## 111.12 Depth

Depth is defined in Algebra, Section 10.72 and further studied in Dualizing Complexes, Section 47.11.

Exercise 111.12.1. Let $R$ be a ring, $I \subset R$ an ideal, and $M$ an $R$-module. Compute $\text{depth}_ I(M)$ in the following cases.

$R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}$,

$R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}/(300)$,

$R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}/(7)$,

$R = k[x, y, z]/(x^2 + y^2 + z^2)$, $I = (x, y, z)$, $M = R$,

$R = k[x, y, z, w]/(xz, xw, yz, yw)$, $I = (x, y, z, w)$, $M = R$.

Here $k$ is a field. In the last two cases feel free to localize at the maximal ideal $I$.

Exercise 111.12.2. Give an example of a Noetherian local ring $(R, \mathfrak m, \kappa )$ of depth $\geq 1$ and a prime ideal $\mathfrak p$ such that

$\text{depth}_\mathfrak m(R) \geq 1$,

$\text{depth}_\mathfrak p(R_\mathfrak p) = 0$, and

$\dim (R_\mathfrak p) \geq 1$.

If we don't ask for (3) then the exercise is too easy. Why?

Exercise 111.12.3. Let $(R, \mathfrak m)$ be a local Noetherian domain. Let $M$ be a finite $R$-module.

If $M$ is torsion free, show that $M$ has depth at least $1$ over $R$.

Give an example with depth equal to $1$.

Exercise 111.12.4. For every $m \geq n \geq 0$ give an example of a Noetherian local ring $R$ with $\dim (R) = m$ and $\text{depth}(R) = n$.

Exercise 111.12.5. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Show that there exists a canonical short exact sequence

\[ 0 \to K \to M \to Q \to 0 \]

such that the following are true

$\text{depth}(Q) \geq 1$,

$K$ is zero or $\text{Supp}(K) = \{ \mathfrak m\} $, and

$\text{length}_ R(K) < \infty $.

Hint: using the Noetherian property show that there exists a maximal submodule $K$ as in (2) and then show that $Q = M/K$ satisfies (1) and $K$ satisfies (3).

Exercise 111.12.6. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $R$-module of depth $\geq 2$. Let $N \subset M$ be a nonzero submodule.

Show that $\text{depth}(N) \geq 1$.

Show that $\text{depth}(N) = 1$ if and only if the quotient module $M/N$ has $\text{depth}(M/N) = 0$.

Show there exists a submodule $N' \subset M$ with $N \subset N'$ of finite colength, i.e., $\text{length}_ R(N'/N) < \infty $, such that $N'$ has depth $\geq 2$. Hint: Apply Exercise 111.12.5 to $M/N$ and choose $N'$ to be the inverse image of $K$.

Exercise 111.12.7. Let $(R, \mathfrak m)$ be a Noetherian local ring. Assume that $R$ is reduced, i.e., $R$ has no nonzero nilpotent elements. Assume moreover that $R$ has two distinct minimal primes $\mathfrak p$ and $\mathfrak q$.

Show that the sequence of $R$-modules

\[ 0 \to R \to R/\mathfrak p \oplus R/\mathfrak q \to R/\mathfrak p + \mathfrak q \to 0 \]

is exact (check at all the spots). The maps are $x \mapsto (x \bmod \mathfrak p, x \bmod \mathfrak q)$ and $(y \bmod \mathfrak p, z \bmod \mathfrak q) \mapsto (y - z \bmod \mathfrak p + \mathfrak q)$.

Show that if $\text{depth}(R) \geq 2$, then $\dim (R/\mathfrak p + \mathfrak q) \geq 1$.

Show that if $\text{depth}(R) \geq 2$, then $U = \mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\} $ is a connected topological space.

This proves a very special case of Hartshorne's connectedness theorem which says that the punctured spectrum $U$ of a local Noetherian ring of $\text{depth} \geq 2$ is connected.

Exercise 111.12.8. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $x, y \in \mathfrak m$ be a regular sequence of length $2$. For any $n \geq 2$ show that there do not exist $a, b \in R$ with

\[ x^{n - 1}y^{n - 1} = a x^ n + b y^ n \]

Suggestion: First try for $n = 2$ to see how to argue. Remark: There is a vast generalization of this result called the monomial conjecture.

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