9

I am looking for an example of an irreducible variety $X$ say over a field $K$ such that the base change $X_\overline K$ to an algebraic closure is no longer irreducible, and has irreducible components of many different dimensions.

For example the curve $x^2+y^2=0$ is irreducible 1 dimensional (in the sense of Krull dimension of the co-ordinate ring) over $\mathbb R$ but splits into two lines $(x+iy)(x-iy)=0$ over $\mathbb C$. But in this case, the components are of all of same dimension 1.

In general, how wild can $X$ be after base change? Is it wilder if we allow $X$ to be an integral scheme? Is there a theory to control it? e.g. if we impose $X$ is affine, projective (scheme), characteristic $0$ etc?

EDIT: The question was resolved in the case of varieties, and I decided to drop the case for schemes, because base change locally corresponds to tensoring with an arbitrary ring, whence it is no longer reasonable to continue without adding many more qualifiers.

Victor Zhang
  • 695

1 Answers1

12

$\newcommand{\pp}{\mathfrak{p}}\newcommand{\mm}{\mathfrak{m}}\newcommand{\qq}{\mathfrak{q}}\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\inv}{^{-1}}\newcommand{\Gal}{\operatorname{Gal}}$ Partial answer: For $X$ integral of finite-type over $k$, all components of $X_{\bar{k}}$ have the same dimension as $X$.

First let's consider the case where $X$ is affine and integral of finite-type over $k$ (i.e. an affine variety). $X=\Spec k[x_1,\ldots,x_n]/\pp$. Now clearly both $k[x_1,\ldots,x_n]$ and $\bar{k}[x_1,\ldots,x_n]$ are both $n$-dimensional. Moreover, since $\bar{k}$ is algebraic over $k$, $\bar{k}[x_1,\ldots,x_n]$ is integral over $k[x_1,\ldots,x_n]$. In fact, being polynomial rings, both are normal domains. Hence we can apply the going up and going down theorems. Take a maximal chain of prime ideals, one of whose terms is $\pp$. It's a fact about polynomial rings that such a thing exists, though I forget where I read it (probably in Atiyah and MacDonald). So we have the chain: $$ \pp_0=0 \subset \pp_1\subset \pp_2\subset \cdots \subset \pp=\pp_k \subset \cdots \pp_n $$ Let $\qq$ be any prime lying over $\pp$. Then by going up and going down, we can find $\qq_0,\ldots,\qq_k=\qq,\qq_n$ such that $\qq_i$ lies over $\pp_i$ for all $i$ and such that they form a chain $$ \qq_0 \subset \qq_1\subset \qq_2\subset \cdots \subset \qq=\qq_k \subset \cdots \qq_n. $$ Since $\bar{k}[x_1,\ldots,x_n]$ has dimension $n$, this chain is maximal. Thus both the varieties $V(\pp)$ and $V(\qq)$ have dimensions $n-k$. Since the primes lying over $\pp$ correspond precisely to the components of $X_{\bar{k}}$, we have that the components of $X_{\bar{k}}$ all have the same dimension.

Since every open subset of an irreducible scheme contains the generic point of the scheme, we can choose an affine open subset $U$, and we'll have that $U_{\bar{k}}=\pi\inv(U)$ if $\pi : X_{\bar{k}}\to X$ is the natural morphism. Thus $U_{\bar{k}}$ contains the generic points of all the irreducible components of $X_{\bar{k}}$, since they all map to the generic point of $X$. Then the dimension of the corresponding component is the same in $U_{\bar{k}}$ as it is in $X_{\bar{k}}$. Hence by the affine case, when $X$ is integral of finite-type over $k$, all the components of $X_{\bar{k}}$ have the same dimension as $X$.

jgon
  • 29,394
  • Thanks! You state this is only partial, but I am happy enough to accept this sometime soon if no one else puts another answer, because my original motivation came from varieties. – Victor Zhang Dec 17 '17 at 07:39
  • Np, I'm also curious about the general case, so I hope someone does add something. – jgon Dec 17 '17 at 07:40
  • nvm, on second thought, the base change of even an affine integral scheme means tensoring with an arbitrary ring. At this point, it seems like it's no longer reasonable to expect someone to answer this without giving a lot more qualifications. – Victor Zhang Dec 17 '17 at 08:04