Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathbb K$ be a number field of degree $n$ over $\mathbb Q$, and let $\alpha_1,\alpha_2, \ldots ,\alpha_n$ be a $\mathbb Q$-basis of $\mathbb K$. Then there are coefficients $(c^{ij}_k)$ (where $i,j,k$ are independent indices between $1$ and $n$) such that

$$ \alpha_i \times \alpha_j = \sum_{k=1}^{n} c^{ij}_k \alpha_k $$

and we have for any indices $i,j,k,l$,

$$ (1) c^{ij}_k=c^{ji}_k \ (\ {\rm commutativity}) $$

$$ (2) \sum_{y=1}^{n}c^{iy}_lc^{jk}_y=\sum_{y=1}^{n}c^{ij}_yc^{yk}_l \ (\ {\rm associativity}) $$

If we take $\alpha_n$ to be $1-\sum_{y=1}^{n-1} \alpha_y$, we also have

$$ (3) \sum_{y=1}^{n}c^{yi}_j=\delta_{ij} $$

where $\delta_{ij}$ is the Kronecker symbol.

Now, let $V$ be the algebraic variety in the variables $( c^{ij}_k)$ defined as the subset of ${\mathbb C}^{n^3}$ satisfying equations (1) to (3). Is the dimension of $V$ (in the sense of algebraic geometry) known ?

share|improve this question
That variety has in general several irreducible components of different dimensions. I'm not sure the maximal dimension of a component is known. –  Mariano Suárez-Alvarez Nov 11 '11 at 12:51

Your Answer


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

Browse other questions tagged or ask your own question.