# Can a subquotient field of a field have a higher transcendence degree?

Let $F$ be an algebraically closed field. Let $$K=F(x_1,x_2,...,x_n)$$ be an extension field of transcendence degree $n$.

Is it possible to find a sub-$F$-algebra $R\subset K$, together with a maximal ideal $m$, such that the quotient field $R/m$ has transcendence degree strictly greater than $n$?

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If $\bar{y}_1, \ldots , \bar{y}_r \in R/m$ are algebraically independent over $F$, then lifting arbitrarily, $y_1, \ldots , y_r \in R \subset K$ are also algebraically independent over $F$, and this implies that $r \leq n$.
This is the obvious thing to try, but it seemed to me that we could very well have an algebraic relation between the $y_1,\ldots,y_r$ that gets killed completely when modding out by $m$, and could leave the $\bar{y}_1,\ldots,\bar{y}_r$ with no (nontrivial) algebraic relations between them. Is it clear that this is impossible? –  Zev Chonoles May 9 '11 at 18:55
This does not happen since the coefficients of such a polynomial are in $F$, and the morphism $F \rightarrow R/m$ is injective since $F$ is a field. –  Plop May 9 '11 at 19:10