I have a question regarding the following exercise:

Let $(A,\mathfrak{m})$ be a Noetherian local ring of dimension $d$. Let $k:=A/\mathfrak{m}$ be its residue field. $f_1, \ldots , f_r \in \mathfrak{m}$. Set $\bar{A}:=A/(f_1, \ldots, f_r)$ and $\bar{\mathfrak{m}} \subset \bar{A}$ denote the image of $\mathfrak{m}$.

Assume $A$ is regular. let $\bar{f_1}, \ldots, \bar{f_r} \in \mathfrak{m}/ \mathfrak{m}^2$ denote the images of $f_1, \ldots, f_r$. Show that:

$\bar{A}$ is regular of dimension $d-r \Leftrightarrow \bar{f_1}, \ldots , \bar{f_r}$ are linearly independent over $k$.

My thoughts:

1.) $\dim_k(\bar{m}/\bar{m}^2) \geq \dim(\bar{A}) \geq d-r$

Because: Take $s = \dim(\bar{A})$. Then we choose $y_1, \ldots, y_s \in A$ whose images $\bar{y_1}, \ldots, \bar{y_s} \in \bar{A}$ form a system of parameters for $\bar{\mathfrak{m}}$. $\bar{\mathfrak{m}}$ is the only prime containing $(\bar{y_1}, \ldots, \bar{y_s}) \Rightarrow \bar{\mathfrak{m}}$ is the only prime containing $(f_1, \ldots , f_r, y_1, \ldots , y_s) \Rightarrow \dim(A) \geq r+s \Rightarrow$ Inequality at the right side.

For the inequality on the left side: Set $r:= \dim_k(\bar{\mathfrak{m}}/ \bar{\mathfrak{m}}^2)$. Suppose $\bar{f_1}, \ldots, \bar{f_r} \in \mathfrak{\bar{m}}$ have the property that their images give a $k$-basis of $(\mathfrak{m}/ \mathfrak{m}^2)$. By Nakayama $\Rightarrow \bar{f_1}, \ldots, \bar{f_r}$ generate $\bar{\mathfrak{m}}$. By Krull's dimension theorem $\Rightarrow d=\operatorname{height}(\bar{\mathfrak{m}})\leq r \Rightarrow$ Inequality on left hand side.

2.) We know that if $A$ is a local ring with $f \in \mathfrak{m}$ and f is a non-zerodivisor, then $\dim(A/(f)) = \dim(A)-1$.

So, if $\bar{f_1}, \ldots, \bar{f_r} \in \bar{\mathfrak{m}}$ are linearly independent, then they are all non-zerodivisors, so $\bar{A}$ is regular of dimension $d-r$, by implementing 2.) iteratively $r$ times. Isn't the reverse then also obvious?

Now, my question is, if 2.) is really correct this way, since I'm not sure, if linear independency of $r$ elements implies the regularity and dimension formula given above.

If not, I need to somehow show in 1.) that $\dim(\bar{A}) \leq d-r$ that I can get the equality.

I would be glad, if someone could help me solve this problem. Thank you in advance!


Your Answer

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

Browse other questions tagged or ask your own question.