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Let $R$ be a ring and M an abelian group such that:

(i)$(r+s)m=rm+sm$, $r,s \in R$, $m \in M$

(ii)$r(m+n)=rm+rn,$ $r \in R, m, n \in M$

(iii) $r(sm) = (rs)m,$ $r,s \in R$, $m \in M$.

Show that $m \cdot r := rm$ defines a right $R$-module on $M$. Apply this result to $M_n(k)$ and $k^n$, where $A \cdot v = A^tv, A \in M_n(k)$.

I already showed that $m \cdot r$ defines a right $R$-module on $M$, but I'm not sure what the question means when it says "Apply this result to $M_n(k)$ and $k^n$, where $A \cdot v = A^tv, A \in M_n(k)$". I was just wondering if anybody could clarify this question, because I'm not really sure what the question is asking for...

Thanks in advance

Community
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    In what context did you encounter this question? – M Turgeon Aug 28 '13 at 21:56
  • This is what I mean. It could have been in a textbook you're self-studying, the textbook for a class, the notes from a professor, the homework he handed out, etc. My point is that, if a professor was involved in the making of this question, you could ask him to clarify. – M Turgeon Aug 28 '13 at 22:04
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    Hmmm, for a right $R$-module, shouldn't we have $(m\cdot r)\cdot s = m\cdot (rs)$? That would mean $(iii)$ ought to be $r(sm) = (sr)m$. And that's what transposing the matrix gives you. – Daniel Fischer Aug 28 '13 at 22:20

2 Answers2

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I think there's a typo in the question, because as it stands it defines a left module structure and will not be applicable to the matrix situation in the last part of the question. I think the equation in condition (iii) should read $r(sm)=(sr)m$. The last sentence of the question asks you to apply the general result (i.e., to verify the hypotheses (i), (ii), (iii) and to see what the conclusion says) for the particular case where $R$ is the ring of $n\times n$ matrices, $M$ is the abelian group of $n$-component (column) vectors, and multiplication is defined not in the usual "obvious" way but rather by first transposing the $n\times n$ matrix and then multiplying in the usual way. (Contrary to azimut's answer, I think this makes good sense as it stands.) The point is that, instead of the version of (iii) in the original question (which produces a left module), you'll have my proposed "twisted" (iii), which gives a right module. The twisting comes from the fact that $(AB)^t$ is not $A^tB^t$ but $B^tA^t$.

Andreas Blass
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  • Thanks. I think you're right, because the question says $r(sm)=(sr)m$ for $(iii)$, but I wrote it wrong when typing. But the wording still seems a bit strange to me. First it asks us to prove that $m \cdot r :=rm$ defines a right $R$-module, and then it asks us to verify the hypothesis for left-modules. I know it makes sense, but I just think it's strange. –  Sep 01 '13 at 15:44
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I think the "Apply this..."-part of the exercise wants you to do the following:

Let $k$ be a field, $n\in\mathbb{N}$, $R = M_n(k)$ the ring of $m\times m$ matrices over $k$ and $M = k^n$ the vector space of $n$-tuples over $k$. Show that $M$ is a right $R$-module with regard to the scalar multiplication $v\cdot A := A^t v$.

Probably, $A\cdot v = A^t v$ is a typo. The transpose sign only makes sense if the multiplication is from the right. This also fits the fact that the first part of the exercise is about right $R$-modules.

azimut
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