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Is it true that a subfield $K$ of $C_t(s)$ (the quotient field of the ring of trigonometric polynomials with complex coefficients) containing a non-constant trigonometric polynomial satisfies that $K=\mathbb C(r)$ for some trigonometric polynomial $r$? (This is Lüroth's theorem for complex trigonometric polynomials.)

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  • $\begingroup$ What's the rationale behind the weird notation $C_t(s)$? What do $t$ and $s$ mean? Excellent question, though. $\endgroup$ Dec 21 '16 at 8:16
  • $\begingroup$ An interesting question. The real challenge is whether we can always choose $r$ to be an element of $\Bbb{C}[\cos z,\sin z]$. $\endgroup$ Dec 21 '16 at 9:21
  • $\begingroup$ Dear @Jyrki, you are right that your question (which I can't answer) is a real and interesting challenge. However my answer only tackled the question (mentioned in the OP's title and in his last sentence) of a Lüroth theorem for trigonometric polynomials. Lüroth's standard theorem never addresses the question whether a generator of a subfield is a polynomial. $\endgroup$ Dec 22 '16 at 17:55
  • $\begingroup$ Dear @Jyrki, I have now added an Edit to my post. $\endgroup$ Dec 23 '16 at 17:45
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The answer is: yes, it is true.

The field you are studying is $K:=\mathbb C(x,y)$ with $x,y$ transcendental over $\mathbb C$ satisfying $x^2+y^2=1$. In the analytic context we have $x=\cos \theta,y=\sin \theta$.
The well known parametrization $x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$ with $t=\frac{y}{1+x}$ shows that $K=\mathbb C(t)$, the rational function field in one indeterminate.
We can thus apply the usual Lüroth theorem stating that any subextension $\mathbb C\subset F\subset K$ is of the form $F=\mathbb C(\phi(t))=\mathbb C(\phi(\frac{y}{1+x}))=\mathbb C(\phi(\frac{\sin \theta}{1+\cos \theta}))$ for some rational function $\phi\in \mathbb C(t)$.
This shows that Lüroth's result also applies to complex trigonometric polynomials.

Edit
Consider the isomorphism invoked above $u:K=\mathbb C(x,y)\stackrel{\cong}{\to} \mathbb C(t):x\mapsto \frac{1-t^2}{1+t^2},y\mapsto \frac{2t}{1+t^2}$ and its restriction $u_\mathbb R:\mathbb R(x,y)\stackrel{\cong}{\to} \mathbb R(t).$
It is proved in the following paper by Cima, Gasull and Mañosas, Lemma 3 (reference friendly provided by @user26857) that the image of $\mathbb R[x,y]$ under $u_\mathbb R$ is the set of rational functions $\frac {f(t)}{(1+t^2)^n}$ with $n\geq 0$ and $f(t)\in \mathbb R[t]$ a polynomial of degree $\leq 2n$.
This allows one to prove that not every subfield of $\mathbb R(x,y)$ is of the form $\mathbb R(P(x,y))$ with $P(x,y)\in \mathbb R[x,y]$ a polynomial rather than an arbitrary rational function.
In fact even $\mathbb R(x,y)$ itself cannot be written in this form!
Indeed, let's transport the problem in terms of $t$:
The only generators over $\mathbb R$ of $\mathbb R(t)$ are fractions of the form $\frac {at+b}{ct+d}$ with $a,b,c,d\in \mathbb R, ad-bc\neq 0$ and it is clear that none of these fractions is of the form $\frac {f(t)}{(1+t^2)^n}$ .
Hence transporting back we see that we cannot write $\mathbb R(x,y)=\mathbb R(P(x,y))$

Although the analogue of the Proposition by Cima et al. is still valid over $\mathbb C$ (with the same proof) the result I just showed is no longer valid over $\mathbb C$ because $1+t^2$ not irreducible over $\mathbb C$.
Indeed $\mathbb C(x,y)$ is generated by the single polynomial $ x+iy$ over $\mathbb C$. Explicitly: $$\mathbb C(x,y)=\mathbb C(x+iy)=\mathbb C(\frac {1+it}{1-it})$$ and $\frac {1+it}{1-it}$ is of the required form $\frac {at+b}{ct+d}$ if you allow complex coefficients $a,b,c,d$.
However I don't know whether an arbitrary subextension $ \mathbb C\subset \mathbb C(x,y)$ can also be written as $\mathbb C(P(x,y))$ with $P(x,y)\in \mathbb C[x,y]$.

