A question about the depth of a ring with respect to some ideal

Question

So here is my question:

I want to compute the depth of $k[x,y]$ with respect to the ideal $(x,y^2)$ where $k$ is a field.

The depth $t_{(x,y^2)}(k[x,y])$ is defined as follows, $$ t_{(x,y^2)}(k[x,y]):=\sup\{r\in\mathbb N:\exists(x_i)_{i=1}^r \text{ a regular sequence in }(x,y^2)\}.$$

Regularity in this context means that $\forall x_i$ one has that $x_i$ is a non zero divisor in $k[x,y]/(\sum_{j=1}^{i-1}(x_j))$.

I know that it holds that $t(k[x,y])\leq \dim(k[x,y])$ where $\dim$ denotes the Krull dimension. But in my case I have to find the depth with respect to an ideal, so i don't know how I can use the upper fact. It would be helpfull if I knew how long the maximal depth could be such that when i start to construct regular sequences I know when to stop. I am sure there must be Theorem which I am missing.

Can someone help me? Thanks.

Answer 1

Let $I=(x,y^2)$. Because $k[x,y]$ is a Cohen-Macaulay ring, we have that $\operatorname{grade} I = \operatorname{height} I$ (this is corollary 2.1.4 in Bruns and Herzog, Cohen-Macaulay Rings). In case you are not aware of the terminology, the grade of $I$ is precisely equal to the length of a maximal regular sequence in $I$. Additionally we have $\operatorname{height} I + \dim k[x,y] / I = \dim k[x,y]$.

The above give you $\operatorname{grade} I = \dim k[x,y] - \dim k[x,y] / I = 2 - \dim k[y]/y^2 = 2 - 0 =2$.

Answer 2

"It would be helpfull if I knew how long the maximal depth could be such that when i start to construct regular sequences I know when to stop. I am sure there must be Theorem which I am missing."

Perhaps the following result is what you are looking for:

Claim: Let $k$ be an algebraically closed field. Let $I$ be any proper ideal of $k[x,y]$. Then $t_{I}(k[x,y]) \leq 2$.

Proof of claim: There exists a maximal ideal of $k[x,y]$, call it $M$, that contains $I$, $I \subseteq M$.

It is obvious from the definition of depth that $t_{I}(k[x,y]) \leq t_{M}(k[x,y])$.

By Hilbert's Nullstellensatz, $M$ is of the form $(x-\lambda, y-\mu)$, for some $\lambda,\mu \in k$; clearly, $\{x-\lambda, y-\mu \}$ is a regular sequence in $M$, so $t_{M}(k[x,y])=2$.

Concluding that $t_{I}(k[x,y]) \leq t_{M}(k[x,y])=2$.

The reason for taking $k$ algebraically closed is to be able to apply Hilbert's Nullstellensatz. However, it seems that the depth of maximal ideals in $k[x,y]$ is two over any field $k$.