Could someone help me give another demonstration of this? The demonstration they give here is very complex and I can not understand it well ...

  • 2
    $\begingroup$ That's the standar proof, I think. I mean, I found this proof (with little variations) in every book where this result is stated. But, what's your problem with the proof? what do you don't understand? $\endgroup$ – Xam Nov 15 '16 at 4:24
  • $\begingroup$ See also math.stackexchange.com/a/654487/589. $\endgroup$ – lhf Nov 17 '16 at 15:58

I don't know another argument; this one is pretty standard. Let me see if I can explain the idea a bit more clearly. The strategy of this argument is to show that $A = \{a_{11}\in R\mid (a_{ij})_{i,j}\in I\}$ is an ideal of $R$, and then to show that the ideal $I$ is actually equal to $M_n(A)$ using simple matrices to ``shift everything around until its in the right spot." Note: I write $M_n(A)$ to mean what the authors of your book meant by $A_n$, and similarly $M_n(R) = R_n$. What the authors call $e_{ij}$ I denote $E^{ij}$. If I write $(m_{ij})$, then $m_{ij}\in R$ for each possible pair $(i,j)$ with $0\leq i,j\leq n$, so $(m_{ij})\in M_n(R)$. Lower indices will denote the entry in the matrix the element is, upper indices will not indicate where you are in your matrix.

  1. $A$ is an ideal: If you take $a_{11}$ and $b_{11}$ in $A$, then they are the ``first slot" in their respective matrices $(a_{ij})$ and $(b_{ij})$ in $I$. Then $((a_{ij}) + (b_{ij}))_{11} = a_{11} + b_{11}$, so the ideal is closed under addition. If $a_{11}\in A$, then $(a_{ij})\in I$, and hence $-(a_{ij}) = (-a_{ij})\in I$, so $A$ has additive inverses. Lastly, take $r\in R$. Then $rI_n(a_{ij}) = (r a_{ij})\in I$, where $I_n$ is the identity matrix, so that $r a_{11}\in A$, and $A$ is an ideal of $R$.
  2. $I = M_n(A)$: Take an arbitrary matrix $(a_{ij})\in I$. We will show that any entry is actually in $A$, so that $I \subseteq M_n(A)$. If $E^{ij}$ is the matrix with $1$ in the $ij$-th entry and $0$s elsewhere, then $$ E^{1r}(a_{ij})E^{s1} = a_{rs} E^{11} $$ (this is a simple computation, if you don't see it, you should try writing it out), so that $a_{rs}$ is an element of $A$: it occurs as the $1,1$-th entry in $a_{rs} E^{11}$, which is in $I$ because $(a_{ij})\in I$ and $I$ is an ideal. Thus $I\subseteq M_n(A)$. To see $M_n(A)\subseteq I$, take $(a_{ij})\in M_n(A)$. We must show that $(a_{ij})\in I$; our strategy will be to express $(a_{ij})$ as a sum of elements we know are in $I$. We know that each $a_{ij}\in A$, so for every $a_{ij}$, we can find a matrix $(b^{ij}_{\ell k})_{\ell,k}$ with $b^{ij}_{11} = a_{ij}$. But then by a similar argument to the above, we can take $E^{i1}(b^{ij}_{\ell k}) E^{1j}$, this will be an element of $I$ because $(b^{ij}_{\ell k})$ is. But if we compute this product, we find $$ E^{i1}(b^{ij}_{\ell k}) E^{1j} = b^{ij}_{11} E^{ij} = a_{ij} E^{ij}. $$ This is an element of $I$ with $a_{ij}$ as its $ij$-th entry and $0$s elsewhere. Doing this for every combination of $i$ and $j$, we can add them all together to get the original matrix $(a_{ij})$, which is now in $I$ because each $a_{ij} E^{ij}$ was. So $M_n(A)\subseteq I$, and thus $M_n(A) = I$.

If you can keep your $p$'s and $q$'s straight to show that $A$ is an ideal of $R$, the key part of the argument is realizing that if you have a matrix $M = (m_{ij})$, then $E^{1r} M E^{s1} = m_{rs} E^{11}$ and $E^{r1} M E^{1s} = m_{11} E^{rs}$: these two identities let you move an entry of $M$ to any other slot, and because $I$ was an ideal, $E^{1r} M E^{s1}$ and $E^{r1} M E^{1s}$ are both in $I$ if $M$ is, so your moving entries around doesn't move you outside of $I$.

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  • $\begingroup$ Just a comment, it's not necessary to prove that "$A$ has additive inverses" because we are dealing with unitary rings. $\endgroup$ – Xam Nov 15 '16 at 4:56
  • $\begingroup$ @Charter Good point! $\endgroup$ – Stahl Nov 15 '16 at 4:58

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