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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 $r \leq n$.

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    Ah, of course. You are correct. +12011-05-09