# Power series ring $\Theta[[X_1,\ldots,X_d]]$ and prime ideals

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

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 $$i, $$\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.