Noether’s Problem – 2. Fischer

See the previous post for the notations and/or definitions used here without reference.

In 1915 Fischer [Fi15] proved that for ${G}$ (finite) abelian the extension ${k(G):=k(x_g\>|\> g\in G)^G}$ is always rational if ${k}$ contains “enough” roots of unity. Let us give the precise and general formulation first.

Theorem 1 (Fischer, [Fi15])

Let ${G}$ be an abelian group with exponent ${e}$ . Assume that ${k}$ is a field containing the ${e^{th}}$ roots of unity and either the characteristic of ${k}$ does not divide ${e}$ or is equal to 0. Then ${k(G)}$ is rational.

This implies, in particular, that ${{\mathbb C}(G)}$ is rational for all abelian ${G}$ !

For our choice of ${k}$ and ${G}$ — we have assumed that the field ${k}$ is of characteristic 0 and that ${G}$ is a finite cyclic group, see the previous post — Fischer’s theorem reduces to:

Theorem 2 (Fischer, special case)

Let ${G={\mathbf C}_n}$ and assume that the field ${k}$ (of characteristic 0) contains the ${n^{th}}$ roots of unity. Then ${k(G)}$ is rational.

This is an important and interesting theorem. Not only does it give us some idea of how we can handle rationality, its proof is also relatively short and elementary. Furthermore, the concepts introduced in the proof will be used throughout the series.

Proof. Let ${\mathcal{B}=\{e_1,\ldots,e_n\}}$ denote the standard basis for the ${n}$ -dimensional vector space ${V=k^n}$ over ${k}$ . In ${V}$ we can look at the “linear action” of ${G}$ on ${k({\mathbf x})=k(x_1,\ldots,x_n)}$ , simply by making the correspondence ${e_i\leftrightarrow x_i}$ . Since ${G=\langle \sigma\rangle}$ acts on ${k({\mathbf x})}$ by ${\sigma:x_1\mapsto\ldots\mapsto x_n\mapsto x_1}$ (by definition), the action of ${G}$ on ${V}$ corresponds with a cyclic permutation matrix ${P}$ permutating the basis vectors ${e_i}$ .

As ${k}$ contains the ${n^{th}}$ roots of unity, the permutation matrix ${P}$ is diagonalizable with linearly independent eigenvectors ${v_i}$ ( ${0\>{\leqslant}\> i\>{\leqslant}\> n-1}$ ) and eigenvalues the ${n^{th}}$ roots of unity: ${P v_i=\zeta_n^i v_i}$ (where ${\zeta_n}$ denotes a primitive ${n^{th}}$ root of unity). These ${n}$ eigenvectors ${v_i}$ correspond in ${k({\mathbf x})}$ with ${n}$ algebraically independent elements ${y_i}$ which we can define explicitely as

$\displaystyle y_i:=\sum_{j=1}^{n} \zeta_n^{ij}x_j \text{ for } 0\>{\leqslant}\> i\>{\leqslant}\> n-1. \ \ \ \ \ (1)$

Hence, we get the equivalent action

$\displaystyle \sigma(y_i)=\zeta_n^iy_i \ \ \ \ \ (2)$

of ${G=\langle\sigma\rangle}$ on ${k(\mathbf{y})=k(\mathbf{x})}$ .

Now set

$\displaystyle \left\{ \begin{aligned} a_0&:=& y_0&\\ a_1&:=& y_1^n&\\ a_i&:=& y_i/y_1^i& \text{ for } 2\>{\leqslant}\> i\>{\leqslant}\> n-1 \end{aligned} \right.. \ \ \ \ \ (3)$

Then I claim that ${\mathbf{a}:=(a_0,\ldots,a_{n-1})}$ is a basis for the extension ${k(G)/k}$ (and hence that ${k(G)}$ is rational).

The ${a_i}$ are fixed by ${\sigma}$ (quick check) — and thus by ${G=\langle\sigma\rangle}$ — such that we have the inclusion ${k(\mathbf{a})\subseteq k(G):=k({\mathbf x})^G}$ . As ${G}$ is the Galois group of ${k(\mathbf{x})/k(G)}$ , we have that ${[k(\mathbf{x}):k(G)]=|G|=n}$ . It is easy to see that ${k(\mathbf{a})(y_1)=k(\mathbf{y})}$ , but — and now comes the essential part — ${y_1}$ is a root of the polynomial ${X^n-y_1^n=X^n-a_1\in k(G)[X]}$ , thus ${[k(\mathbf{y}):k(\mathbf{a})]\>{\leqslant}\> n}$ . Or in a diagram:

We conclude ${k(\mathbf{a})=k(G)}$ and the claim is proved. $\Box$

First of all, note that not only did we prove that ${k(G)/k}$ is rational, we also have an explicit basis ${\mathbf{a}}$ , defined in Eq. 3, for it! Secondly, the proof could be made somewhat shorter by just giving the basis ${\mathbf{a}}$ and checking that it is indeed a basis for ${k(G)}$ , as in the last paragraph. Personally, I think this longer version provides some meaningful context.

