-1
$\begingroup$

Ring

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

$\endgroup$
  • 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
3
$\begingroup$

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 $I$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.