Let $F$ is a field, and $R = M_n(F)$. Prove that  (i) Every minimal left ideal of $R$ are isomorphic to each other  (ii) Prove every simple $R$-module are isomorphic to each other  I have if $M$ is a simple $R$-module, then there is a maximal left ideal $I$ of $R$ such that $M \cong R/I$. Now the big problem of mine is that I can't determine how the maximal ideal of $R$ look like? Why $R$ is semi-simple?
12199
I've a basic question.      $R$ is a simple left artinian ring. I want to show that all simple left $R$-modules are isomorphic.    A simple $R$-module $M$ is isomorphic to $R/J$ where $J$ is a maximal left ideal of $R$. Does this imply that $M\cong Ra$ for some $a\in R$, $a\neq 0$?   Many thanks.
7149
This is a very simple question but I believe it's nontrivial.  I would like to know if the following is true:      If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic.    Thanks!  If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true.
15106
Let $M,N$ be $R$-modules with $M$ finitely generated and let $S\subseteq R$ be multiplicatively closed. Then there exists a module isomorphism   $$S^{-1}\text{Hom}_R(M,N) \xrightarrow{\sim} \text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N).$$   As homework, I found a 'proof' of this statement. However,  claims to prove me wrong. I am having trouble in finding the error in my proof. Could someone help me out?  I did prove there exists a map $\phi:S^{-1}\text{Hom}_R(M,N) \to\text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ defined by $f/s\mapsto(m/t\mapsto f(m)/st)$, and I am fairly confident that it contains no mistake.     Proof: Let $m_1,\dotsc,m_n$ be the generators for M. Let $f\in \text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ be given.   Choose $n_i\in N$ and $s_i\in S$ such that $f(m_i/1)=n_i/s_i$.   Define $\pi_f = s_1\dotsm s_n$. Then   $$\pi f(m/1) = \pi f\left( \sum_{i=1}^n r_i m_i \right) = \pi \sum_{i=1}^n r_i f(m_i) = \pi \sum_{i=1}^n r_i n_i/s_i = \left(\sum_{i=1}\left(\prod_{j=1, j\neq i}^n s_i\right)r_in_i\right)/1,$$   so $\pi f(m/1) \in N$ using the canonical embedding in $S^{-1}N$, so $m\mapsto \pi f(m/1)$ is a map $M\to N$ and is clearly an homomorphism.   If we define the map $\psi:\text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)\to S^{-1}\text{Hom}_{R}(M,N)$ as $$ f \mapsto (m\mapsto \pi_f f(m)) / \pi_f,$$ then we clearly have a homomorphism if it is well-defined.      Proof: It remains to check that the element we map to does not depend on choice of $\pi_f$.   But if we have both $\pi_f$ and $\pi_f'$ for which $\pi_f f$ and $\pi_f' f$ are homomorphisms to $N$, then   $$(\pi_f f(m))/\pi_f = (\pi_f\pi_f' f(m))/(\pi_f\pi_f') = (\pi_f' f(m))/(\pi_f')$$   for all $m\in M$, so $\psi$ is in fact well defined.   It is now easy to check that $\psi$ and $\phi$ are inverses, which makes them homomorphisms.
21427
Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?     If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.   If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.  I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.
97139
Let $R \simeq S$ be isomorphic commutative rings with unity. Prove the following:  a). If $R$ is an integral domain then $S$ is an integral domain.  For this, I said that If I let $f: R \to S$ be an isomorphism and $(R, + , \circ), (S, \oplus, \ast)$, then let $x, y\in R$,  such that $x \circ y = 0_R$, because $f$ is an isomorphism, $f(x \circ y) = f(x) \ast f(y) = f(0_R)$. If I let $R$ be an integral domain then, as per the definition $R$ is commutative, has unity, and no zero divisors.   So $f(x \circ y) = f(x) \ast f(y) = f(0_R)$ where we have either $f(x \circ y) = f(0_R) \ast f(y) = f(0_R)$ or $f(x \circ y) = f(x) \ast f(0_R) = f(0_R)$.   Since $f$ is an isomomorphism, let $s = f(x), t = f(y)| s,t \in S$. Then $s \ast t = f(0_R)$. By definition, $S$ also has unity so $0_S$ exists. If we map $0_R \to 0_S$. Then either $s = 0_S$ or $t = 0_S$.   So then, since $S$ has unity, is commutative, and has a no zero divisors as we showed above, it is also an Integral domain.  I'm not sure I have this correctly.  b). If $R$ is a field then $S$ is a field. I don't really understand how to do this one. Since $S$ is an integral domain, it has a $0_S$. So I need to show that $S$ distributes $\ast$ over $\oplus$.  Any help with these is appreciated
121070
I want to understand why the following is true:     Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.   So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:  1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.  2) Why exactly at most $k$ ?  Can you please explain?
34488
I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over.     Let $R$ be a commutative ring with identity and let $U$ be an ideal that is maximal among non-finitely generated ideals of $R$. I wish to show that $U$ is a prime ideal.   Assume that $U$ is not prime. Let $x, y\not\in U$ be such that $xy\in U$. $U$ is contained in a maximal ideal $M$ and $xy\in M$, so either $x$ or $y$ is in $M$; assume $x\in M$. The condition $U\subset M$ then implies that there is a ring homomorphism $$\varphi: R/M\to R/U$$  Since $R/M$ is a field, $\varphi$ is injective. Hence, $\varphi(x)\in U$. This is a contradiction, so $U$ must be prime.  The thing that worries me is that I never explicitly used the hypothesis that $U$ was not finitely generated or the result that $M$ must be finitely generated.
