3

Is it true that $k(X)$ and $k(X,Y)$ are not isomorphic as fields when $k=\Bbb R$ or $k=\Bbb C$?

My thoughts :

— This is true for $k=\Bbb Q$ since a field isomorphism $\Bbb Q(X) \to \Bbb Q(X,Y)$ is a $\Bbb Q$-algebra isomorphism, in this case, so that it should preserve the transcendence degree over $\Bbb Q$, but $2 \neq 1$.

The transcendence degree of $\Bbb C$ (or $\Bbb R$) over $\Bbb Q$ is $2^{\aleph_0}$, so such an argument fails. We only know that $\Bbb C(X)$ and $\Bbb C(X,Y)$ are not isomorphic as $\Bbb C$-algebras (nor as $\Bbb R$-algebras).

— I know that there is no field morphism $\Bbb R(X) \to \Bbb R$ (see here). Assume there is an isomorphism $f:\Bbb R(X,Y) \to \Bbb R(X)$. Either $X$ (or $Y$) is mapped to some real number, so that $f\vert_{\Bbb R(X)}$ should have a range included in $\Bbb R$, which is impossible. Or $X$ and $Y$ are mapped respectively to non-constant rational fractions $P(X)$ and $Q(X)$, and maybe this would contradict the injectivity of $f$, but I wasn't sure how to find a suitable relation between $P$ and $Q$ (this might not work in general, I believe).

— Notice that there is a field morphism $\Bbb C(x) \to \Bbb C$. Moreover, the field $F=\Bbb R(X_1,X_2,\dots)$ having cardinality $2^{\aleph_0}$, its algebraic closure is isomorphic to $\Bbb C$, so we see that there is a subfield $F' \cong F$ of $\Bbb C$ such that $F(X) \cong F(X,Y) \;(\cong F)$ (see also here).

I thought to a geometric argument (because $\Bbb C$ is algebraically closed… otherwise I'm not sure it works) : $\Bbb C(X)$ is the function field of the affine variety $Y=0$ in $\Bbb A^2(\Bbb C)$, while $\Bbb C(X,Y)$ is the function field of the affine variety $\Bbb A^2(\Bbb C)$, if I'm not mistaken. But $\Bbb C(X) \cong \Bbb C(X,Y)$ would yield an isomorphic of affine varieties between $\Bbb A^1(\Bbb C)$ and $\Bbb A^2(\Bbb C)$. In particular, this would be an homeomorphism between the spaces $\Bbb C$ and $\Bbb C^2$ w.r.t. to usual topologies (since polynomials are continuous maps), but this is not possible. Is it right, and is there an easier way to show it?

From this question, I think we can conclude that $\Bbb C(X)$ and $\Bbb C(X,Y)$ are not isomorphic as fields, but it seems too complicated.

– Moreover, I think that we have in general $k(X_1,\dots,X_n) \cong k(Y_1,\dots,Y_m) \implies n=m$ when $k=\Bbb R$ or $\Bbb C$.

Any comment would be appreciated.

Community
  • 1
Watson
  • 23,793
  • 1
    Your claim about affine varieties is not true, for two reasons. First, an isomorphism of function fields only implies that two varieties are birational, not isomorphic. Second, for this to be true the isomorphism of function fields needs to be an isomorphism over the ground field (here $\mathbb{C}$). – Qiaochu Yuan Nov 22 '16 at 08:10
  • @QiaochuYuan : ah yes… I should have worked with projective varieties, maybe. When you say "an isomorphism over the ground field", do you mean a $\Bbb C$-algebra isomorphism? In this case, such an argument is useless. – Watson Nov 22 '16 at 08:12
  • 1
    Yes, that's what I mean. And whether you work with affine or projective varieties does not affect the point about birationality; it's very special to the case of smooth projective curves that they're birational iff they're isomorphic. – Qiaochu Yuan Nov 22 '16 at 08:13
  • Related: http://mathoverflow.net/questions/96777 – Watson Feb 02 '17 at 17:35
  • This answer provides an answer to my question. The claim is: if for some $m>1$, the $m$-th power map on $K$ is onto (e.g. $K=\Bbb C,K=\Bbb R$), then $$\exists n \geq 1,; K(x_1,\dots,x_n) \cong L(x_1,\dots,x_n) \implies K \cong L$$ We can apply this to $L=K(X)$, and we then use $\Bbb C(X) \not\cong \Bbb C$ (the first is not algebraically closed, the second one is) and $\Bbb R(X) \not\cong \Bbb R$, see here. – Watson Feb 02 '17 at 20:47

1 Answers1

2

Yes. It suffices to find the copy of $k$ inside these fields in each case, since then we can distinguish based on transcendence degree. When $k = \mathbb{C}$ we can hope that $k$ is the largest algebraically closed subfield, and similarly when $k = \mathbb{R}$ we can hope that $k$ is the largest real closed subfield.

Both of these are true and this is straightforward to see. When $k = \mathbb{C}$, let $L$ be a subfield of either $k(X)$ or $k(X, Y)$. If $L$ contains a nonconstant rational function, then this rational function cannot have $n^{th}$ roots in $k(X)$ or $k(X, Y)$ (and hence not in $L$) for sufficiently large $n$, just by considering degrees. Hence $L$ cannot be algebraically closed. Similarly, when $k = \mathbb{R}$ you can run the same argument only considering odd $n$ (in a real closed field, every polynomial of odd degree has at least one root).

The same argument applies to $k(X_1, \dots X_n)$ and so we can distinguish these by transcendence degree as hoped.

Qiaochu Yuan
  • 419,620
  • Thank you for your answer! 1) I don't understand the argument for $k=\Bbb R$. Your claim is that a subfield $L \subset \Bbb R(X)$ that is real closed can't contain any nonconstant rational function, is it right? But how is it related to odd degree polynomials or $n$-th roots? I'm a bit confused. – Watson Nov 22 '16 at 09:54
  • We have proved that any isomorphism $f : k(X) \to k(X,Y)$ satisfies $f(k) \subset k$. Does this imply that $f$ is a $k$-algebra homomorphism? Then we would be done, but I'm not sure how to conclude.
    • – Watson Nov 22 '16 at 09:56
  • (Anyway, your argument for $k=\Bbb C$ is very nice, it can be generalized to any algebraically closed field, if I understood well). – Watson Nov 22 '16 at 09:57
  • 1
    @Watson: 1) A field $L$ has $n^{th}$ roots iff all polynomials of the form $x^n - a, a \in L$ have roots in $L$, and real closed fields have $n^{th}$ roots for $n$ odd. 2) it doesn't, but that isn't necessary. The point is that there is an invariant we can write down called "the transcendence degree of this field over its largest algebraically closed subfield" and this is already enough to distinguish $\mathbb{C}(X_1, \dots X_n)$ for different values of $n$. – Qiaochu Yuan Nov 22 '16 at 20:56
  • OK, thank you. 2) I understand that this invariant deals with the case $k=\Bbb C$, but for the case $k=\Bbb R$, we would have an invariant called "the transcendence degree of this field over its largest real closed subfield", is that right (we just have to show that this is indeed an invariant)?
    • – Watson Nov 22 '16 at 21:02
  • 1
    @Watson: yes, that's right. The only slight subtlety is that there may not be a largest algebraically closed or real closed subfield (I haven't thought very hard about this), but if there is one it's unique ("largest" here means "containing every other algebraically closed resp. real closed subfield"). – Qiaochu Yuan Nov 22 '16 at 21:05
  • OK, thank you very much! I will think about it a little bit and will accept your answer later (after spending some time on it). – Watson Nov 22 '16 at 21:08