# Dominant Morphism on Affine Varieties

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 $$X\rightarrow \mathbb{A}_{k}^{n}$$?

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

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: $$mm+ht(I(X)) \Rightarrow n>m$$