I'm solving exercises of noncommutative ring theory and I have find across the following problem.     If $D$ and $D'$ are division rings and $M_m(D)\simeq M_n(D')$, show that $D\simeq D'$ and $m=n$.   I have already tried to attack in various ways and I understand that this exercise says that, by the Wedderburn-Artin theorem, a simple Artinian ring $R$ is a ring of matrices over a division ring $D$ unique up to isomorphism.  Any suggestion is appreciated. Thank you.
123855
I am studying Artin-Wedderburn structural decomposition theorem for semisimple rings. I understand that it says that any semisimple ring, $R$ is isomorphic (as rings) to $M_{n_1}(D_1) \times M_{n_2}(D_2)\times\cdots\times M_{n_k}(D_k)$ for some $n_i$ and division rings $D_i$. Is this decomposition unique, i.e., suppose $R$ is also isomorphic to $M_{n'_1}(D'_1) \times M_{n'_2}(D'_2)\times\cdots\times M_{n'_{k'}}(D'_{k'})$, then is $k=k'$ and $n_i$ equal to $n'_i$ and $D_i \approx D'_i$ up to some permutation? If so, how?
113234
I can't seem to find a reference for the following statement:     A quotient ring of a (left) semi-simple ring (left artinian with no nonzero nilpotent ideals) is a semi-simple ring.    Does this follow from quotients of semi-simple modules being semi-simple?  A quotient ring of a left artinian ring is left artinian. So we only have to show that the quotient contains no nonzero nilpotent ideals. I'm looking for something more low-level than $\text{rad}(R/I) = \text{rad}(R) + I$, which I seem to remember to be true. Can I somehow use the characteristic that every left ideal is generated by an idempotent element?
53094
How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?
49646
This is question #66 from        Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur \in U$ for all    $u \in U$ and for all $r \in R$ , then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of    the following must be true?      I. $R$ is commutative.   II. $R$ is a division ring (that is, all elements except the additive identity have multiplicative inverses).   III. $R$ is infinite.     Here is my reasoning: Because $R$ is a ring, $R$ is also a additive group with some identity element $0$. We have a theorem that says $0r = 0 $ in any ring, so $\{0\}$ is a right ideal of $R$. Also, $R$ is a right ideal of $R$. Now I have found two different right ideals and there mustn't be any more.     Edit: As mentioned in the comments, the example below is not a ring, so it is not applicable to the problem. I could not fix it by taking additive closure because that introduced more than two ideals.  A possible candidate for $R$ could be the set of $2\times2$ matrices $\{0,I,-I,a,-a\}$ where $a = [[^1_0] ,[^0_0]]$. The only right ideals are $R$ and $\{0\}$. This ring satisfies only property I, but the answer key says that II is the correct answer.
114093
Show that a division ring is simple.   With a division ring I mean a ring such that all onzero elements are invertible. And with a simple ring I mean a ring which has exactly two two-sided ideals.  I'm doing some exercises to prepare for a new course that start next week. I'm trying to get my ring theory neurons back in the shape they used to be 6 months ago
103493
Suppose that $I$ and $J$ are ideals of $\mathbb{Z}$, with $I=(m)$ and $J=(n)$. This question has two parts:  1) Let r be the least common multiple of $m$ and $n$. Show that $I\cap J = (r)$.  2) Let $d=(m,n)$.  Show that $I+J = (d)$.
117131
Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$  where $I$ is an ideal of $R$.   I have spent 30 minutes on this question and I still got nowhere. Can anyone give some hints ?
82316
If $S$ is a ring with the property that $s=s^2$ for each $s \in S$, then is $S$ commutative?   I know $st=(st)^2=stst$. On the other hand, $st=s^2t^2$. Hence, $$stst=sstt$$  If $S$ is a division ring, then by cancellation we can have $st=ts$.  But how to make sure that any non-zero element in $S$ is not a zero-divisor?
66670
I recently read the concept of semi-simplicity of a (not necessarily commutative) ring.  A ring $R$ is said to be semi-simple if $R$ as a left module over itself is a semi-simple module (This in turn means that each submodule (i.e, left ideal) of $R$ is a direct summand for $R$.)  I am grappling with the following question:     If $R$ is semi-simple, then so is the matrix ring $M_n(R)$.   It is a theorem that each left module of a semi-simple ring is a semi-simple module. Thus $M_n(R)$ is a semi-simple module of $R$. But this doesn't help me.  I need to show that each left ideal of $M_n(R)$ is a direct summand for $M_n(R)$. Is there a characterization  of the left ideals of $M_n(R)$ in terms of the left ideals of $R$?  Lastly, does the converse also hold?     If $M_n(R)$ is a semi-simple ring, then is $R$ necessarily a semi-simple ring?   Thanks.
