1
$\begingroup$

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 ${\mathrm{ht}}(\Lambda_d) = d$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.