Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$?

share|cite|improve this question
up vote 5 down vote accepted

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 $r \leq n$.

share|cite|improve this answer
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
Ah, of course. You are correct. +1 – Zev Chonoles May 9 '11 at 19:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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