88343
I'm trying to find a ring which has an element that has many(more than two) distinct left inverses.  I first tried the (square) matrices but then realized that if a matrix is a left unit, then it has determinant greater than 0, thus it is also a right unit and has unique inverse.  And then next I tried to do it with functions with usual addition and composition, and I found this example(in the first answer, by user23211):   But I realized that the space of functions does not form a ring since the distribution law fails.  To make distribution law hold, we can only have homomorphisms; while the elements in the above example are not.  I also have seen that we can construct a similar examples with infinite vector spaces, but they are not rings.  Is there any ring like this?
121367
This is just a textbook problem from Dummit and Foote, but the issue is that our class barely touched on PIDs and the preceding material, so I don't really know or understand much.  Anyway,     Let $R$ be an integral domain.  If $R$ satisfies the following:           i. any two nonzero $a,b \in R$ have a gcd in the form of $d=ra+sb, r,s \in R$          ii. If $a_1, a_2,...$ are nonzero in $R$ such that $a_{i+1}$ divides $a_i$ for every $i$ then $\exists N \in \mathbb{Z}$ such that $a_n$ is an unit times $a_N$  $\forall n \ge N$,          prove that $R$ is a principal ideal domain.   From what I've gathered, the second condition means that $(a_1) \subset (a_2) \subset...$ and $(a_N) = (a_n)$ $\forall n \ge N$.  However, I'm not sure what I should do now. I'm worried that I'm lacking the necessary context here especially since we used Artin pretty much all semester.  In short, my question is: how should I proceed with this proof given what I know?
65375
I'm trying to do a few proofs about rings: $(1)$ I want to show that for a ring $R$ and the ring of matrices $M_n(R)$, if $I\lhd R\implies M_n(I)\lhd M_n(R)$ and that all ideals of $M_n(R)$ are of the form $M_n(I)$. $(2)$ Prove that if $R$ is a ring with division then $M_n(R)$ is a simple ring.  The relation $I\lhd R$ implies that with $r\in R$ and $x\in I$ we have $rx\in I, xr\in I$. But how can I prove the statement for matrices?, here's what I would do: Consider $\psi \in M_n(R)$, if is given any matrix $\phi\in M_n(I)$ it must be $\psi \phi \in M_n(I),\phi \psi \in M_n(I)$, if we denote the element $(i,j)$ of each matrix by $\phi_{ij}$ and $\psi_{ij}$, then $(\psi \phi)_{ij}=\sum_{k=1}^n \psi_{ik}\phi_{kj}$ and $(\phi \psi)_{ij}=\sum_{k=1}^n \phi_{ik}\psi_{kj}$.  Since $I\lhd R$ is $\psi_{ik}\phi_{kj}\in I$ and $\phi_{ik}\psi_{kj}\in I$, which means that $I\lhd R\implies M_n(I)\lhd M_n(R)$.  To show that every ideal of $M_n(R)$ are of the form $M_n(I)$ with $I$ ideal: Suppose $I$ is not an ideal of $R$, then exists $r\in R$ such that $rx\notin I$ or $xr\notin I$, then with taking a matrix $\phi\in M_n(I)$ such that $x$ is an entry of $\phi$ and a matrix $\psi$ such that $r$ is an entry of $\psi$ we get that the matrices $\phi$ and $\psi$ have an entry $xr$ or $rx$, in either case this is not an element of $I$ then the matrix is not in $M_n(I)$. This concludes my proof (is it?) for $(1)$.  For $(2)$ I'd like to proceed by absurdum . Let's say that $M_n(R)$ is not a simple ring, then it has at least one ideal. Denote by $\varphi$ this ideal in $I$, then for any matrix $\psi$ we have $\varphi\psi \in M_n(I)$ and $\psi \varphi \in M_n(I)$.  Now, can $R$ be a ring with division?.I'm not sure how to prove that it can't. I believe I could use this theorem: "If $I\lhd R$ and $1\in I$ then $I=R$. The case $I\neq R$ implies $1\neq R$ and since $\varphi \psi$ and $\psi \varphi$ are in $M_n(I)$ then there are at entries in the product matrix which belong to $I$ (using $I \lhd R$) but are not $1$ (1\notin I); then there are elements in $I$ (and therefore in $R$ by $I\subset R$) that are not invertible, follows that $R$ is not a ring with division. But if $I=R$?, here I can't see a why $R$ can't be a ring with division.
