There is a ring $R$ and its subring $K$ with unit. We have a matrix $A$ of order $n$ over $R$. Someone said, that if $A^m$ for $m=1,...,n$ can be represented as a sum of matrices over $R$ which a integral over $K$, than $A$ is integral over $K$. Why is that?
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$\begingroup$ I forgot to mention, that char(R)=0 or char(R)>n. $\endgroup$– MilenaCommented Jun 23, 2013 at 15:11
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$\begingroup$ Cross-posted on MSE: math.stackexchange.com/questions/427587/… $\endgroup$– JulienCommented Jun 23, 2013 at 15:39
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$\begingroup$ Yes, I posted it there about 10 minutes ago. $\endgroup$– MilenaCommented Jun 23, 2013 at 15:43
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2 Answers
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A matrix is integral iff its eigenvalues are integral. Thus A is integral if and only if A^m is integral.
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$\begingroup$ What eigenvalues? What if the characteristic polynomial has no roots in $R$? $\endgroup$ Commented Jun 23, 2013 at 20:21
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$\begingroup$ Hmm? Please see my "answer" for another reservation... $\endgroup$ Commented Jun 24, 2013 at 7:27
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$\begingroup$ A is integral, if the minimal polynomial is integral. That means the eigenvalues are integral. Of course, the entries might not be integral. $\endgroup$– guestCommented Jun 24, 2013 at 19:43
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$\begingroup$ I can't understand a word from what you say: minimal polynomial is integral? What is this supposed to mean? And again: what eigenvalues are you talking about when work with matrices over a commutative ring, not over a field? $\endgroup$ Commented Jun 25, 2013 at 20:30
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--- This should go as a comment at @guest 's answer. ----
Let $$ E = \small \begin{bmatrix} 1 & . & . & . & . & . & . & . \\\ 0 & 1 & . & . & . & . & . & . \\\ 0 & 1/2 & 1 & . & . & . & . & . \\\ 0 & 1/6 & 1 & 1 & . & . & . & . \\\ 0 & 1/24 & 7/12 & 3/2 & 1 & . & . & . \\\ 0 & 1/120 & 1/4 & 5/4 & 2 & 1 & . & . \\\ 0 & 1/720 & 31/360 & 3/4 & 13/6 & 5/2 & 1 & . \\\ 0 & 1/5040 & 1/40 & 43/120 & 5/3 & 10/3 & 3 & 1 \end{bmatrix} $$ (which is the factorially scaled matrix of Stirling-numbers 2nd kind), then let's define $$ M = E \cdot diag(1,2,4,8,16,2^5,2^6,2^7) \cdot E^{-1} $$ Then M has integral eigenvalues but has fractional entries: $$ M= \small \begin{bmatrix} 1 & . & . & . & . & . & . & . \\\ 0 & 2 & . & . & . & . & . & . \\\ 0 & -1 & 4 & . & . & . & . & . \\\ 0 & 1 & -4 & 8 & . & . & . & . \\\ 0 & -13/12 & 5 & -12 & 16 & . & . & . \\\ 0 & 5/4 & -19/3 & 18 & -32 & 32 & . & . \\\ 0 & -541/360 & 49/6 & -26 & 56 & -80 & 64 & . \\\ 0 & 223/120 & -961/90 & 37 & -272/3 & 160 & -192 & 128 \end{bmatrix} $$
So is there something that I misunderstood in the question/your answer?
answered Jun 24, 2013 at 7:26
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$\begingroup$ The actual criterion is that a matrix is conjugate to an integral matrix if and only if its eigenvalues are integral. $\endgroup$ Commented Jun 24, 2013 at 22:23
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$\begingroup$ Or rather that this conjugate criterion is what is meant by an integral matrix. Otherwise, a sum of integral matrices would just be an integral matrix. $\endgroup$ Commented Jun 24, 2013 at 22:27