To conclude, let us make the proof of Fischer’s theorem a little more explicit by considering an example.

Example 1 ( ${{\mathbb Q}(\zeta_3)({\mathbf C}_{3})}$ )

Let ${k={\mathbb Q}(\zeta_3)}$ and ${G={\mathbf C}_3}$ . As in the previous post, ${G=\langle\sigma\rangle}$ works on ${k({\mathbf x})=k(x_1,x_2,x_3)}$ by definition of ${\sigma:x_1\mapsto x_2 \mapsto x_3\mapsto x_1}$ . The cyclic permutation matrix ${P}$ defined in the proof is given by

$\displaystyle P:=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right),$

which acts on the standard basis ${\mathcal{B}=\{e_1, e_2, e_3\}}$ corresponding with the algebraically independent elements ${\{x_1,x_2,x_3\}}$ . Such that, for example, we have the correspondence

$\displaystyle x_1-x_2 \leftrightarrow e_1-e_2 =\left( \begin{array}{c} 1 \\ -1 \\ 0\end{array} \right).$

The eigenvectors ${v_i}$ of ${P}$ , satisfying ${P v_i=\zeta_3^i v_i}$ , are given by

$\displaystyle v_0:=\left( \begin{array}{c} 1 \\ 1 \\ 1\end{array} \right), v_1:= \left( \begin{array}{c} \zeta_3^2 \\ \zeta_3 \\ 1\end{array} \right), v_2:=\left( \begin{array}{c} \zeta_3 \\ \zeta_3^2 \\ 1\end{array} \right).$

From the eigenvectors we can read off the corresponding algebraically independent elements

$\displaystyle \left\{ \begin{aligned} y_0&:=& x_1 & + & x_2 & + & x_3\\ y_1&:=& \zeta_3^2 x_1 & + & \zeta_3 x_2 & + & x_3\\ y_2&:=& \zeta_3 x_1 & + & \zeta_3^2 x_2 & + & x_3 \end{aligned} \right.. \ \ \ \ \$

and we can easily check that they indeed satisfy ${\sigma(y_i)=\zeta_3^iy_i}$ . For example, $\sigma(y_1)=\sigma( \zeta_3^2 x_1 + \zeta_3 x_2 + x_3)=\zeta_3^2 \sigma(x_1) + \zeta_3 \sigma(x_2) + \sigma(x_3) = \zeta_3^2 x_2 +$ $\zeta_3 x_3 + x_1 = \zeta_3 y_1.$

Finally, for computational reasons we replace the basis ${\mathbf{a}=(a_0, a_1, a_2)}$ , defined in general as Eq. 3, by the equivalent basis ${(a_0, a_1, a_1a_2)}$ . (We have replaced ${a_2:=y_2/y_1^2}$ by ${a_2a_1=(y_2/y_1^2)y_1^3=y_1 y_2}$ .) We get

$\displaystyle \left\{ \begin{aligned} &a_0&:=& y_0 &=& \>\>x_1 + x_2 + x_3&\\ &a_1&:=& y_1^3 &=& \>\>x_1^3 + x_2^3 + x_3^3 + & \\ && & & & \>\> 3\zeta_3(x_1 x_2^2 + x_1^2 x_3 + x_2 x_3^2) +&\\ && & & & \>\>3\zeta_3^2(x_1^2 x_2 + x_1 x_3^2 + x_2^2 x_3) + 6 x_1 x_2 x_3&\\ &a_1a_2&:=& y_1 y_2 &=& \>\> x_1^2 + x_2^2 + x_3^2 - (x_1x_2 + x_2x_3+x_1x_3) & \end{aligned} \right.,$

with which we have ${{\mathbb Q}(\zeta_3)({\mathbf C}_{3})} ={\mathbb Q}(\zeta_3)(a_0, a_1, a_1a_2)$ .

In the next post we will use Fischer’s theorem to formulate the basic idea on which we will build our “complete solution” for ${k={\mathbb Q}}$ and ${G={\mathbf C}_p}$ ( ${p}$ prime). We will also prove an interesting theorem from Masuda that will allow us to prove the rationality for some small order groups.

[Fi15] Fischer, E.; Die Isomorphie der Invariantenkorper der endlichen Abelschen Gruppen linearen transformationen, 1915.

Noether’s Problem – 2. Fischer

See the previous post for the notations and/or definitions used here without reference.

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Noether’s Problem – 2. Fischer

See the previous post for the notations and/or definitions used here without reference.

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