6

I'm currently reading Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, and he defines a system of parameters in Section 10.1.

He says that, geometrically, if $x_1,\ldots,x_d$ form a system of parameters for the local ring of a point $p$ on an algebraic variety, then the values of the functions $x_i$ determine points near $p$ up to a "finite ambiguity".

The example given is that $y$ and $y-x^2$ form a system of parameters for $k[x,y]_{(x,y)}$, and he points out that only finitely many points lie in the intersections of $y-x^2=\delta$ and $y=\varepsilon$ for small $\delta$ and $\varepsilon$.

Is the "finite ambiguity" in this sense that each point near $(0,0)$ is determined by its values on these functions up to the sign of the $x$-coordinate? For example the point $(1,0)$ isn't quite defined uniquely by these values, but there are only finitely many other points which share the same values, namely just $(-1,0)$.

If that is the case, then I'm wondering if this characterisation is necessary and sufficient, but don't really know how to begin proving it. I've taken the notion of points "near" each other to mean they lie in some Zariski open neighbourhood.

Let $V$ be an affine variety, and let $R$ be the local ring of a point $p$ on $V$. Say that $\dim R=d$, and let $x_1,\ldots,x_d$ be functions on our variety. Suppose that there exists some Zariski open neighbourhood $U\subseteq V$ with $p\in U$ such that, for every point $q\in U$, we have that the set $$\{r\in U:x_i(r)=x_i(q)\text{ for }1\leq i\leq d\}$$ is finite. Then is this criteria necessary and sufficient for $x_1,\ldots,x_d$ to be a system of parameters for $R$?

Edit:

It seems like there are trivial counterexamples to my criteria as written. For example, if we localise at the point $(1,1)$ instead of $(0,0)$ as in his example, then clearly no power of $(x-1,y-1)$ lies in $(y,y-x^2)$. Then $y,y-x^2$ cannot be a system of parameters for this new ring, despite satisfying my conditions.

However if we specify additionally that $p$ lies in the subvariety $V(x_1,\ldots,x_d)$, then this rules out such counterexamples. This is also consistent with his example.

Update:

I've added a potential answer based on another reading of his example. If anybody has any thoughts, or thinks that the argument is faulty, then please let me know.

Dave
  • 1,413

1 Answers1

1

I now think that what he means when he says that the intersection is finite near $p$ is that on some Zariski open neighbourhood we have that $V(\mathfrak{q})$ is finite, where $\mathfrak{q}=(x_1,\ldots,x_d)$. If this is the case, then I believe the following gives an answer:

Since any affine open set is a union of open sets of the form $D(f)$, we may assume that $U$ is of this form since $p$ must belong to one of them. Furthermore, the intersection of an affine variety with $D(f)$ is also an affine variety, and so we may assume that our property holds for every point on our variety $V$.

Since $x_i(p)=0$ for all $i$, there can be only finitely many other points in $V(\mathfrak{q})$, and so $V(\mathfrak{q})$ is a finite subvariety of $V$.

He says in Section 10.1 that $x_1,\ldots,x_d$ being a system of parameters is equivalent to $R/\mathfrak{q}$ being of finite length. Since $V(\mathfrak{q})$ is finite, it has only finitely many subvarieties, and so $R/\mathfrak{q}$ must be of finite length. Then the property is sufficient.

Conversely, if there doesn't exist a neighbourhood containing $p$ where $V(\mathfrak{q})$ is finite, then we can always construct a chain of subvarieties of $V(\mathfrak{q})$ of arbitrary length, and so $R/\mathfrak{q}$ is not of finite length. Then the property is necessary.

Dave
  • 1,413