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bio website math.unice.fr/~cazanave
location Nice, France
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visits member for 4 years, 6 months
seen 12 hours ago

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).