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    $\begingroup$ Maybe I'm missing something: but the more complicated point in the problem seems to be, that the subfield should be generated by a trigonometric polynomial. OK, sorry I just saw that Jyrki already mentions this point. $\endgroup$
    – Hagen Knaf
    Dec 21 '16 at 10:27
  • $\begingroup$ In Georges Elencwajg's prove it is not clear at all that ϕ(t) is a complex trigonometric polynomial $\endgroup$
    – Claudia X
    Dec 22 '16 at 8:56
  • $\begingroup$ @Claudia: I have written explicitly that $\phi \in \mathbb C(t)$ is a rational fraction. Lüroth's theorem does not mention polynomials at all. $\endgroup$ Dec 22 '16 at 17:48
  • $\begingroup$ Dear @Hagen, please see my answer to Jyrki. $\endgroup$ Dec 22 '16 at 17:57
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    $\begingroup$ Dear @user26857, thanks a lot for your very interesting (as always!) comments and for the article you link to, which I would never have come across without you. I had indeed tried to solve the problem by considering $\mathbb C[X,X^{-1}]$, a UFD with trivial class group, and my frustration came from my being unable to handle such a simple situation... $\endgroup$ Dec 22 '16 at 23:12
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An attempt to prove that the answer to Claudia X's question is Yes - but with a gap:

Under the link

https://commons.wikimedia.org/wiki/File:Another_elementary_proof_of_Luroth%27s_theorem-06.2004.pdf

one can find a proof for Lüroth's theorem, that tells us something about the generator of the intermediate field (I didn't check the proof!). So let $M$ be subfield of the rational function field $k(X)$, $M\neq k$, $M\neq k(X)$. Let $p$ be the minimal polynomial of $X$ over $M$. Then in the article mentioned above it is shown that $M$ is generated by any of the coefficients of $p$, that does not lie in $k$.

In our situation $k=\mathbb{C}$. Moreover the ring of trigonometric functions with complex coefficients equals $\mathbb{C}[x,y]$, where $x^2+y^2=1$ as Georges has already explained. The algebraic curve over $\mathbb{C}$ given by this equation has no singularities, therefore the ring $\mathbb{C}[x,y]$ is integrally closed.

Due to the equation $(x+iy)(x-iy)=1$, the fraction field $\mathbb{C}(x,y)$ can be generated by the trigonometric polynomial $x+iy$.

Now let $M\neq\mathbb{C}$ be a proper subfield of $\mathbb{C}(x,y)$, then $\mathbb{C}(x,y)|M$ is a finite extension and the ring $R:=M\cap\mathbb{C}[x,y]$ is integrally closed in $M$.

If we would know that $\mathbb{C}[x,y]$ is integral over $R$, then we are done: the generator $x+iy$ is integral over $R$, hence every coefficient lies in $R$, which proves the assertion.

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  • $\begingroup$ Are you refering to Theorem 6 in the paper you mention? Then there is one case, in which the problem is still open: in that case the subfield is generated by a tangens function. After all looking at my attempt I (also) feel like the question of Claudia X does not always have a positive answer. $\endgroup$
    – Hagen Knaf
    Dec 22 '16 at 23:08
  • $\begingroup$ Dear Hagen, I now see, after having posted my Edit, that you had proved that $\mathbb C(x,y)$ is generated by $x+iy$ before me and in a more elementary way! And +1. $\endgroup$ Dec 23 '16 at 18:35
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Strictly speaking the answer to the question is no, since $K$ has countable subfields: $\mathbf Q(\sin z)$.

