1)$-syzygies are generated by the trivial ones. Consider an $\ell$-syzygy (1). The assumptions on the $\theta_i$ imply that they are a regular sequence in $\Theta$. Hence $a_\ell\in\Theta_{\ell-1}$. Write $a_\ell=\sum\limits_{i=1}^{\ell-1}b_i\theta_i$, $b_i\in\Theta$. Then $a-\sum\limits_{i=1}^{\ell-1}b_is_{i\ell}$ is an $(\ell-1)$-syzygy and hence is generated by the $s_{ij}$ by the induction assumption. This proves the claim.

The Claim implies that for every element $\sum\limits_{i=1}^\ell b_i\lambda_i\in\Lambda_\ell$ its leading monomial belongs to the ideal generated by $(\theta_1,\dots,\theta_\ell)$. Thus $\theta(\Lambda_\ell)=(\theta_1,\dots,\theta_\ell)$. Since $\theta(\Lambda_\ell)$ is a prime ideal, so is $\Lambda_\ell$ itself. Since for different values of $\ell$ the corresponding ideals $\Lambda_\ell$ are distinct, we have $height(\Lambda_\ell)\ge d$. By the noetherian hypothesis and since $\Lambda_d$ is generated by $d$ elements, this inequality is, in fact, an equality.

Note: only the last sentence of the proof uses the noetherian hypothesis. Thus the statements that $\Lambda_d$ is prime and that its height is at least $d$ is unconditional.


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