bio | website | math.unice.fr/~cazanave |
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location | Nice, France | |
age | ||
visits | member for | 4 years, 6 months |
seen | 12 hours ago | |
stats | profile views | 463 |
Sep 26 |
comment |
Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$
I wanted to edit the previous comments, but it was not possible, so I (stupidly) erased it. You are right: $V$ is semi-simple iff $m_A$ has no square factors, so the double centralizer theorem applies only in this situation. |
Sep 26 |
comment |
Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$
I don't easily see how one can deduce the general case from the double centralizer theorem. However, if you know the proof over an algebraically closed field, you can deduce it over $F$. (Take an algebraic closure $F \subset \overline{F}$, and remark that $\overline F[A]\cap M_n(F)= F[A]$.) |
Sep 15 |
revised |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
added 1 character in body |
Sep 8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
In general, p/q=pq/q^2 and use Lagrange's theorem to decompose a the (positive) numerator as a sum of at most 4 squares. |
Sep 8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
1/7=7/49 and 7=2²+1²+1²+1², so 1/7=(2/7)²+(1/7)^2+(1/7)^2+(1/7)^2. |
Sep 8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
Yes, that's more or less implicit in my "note n°2". |
Sep 8 |
asked | Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)? |
Aug 29 |
comment |
Upper bound on the number of ismorphism classes of bilinear forms on $\mathbb{Z}^n$
@NoamD.Elkies: A priori the question refers to all symmetric bilinear forms (not only to even ones)... |
Jul 22 |
awarded | Self-Learner |
Jul 22 |
accepted | Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$ |
Jul 22 |
answered | Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$ |
Jul 2 |
awarded | Curious |
Jul 1 |
comment |
Units of $\mathbf Z[X,Y]/(P(X,Y))$
Note that this is true only for reduced algebras. The "standard" counterexample is: $\mathbf Z[X,Y]/(X^2)$. |
Jun 25 |
comment |
Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$
@few_reps: is there an analogous command in pari/gp? (I know that there is an online Magma calculator, but for various reasons, it would be more convenient for me to do it in pari/gp). |
Jun 24 |
comment |
integral equivalence classes of quadratic forms
Isn't the example of disagreement in dimension 16 due to Witt instead (and used by Milnor later for is counteraxample)? |
Jun 24 |
comment |
Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$
All this seems like a "rather complicated task". I mean: I am in front of a $16 \times 16$ matrix, with not so small integers. Pari/gp can list all 480 minimal length vectors, but analyzing this data will be tedious. I would have hoped for a trick. |
Jun 24 |
asked | Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$ |
Jun 21 |
awarded | Yearling |
May 27 |
comment |
Proportion of irreducible polynomials $P$ such that $\mathbf Z[X]/(P)$ is the ring of integers of $\mathbf Q[X]/(P)$
Actually the reference by Ash--Brakenhoff--Zarrabi answers exactly my question. |
May 27 |
comment |
Proportion of irreducible polynomials $P$ such that $\mathbf Z[X]/(P)$ is the ring of integers of $\mathbf Q[X]/(P)$
In case it may interess someone: the proportion 60% found is conjectured by Lenstra to be $\frac 6 {\pi^2}$ (cf. the paper of Ash--Brakenhoff--Zarrabi mentionned in David Speyer answer). |