4
$\begingroup$

Let $X,Y\in \mathbb{A}^{n}_{k}$ affine varieties, I know that a morphism $f:X\rightarrow Y$ is dominant iff the correspondent morphism $\phi:k[Y]\rightarrow k[Y]$ is injective. How can I show from here that $\dim X$ is the largest number $n$ such that exists a dominant morphism $X\rightarrow \mathbb{A}_{k}^{n}$?

$\endgroup$
1
  • 3
    $\begingroup$ Do you know the relationship between Krull dimension and transcendence degree? $\endgroup$ – Mariano Suárez-Álvarez May 29 '16 at 3:09
0
$\begingroup$

Let be $X\subseteq \mathbb{A}^n$ an affine variety. Let be $C=\{r\in\mathbb{N}:\exists\hspace{0.1cm} X\longrightarrow \mathbb{A}^r \hspace{0.1cm}dominant \hspace{0.1cm}morphism\}$.

We are going to show dim(X)=sup(C).

  1. We are goint to see $dim(X)\geq r$ $\forall$ $r\in C$:

Let be $r\in C$, then $\exists$ $X \longrightarrow \mathbb{A}^r$ dominant morphism, then $\exists$ $\mathbb{K}(X_1,...,X_r) \longrightarrow K(X)$ an inyective homomorphism of fields, or equivalently, $K(X)/\mathbb{K}(X_1,...,X_r)$ is an extension of fields.

So we have a tower of fields $\mathbb{K}\subset\mathbb{K}(X_1,...,X_r)\subset K(X)$, then:

$$trdeg(K(X)/\mathbb{K}) = trdeg(K(X)/\mathbb{K}(X_1,...,X_r)) + trdeg(\mathbb{K}(X_1,...,X_r)/\mathbb{K})=trdeg(K(X)/\mathbb{K}(X_1,...,X_r)) + r$$ Now we compute dim(X):

$$dim(X) = dim(A(X)) = trdeg(q.f(A(X))/\mathbb{K})=$$ $$=trdeg(K(X)/\mathbb{K})=trdeg(K(X)/\mathbb{K}(X_1,...,X_r)) + r\Rightarrow dim(X)\geq r$$

  1. We are going to see $dim(X)\leq sup(C):=m$.

We suppose $dim(X)>m$ and find a contradiction:

We know the identity map $X\longrightarrow \mathbb{A}^n$ is a dominant morphism, so n>m would be a contradiction.

So it is sufficient to show n>m:

$X$ is a variety, then X is irreducible, then I(X) is a prime ideal, then

$$dim(\mathbb{K}[X_1,...,X_n]/I(X)) = dim(\mathbb{K}[X_1,...,X_n]) -ht(I(X)) = n-ht(I(X))$$

So: $$ m<dim(X) = dim(A(X)) = n-ht(I(X)) \Rightarrow n>m+ht(I(X)) \Rightarrow n>m$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.