53422
Let $f :R \to S$ be a surjective ring homomorphism , $M$ be a maximal ideal of $S$ , I am writing a proof showing $f^{-1}(M)$ is a maximal ideal of $R$ , Please verify whether it is correct or not .   Proof :- Let $J$ be an ideal such that $f^{-1}(M) \subset J \subseteq R$ , we want to show $J=R$ i.e. $R \subseteq J$. Now as  $f$ is surjective , $M=f(f^{-1}(M)) \subseteq f(J) \subseteq S$ . As $f$ is surjective , $f(J)$ is an ideal of $S$ . Now if it were possible that $M=f(J)$ , then $x \in J \implies f(x) \in f(J)=M \implies x \in f^{-1}(M)$ , so $J \subseteq f^{-1} (M)$  ,contarry to our assumption $f^{-1}(M) \subset J$ . Thus $M \ne f(J)$ , $f(J)$ is an ideal of $S$ containing $M$ ; since $M$ is maximal ideal in $S$ , so $f(J)=S$ . Then $x \in R \implies f(x) \in S=f(J) \implies \exists j \in J : f(x)=f(j) \implies f(x-j)=0_S \in M $  $\implies x-j \in f^{-1} (M) \subset J \implies x=x-j+j \in J$ , so $R \subseteq J$   Is the proof correct . Please comment .
99175
Given $F$ a field and $H$ a finite group and $H_n=\sum_{h \in H}h \in F[H]$, I'm trying to show that the left ideal $(H_n)\subset F[H]$ is a simple submodule.  Now I'm not really sure what the ideal looks like, I think it will be just the elements of $H$ but I don't see how simplicity follows.
72488
If $R$ is a commutative ring with $1\ne 0$ and $R^m \cong R^n$ as direct sums of $R$-modules, then $m=n$.   There is a hint to this problem (I don't know how to use it): Reduce this to the case of a field.  I noticed that there was . But as a beginner of commutative algebra, I don't know if my question is closely related to that one. Thanks in advance!
80299
I want to prove the following result:      Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.    I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:   $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal; If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification); The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.   I'll appreciate any comment.
12858
Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring?    This is a well known result for $k = 1$, as $R/m$ is a field. It seems to be true in other cases, e.g. $p\mathbb{Z}\subset \mathbb{Z}$ for prime $p$ and for $(x,y) \subset \mathbb{F}[x,y]$, the maximal ideal generated by $x$ and $y$ in a polynomial ring over the field $\mathbb{F}$.   Equivalently, if an element $x\notin m$, is $x + m^k$ an invertible element in $R/m^k$?
54222
Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R.$ Prove that if $I$ is maximal then $R/I$ is a field.   This problem is probably trivial for most of you abstract algebra fans out there, but remember that I'm new to the field, so please don't use some next-level math that I can't understand at all. I think this problem could be solved using next-level math very quickly though.   For this problem, I know that fields are commutative, unital, and every nonzero element is a unit/invertible. Also, $I$ is maximal wrt subset, so if there is an ideal that contains $I,$ then it must be $R.$
60282
Let $R$ be a commutative ring with $1$ and $M_1,...,M_n$ be distinct maximal ideals in $R$. What I want to show is $M_1\cap\cdots \cap M_n=M_1\cdots M_n$. If I can show that $M_1\cdots M_n$ is radical, then the proof ends since $M_1\cap\cdots\cap M_n=rad(M_1\cdots M_n)$. How can I prove this?
73093
Let $f:R\to S$ be a surjective homomorphism, where $R$ is a commutative ring and $S$ is a field. Prove that $\ker(f)$ is a maximal ideal.  I already know that $\ker(f)$ is an ideal of $R$. I tried to consider some ideal $J$ of $R$ such that $\ker(f) \subset J$. If we can show that for arbitrary $y\in J$, $f(y)=0$ then we are good. But I don't know how to show that. Specifically, how does being surjective come into play?  Another theorem I know is that if $f$ is a surjective homomorphism, then quotient ring $R/\ker(f)$ is isomorphic to $S$. Don't know if that's gonna help.
88321
Let $R$ be a (not necessarily commutative) ring and suppose $R$ has a maximal left ideal $I$ consisting of nilpotent elements, that is for $x\in R$ there exists some $n\in\mathbb{Z}_{\geq 0}$ such that $x^{m}=1$. I want to show that in this case, $I$ is the unique maximal left ideal.  To attempt to prove this, assume $J$ is some other maximal left ideal of $R$. Since $x^{m}=0$, we have $x^{m}\in J$. This is where I get stuck.  In the commutative case we can note that $I$ is necessarily a prime ideal, so either $x\in J$ or $x^{m-1}\in J$. Iterating this process shows that $x\in J$, then $I=J$ by maximality.  In the non-commutative case I can't seem to proceed. This is probably really easy but I'd appreciate a hint.
55550
My first instinct is to go to the 1st isomorphism theorem and say that the two rings are isomorphic if there is a field where $(x^2)=(x^2-1)$, which doesn't seem terribly promising.  But then it seems to me that $F[x]/(x^2)$ is the ring you get when you take $F$ and adjoin an element that satisfies the relation $x^2=0$, which would be $0$ and every field's got one.  Same with $F[x]/(x^2-1)$ which is $F$ adjoined by an element satisfying $x^2=1$.  $1$ does the trick and every field's got one of those too.  Any hints (or answers!) would be appreciated.  Thanks...