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The answer to Claudia X's question is Yes.

Let $K$ be an algebraically closed field of characteristic $\neq 2$. Consider the function field $F=K(x,y)$ with $x^2+y^2=1$ and the ring $T:=K[x,y]$. Let $z:=x+iy$, $i^2=-1$. Then $F=K(z)$ and $T=K[z,z^{-1}]$; in particular every $t\in T$ has the form $\frac{f}{z^m}$ for some $f\in K[z]$ not divisible by $z$.

Claim 1: A subfield $E\neq K$ of $F$ is generated by an element $t\in T\setminus K$ if and only if it is generated by an element of the form

$\frac{af+bz^m}{cf+dz^m},\; f\in K[z],\; a,b,c,d\in K, ad-bc\neq 0$.

Proof: $\Rightarrow$: let $E=K(t)$, $t=\frac{f}{z^m}$, $f\in K[z]$ not divisible by $z$. Every generator of $E$ then has the form

$ \frac{at+b}{ct+d}=\frac{a\frac{f}{z^m}+b}{c\frac{f}{z^m}+d}=\frac{af+bz^m}{cf+dz^m} $

with $a,b,c,d\in K$, $ad-bc\neq 0$.

$\Leftarrow$: let $ g:=\frac{af+bz^m}{cf+dz^m} $ be a generator of $E$. Then $ \frac{1}{ad-bc}\frac{dg-b}{-cg+a}=\frac{f}{z^m}\in T $ is a generator of $E$.

Claim 2: A subfield $E\neq K$ of $F$ is generated by an element $t\in T\setminus K$ if and only if $T\cap E\neq K$.

Proof: let $t=\frac{f}{z^m}\in E\setminus K$, $f\in K[z]$ not divisible by $z$. Let $E=K(\frac{p}{q})$ with coprime polynomials $p,q\in K[z]$. Then there exist coprime polynomials $G,H\in K[X]$ such that

$ \frac{f}{z^m}=\frac{G(\frac{p}{q})}{H(\frac{p}{q})}. $

Since both polynomials $G$ and $H$ split into linear factors $X-\gamma$, $\gamma\in K$, and since $ \frac{p}{q}-\gamma=\frac{p-\gamma q}{q} $ one gets

$ \frac{f}{z^m}=q^n \frac{\prod\limits_{i=1}^r(p-\alpha_i q)^{e_i}}{\prod\limits_{j=1}^s(p-\beta_j q)^{f_j}},\; n\in\mathbb{Z},\quad (1) $

where the elements $\alpha_i$ and $\beta_j$ are pairwise distinct.

The following properties are present:

  • By assumption about $p$ and $q$, the polynomials $p-\alpha_i q$ und $p-\beta_j q$ possess no zero in common with $q$.
  • Two polynomials $p-\alpha q$ und $p-\beta q$ with $\alpha\neq\beta$ possess no common zero.

Due to these properties in equation (1) no linear factors can cancel out. Since the denominator of the left-hand-side has $z$ as its only linear factor, only the following cases are possible:

Case 1: $m=0$.

Case 2: $s=1$.

Case 3: $H$ is constant.

In the first case the polynomial $H$ must be constant, thus one actually is in the third case.

In the second case case $p-\beta_1 q=z^\ell$ and therefore $ \frac{p}{q}=\frac{\beta_1 q +z^\ell}{q}. $ The first claim then yields that $\frac{q}{z^\ell}\in T$ is a generator of $E$.

In the third case $n<0$ hence $q=z^\ell$ must hold, which gives $\frac{p}{q}\in T$.

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  • $\begingroup$ I haven't checked your proof, but how I implied few days ago the subfield generated by tangents don't show up in the complex case since $1+x^2$ is reducible over $\mathbb C$. $\endgroup$
    – user26857
    Dec 29 '16 at 9:40

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