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Let $A$ be a non-singular $n$-by-$n$ matrix with integer entries. Assume that $p^r\nmid \det(A)$. Does it follow that $A$ has an $(n-r+1)$-by-$(n-r+1)$ minor that is non-singular modulo $p$?

If the answer is no: what if we put some additional conditions -- say, $p$ large compared to $n$, and/or $A$ having bounded entries?

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  • $\begingroup$ This example does not meet your specifications, but comes close to a counterexample. Pick prime p and n=p+2. Set A=J-I and set B= (A with first column and first row multiplied by p). Det of B is p^2(p+1). Every order n-1 minor of B is a multiple of p. I don't see yet how to alter it to get every order n-2 minor a multiple of p while leaving the determinant alone. Gerhard "Maybe You Will See It" Paseman, 2019.07.17. $\endgroup$ Jul 17 '19 at 15:37
  • $\begingroup$ What do you mean by J-I? (Perhaps $a_{i,j}=j-i$?) $\endgroup$ Jul 17 '19 at 15:42
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    $\begingroup$ @HAHelfgott: It is a fairly common practice to let $J$ denote the (square) matrix with every entry $1$. $\endgroup$ Jul 17 '19 at 15:50
  • $\begingroup$ Sorry . J is all ones, I is identity, and both are order n 0-1 matrices. (Well, J has no 0.) Looking at a cofactor matrix I'm now thinking that every n-1 minor having determinant divisible by p means det is divisible by p^2. Gerhard "Don't Have A Proof Yet" Paseman, 2019.07.17. $\endgroup$ Jul 17 '19 at 15:51
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Let $s = \operatorname{ord}_p(\det(A))$, i.e. $p^s\ \| \det(A)$. Note that $|\det(A)| = \# \operatorname{coker}(A \colon \mathbf Z^n \to \mathbf Z^n)$. Right exactness of the tensor product shows that $\operatorname{coker}(A \otimes \mathbf F_p) \cong \operatorname{coker}(A) \otimes \mathbf F_p$, so $\operatorname{coker}(A \otimes \mathbf F_p)$ has cardinality at most $p^s$, hence dimension at most $s$. This should answer your question (affirmatively).

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  • $\begingroup$ Ah, I see. In very concrete terms: let $L$ be the image of $A$. Evidently, $(Z^n/(L+pZ^n)) | (Z^n/L) = \det(L)$, so $Z^n/(L+pZ^n)\leq p^s$. Hence the rank of $L \mod p$ is at least $k-s$, and so $L\mod p$ does have a non-singular $(k-s)$-by-$(k-s)$ minor. $\endgroup$ Jul 17 '19 at 16:01
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I think this is all about Smith Normal Form. Write $XAY = {\rm diag}(d_{1},d_{2},\ldots, d_{n})$ where $X$ and $Y$ are unimodular and where the $d_{i}$ are integers such that $d_{i} | d_{i+1}$ for each $i$. There are various terminologies for the $d_{i}$ sometimes conflicting. Let me call them the determinantal divisors (sometimes they are called elementary divisors sometimes invariant factors). In any case, they are unique up to sign and they determine and are determined by the structure of the Abelian group $\mathbb{Z}^{n}/{\rm Im}A$ when $A$ acts by multiplication on $\mathbb{Z}^{n}$ identified with $n$ long integer column vectors.

Your hypotheses imply that $p \not | d_{i}$ for $1 \leq i \leq n-(r-1).$ On the other hand, it is "well-known" that for $1 \leq i \leq m$, the product $d_{1}d_{2} \ldots d_{m}$ is the gcd of all the $m \times m$ minors of $A$. Hence your assumptions imply that there is indeed some $(n-r +1) \times (n-r+1)$ minor of $A$ which is not divisible by $p$.

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  • $\begingroup$ Nice - but where do I find that well-known fact? $\endgroup$ Jul 17 '19 at 16:11
  • $\begingroup$ Most texts that treat Smith Normal Form. Actually, it is easier just to argue that the $p$-rank of $XAY$ is at least $n-(r-1)$ because a diagonal $(n-r+1) \times (n-r+1)$ submatrix of $XAY$ has determinant prime to $p$, while clearly the $p$-rank of $XAY$ is less than or equal to the $p$-rank of $A$. $\endgroup$ Jul 17 '19 at 16:16
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    $\begingroup$ @HAHelfgott : To your question above: See Theorem 2.4 of www-math.mit.edu/~rstan/papers/snf_survey.pdf $\endgroup$ Jul 17 '19 at 17:55

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