95056
Well i tried to prove it but failed. They are both commutative and have the same set of units. So i can't see a property of one ring which the other ring does not possess. Thank you for any help.
106261
I'm having trouble with this homework problem (from Algebra by Hungerford).     If $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring.   We've just proven in the previous part that if $S$ is a ring with identity and with no proper left ideals, then $S$ is a division ring. So the strategy is to suppose $S^2 \neq 0$ and prove that $S$ has an identity. The problem also has a hint:     Hint: Show that $\{a \in S \mid Sa = 0\}$ is an ideal. If $cd \neq 0$, show that $\{r \in S \mid rd = 0\} = 0$. Find $e \in S$ such that $ed = d$ and show that $e$ is a two-sided identity.     Here's what I have so far, with some boring details removed:  Note that $S_0 = \{a \in S \mid Sa = 0\}$ is an ideal, because [proof it's an ideal].  Since $S$ has no proper left ideals, we must have $S_0 = 0$ or $S_0 = S$. This means that either $S^2 = 0$ (if $S_0 = S$) or for every non-zero $a \in S$, there is some $b \in S$ with $ba \neq 0$ (if $S_0 = 0$). From now on suppose $S_0 = 0$, and we will show that $S$ is a division ring.  Note also that $A(d) = \{r \in S \mid rd = 0\}$ is a left ideal, because [proof it's a left ideal].  Now if $cd \neq 0$ for some $c, d \in S$, then $A(d) \neq S$ because $c \notin A(d)$; but $S$ has no proper left ideals, so $A(d) = 0$. Then $rd = sd$ if and only if $r = s$, for otherwise $(r-s)d = 0$ contradicts $A(d) = 0$.  Consider the left ideal $I_a = \{ra \mid r \in S\}$. [proof that it's a left ideal]. Since $S$ has no proper left ideals, we must have $I_a = 0$ or $I_a = S$ for all $a \in S$.  If $cd \neq 0$ then $I_d = S$ (so $I_d = S$ for all non-zero $d \in S$, since we showed that for all $d$ there is a $c$ with $cd \neq 0$), and therefore $d \in I_d$ so there is some $e \in S$ with $ed = d$. As we observed before, $rd = sd$ iff $r = s$, and $(re)d = r(ed) = rd$, so $re = r$ for each $r \in S$.    In summary, for every $a \in S$ we have some $e \in S$ (specific to this $a$) such that $e a = a$ and $r e = r$ for all $r \in S$. I feel like it should be very easy to show that there is a two-sided identity (e.g. by showing that all of these $e$ are in fact the same), but I can't seem to get there.  Thanks for any help!
95653
In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?  I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated
116313
In answering  I was surprised how hard it was to find anything on the Web about the generalization of the following fact to commutative rings:     A square matrix over a field has trivial kernel if and only if its determinant is non-zero.   As Bill demonstrated in the above question, a related fact about fields generalizes directly to commutative rings:     A square matrix over a commutative ring is invertible if and only if its   determinant is invertible.   However, the kernel being trivial and the matrix being invertible are not equivalent for general rings, so the question arises what the proper generalization of the first fact is. Since it took me quite a lot of searching to find the answer to this rather basic question, and it's  to write a question and answer it to document something that might be useful to others, I thought I'd write this up here in an accessible form.  So my questions are: What is the relationship between the determinant of a square matrix over a commutative ring and the triviality of its kernel? Can the simple relationship that holds for fields be generalized? And (generalizing with a view to the answer) what is a necessary and sufficient condition for a (not necessarily square) matrix over a commutative ring to have trivial kernel?
69528
I have to do the following question:  Given a unital ring $R$ and a natural number $n$ $\geq 1$, identify (with proof) the prime ideals of the matrix ring $\mathbb{M}_{n}(R)$.  I understand the theory, but putting it all together is proving difficult.   I understand that the prime ideals of $\mathbb{M}_{n}(R)$ should be $\mathbb{M}_{n}(\mathbb{P})$.  Can anyone suggest any steps as to how to prove this?