1

This is exercise $6.34$ from Lang's book:

Give an example of a field $K$ which is of degree $2$ over two distinct subfields $E$ and $F$, respectively, but such that $K$ is not algebraic over $E\cap F$.

So, I've nearly proven the existence of such a field, but there is one part I am struggling with, and if anyone could provide an actual example, I'd greatly appreciate it! My proof goes as follows:

Consider the diagram consisting of $k(s,t)$ at the top, $k(t+t^{-1},s)$ on the lower middle left, and $F$ on the lower middle right, where $F$ is the fixed field of the map $\sigma$ with $\sigma(s)=st$ and $\sigma(t)=t^{-1}$. It's easy to see that $x^2-(t+t^{-1})x+1$ is an irreducible polynomial over the left field, hence it is of degree $2$. Now, I've made the claim that $s\notin F$ since $\sigma^n(s)=st^n$ is an infinite chain, but I'm not sure why this works, nor am I sure why $F$ is of degree $2$. After these two steps, then it's clear that the intersection can't contain $s$, hence $k(s,t)$ can't be algebraic over $k(t+t^{-1},s)\cap F$. Can anyone help me fill in these two gaps?

Clayton
  • 25,087
  • 7
  • 60
  • 115
  • 3
    What Lang's book, exactly? He wrote more than 50... – Daniel Robert-Nicoud Aug 18 '13 at 20:25
  • I think that an example like this is studied in this question. I am not going to call it a duplicate, as you are more interested in filling details in you own example. A general idea could be to find two automorphisms of $K$, both of order two, but such that their product is of infinite order. As you see a univariate rational function field works at least if it has characteristic zero. – Jyrki Lahtonen Aug 18 '13 at 20:25
  • 1
    @DanielRobert-Nicoud: Algebra, sorry, I figured that much would be able to be discerned from the nature of the question. – Clayton Aug 18 '13 at 20:26
  • Thanks for the example @JyrkiLahtonen! I hadn't noticed that page :) – Clayton Aug 18 '13 at 20:29
  • 1
    I recommend that anyone visiting this question also checks out Michael Zieve's new answer to the question I linked to above. IMHO interesting stuff about generalizations of this problem. Getting to research level though. – Jyrki Lahtonen Aug 19 '13 at 17:35

2 Answers2

5

A simpler example: $K=\mathbf{Q}(x)$ has order-$2$ automorphisms $\sigma:x\mapsto -x$ and $\tau:x\mapsto 1-x$ for which $\sigma\tau$ has infinite order (since it maps $x\mapsto x+1$). I leave it to you to determine $E$ and $F$.

Michael Zieve
  • 1,507
  • This is, indeed, simpler: 1) only needs a field of univariate rational functions, 2) it is "geometrically" obvious that the generated group of automorphisms is the infinite dihedral group. It is also the example studied in the linked question. – Jyrki Lahtonen Aug 19 '13 at 03:47
  • @Jyrki: thanks, I hadn't noticed the linked question. Now that you mentioned it, I posted another answer to that other question, which also includes remarks on generalizations of these questions. – Michael Zieve Aug 19 '13 at 16:09
  • Just to make sure if I understand you correctly the fixed field of $\sigma$ is the set of all even rational functions of $x$ and the other fixed field is those rational functions which satisfy $f(x) =f(1-x)$. Their intersection is the periodic functions with period $1$ which in this case must be constant so that the intersection is just $\mathbb{Q} $. – Paramanand Singh Apr 20 '17 at 04:29
  • @Paramanand: yes, that's correct. – Michael Zieve Apr 20 '17 at 07:13
3

I would fill in the details along the following lines. Essentially you have introduced two automorphisms of the field $K=k(s,t)$. The first automorphism $\tau$ is defined by the formulas $\tau(t)=t^{-1}$, $\tau(s)=s$. The other automorphism $\sigma$ is similarly defined by setting $\sigma(s)=st$, $\sigma(t)=t^{-1}$.

It is easy to see that both $\tau$ and $\sigma$ are of order two. Therefore Galois theory tells us their respective fixed fields $E$ and $F$ have the property $$[K:E]=\operatorname{ord}(\tau)=2=\operatorname{ord}(\sigma)=[K:F].$$ As you have observed, $E=k(s,t+t^{-1})$. Unless I made a mistake, we also get that $F=k(s+st,t+t^{-1})$, as $t$ is algebraic of degree two over $k(s+st,t+t^{-1})$ and $K=k(s+st,t+t^{-1})(t)$.

Anyway, the remaining task is to show that $[K:E\cap F]=\infty.$ Anything in $E\cap F$ is fixed under both $\sigma$ and $\tau$. Conversely if an element of $K$ is fixed by the group of automorphisms $G=\langle\sigma,\tau\rangle$, then it is in the intersection $E\cap F$. If the extension $[K:E\cap F]$ were finite, then by basic Galois theory, the group $G$ would have to be finite. But this is not the case, as the automorphism $\alpha=\sigma\circ\tau$ maps the generators like $\alpha(s)=st$, $\alpha(t)=t$, and hence has infinite order. This settles the claim.

I would guess that the fixed field of $G$ is actually $E\cap F=k(t+t^{-1})$. I haven't checked that this is the case though.

Jyrki Lahtonen
  • 140,891
  • Jyrki, this is perfect! I am curious, is there a standard technique to checking the fixed field of some automorphism? How did you "guess" that the fixed field of $\sigma$ is $k(s(1+t),t+t^{-1})$? – Clayton Aug 19 '13 at 00:25
  • 1
    @Clayton: note that $s+st=s+\sigma(s)$ and $t+t^{-1}=t+\sigma(t)$ are fixed by $\sigma$. This means that the fixed field contains $k(s+st,t+t^{-1})$, and then you check degrees to see that this containment is equality. In general, if you have an order-$n$ automorphism $\sigma$ of a field $K$, and an element $\alpha\in K$, then $\sigma$ fixes the sum of the $\sigma^i(\alpha)$, and more generally $\sigma$ fixes all coefficients of $\prod_{i=1}^n (t-\sigma^i(\alpha))$. Usually such elements generate the fixed field. For an interesting example of this, see the proof of Luroths thm in vdWaerden – Michael Zieve Aug 19 '13 at 02:14