The local ring at a point satisfies $\dim \mco_P = \dim X$

If $P$ is a point on a variety $X$, show that $\dim \mco_P = \dim X$.

Hint

Reduce to the affine case and use (3.2c).

Solution

Choose an affine open neighborhood $U\subseteq X$ of $P$.

1. The local rings of $X$ and $U$ at $P$ are canonically isomorphic:\[ \mco_{P,X}\cong\mco_{P,U}. \]

A germ of a regular function at $P$ is represented by a regular function on some open neighborhood of $P$. Intersecting that neighborhood with $U$ does not change the germ, and every open neighborhood of $P$ inside $U$ is also open in $X$ because $U$ is open. Thus the two germ constructions have the same representatives and the same equivalence relation.

2. The affine open subset $U$ has the same dimension as $X$.

The variety $X$ is irreducible, and $U$ is a nonempty open subset. The function fields are therefore the same: \[ K(U)=K(X). \] For a variety over $k$, dimension equals the transcendence degree of its function field [@Har10a, Chapter I, §1]. Hence \[ \dim U=\operatorname{trdeg}_k K(U) =\operatorname{trdeg}_k K(X)=\dim X. \]

3. The local ring at $P$ has dimension $\dim X$.

Since $U$ is affine, Hartshorne’s Theorem I.3.2(c) gives \[ \dim\mco_{P,U}=\dim U. \] Combining steps <1>1–<1>2 yields \[ \dim\mco_{P,X}=\dim\mco_{P,U}=\dim U=\boxed{\dim X}. \]

4. Q.E.D.

Step 3 is the required equality.

Source Collections

Hartshorne, Algebraic Geometry, I.3.12

Backlinks

P-AGH320EXTENDREGExtending a regular function across a normal point in dimension at least twouses