I'm trying to figure out all the possible right ideals of $M_{n\times n}(\mathbb{F})$.   I know that the two-sided ideals are only the trivial ones, and the proof uses the fact that we can operate on the rows and line of a matrix to make $E_{ij}$ - the matrix that has one in ij and zero in the rest of it.  Does anyone have any idea(l)?  Thanks.
4257
What are the left and right ideals of matrix ring? How about the two sided ideals?
63833
Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.     Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.   This is what I tried.  Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$. This is a well-defined bilinear map.  By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.  This map is surjective.  So all we need to show is that the kernel of $F$ is $0$.  But I am unable to do this.
55082
How do you construct, for each $n\geq 1$, an ideal in $\mathbb Z[x]$ of the form $(a_1,a_2,\dots,a_n)$ with $a_i\in \mathbb Z[x]$ such that it is impossible to have $(b_1,b_2,\dots,b_m)=(a_1,a_2,\dots,a_n)$ with $m&lt;n$ and $b_j\in\mathbb Z[x]$?   Thank you!
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.
72544
Let $A$ be an $n \times n$ matrix with integer coefficients and nonzero determinant. Can we say something about $ \mathbb{Z}^n /\ \text{im}( \phi )$ (here $\phi : v \mapsto Av$ )?    This problem has arised as I was solving some problem in homology theory.
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
59473
I wanted to find a quick proof of the following well-known fact. Since I couldn't easily find a reference, I provide a proof below.     Let $H$ be a separable Hilbert space, and $J\subset B(H)$ be a (not necessarily closed) two-sided ideal. If $T\in J$ is non-compact, then $J=B(H)$.    I'm also interested to know if there is a shorter/simpler argument.
21262
Let $X\in\mathbb{A}^n$ and $Y\in\mathbb{A}^m$ be affine varieties (irreducible algebraic sets). Then, if we denote by $i_X$ and $i_Y$ their ideals in their respective affine spaces, we define their product in $\mathbb{A}^{n+m}$ as the set $$ X\times Y = \{(x,y):x\in X\text{ and } y\in Y\} = \{(x,y):x\in Z(i_X)\text{ and }y\in Z(i_Y)\}. $$ Using the second characterization it's easy to show that in fact $X\times Y$ is closed in $\mathbb{A}^{n+m}$ and particularly that $X\times Y = Z(I_X+I_Y)$, where $I_X$ and $I_Y$ are the vanishing ideals of $X\times\mathbb{A}^m$ and $\mathbb{A}^n\times Y$ (these ideals clearly are in bijective correspondence with $i_X$ and $i_Y$, respectively).  Then, by the Nullstellensatz, we get $I(X\times Y) = \sqrt{I_X+I_Y}$. The problem is that the references I've seen claim that in fact $I(X\times Y) = I_X+I_Y$ (, example 2.3.9 in page 25, and Georges' answer in , specially the last display).   This would only be true if $I_X+I_Y$ is radical, which I haven't been able to show for this case and know isn't true in general (see ). Are my references wrong, or what am I missing?
17207
The question is this:      Let $f:\mathbb C[x,y]\rightarrow\mathbb C[t]$ be the homomorphism that sends $x\mapsto t+1$ and $y\mapsto t^3-1.$ Determine the kernel $K$ of $f$, and prove that every ideal $I$ of $\mathbb C[x,y]$ that contains $K$ can be generated by two elements.   Solution:  I have shown that $((x-1)^3-y-1)=K.$ However, I am having trouble showing the second part. Any hints??
106259
Dummit and Foote, 7.4.33(a): Let $R$ be the ring of all continuous functions $[0,1] \to \mathbb{R}$ and let $M_c$ be the kernel of evaluation at $c \in [0,1]$, i.e. all $f$ such that $f(c) = 0$. Show that if $M$ is a maximal ideal in $R$ then $M = M_c$ for some $c$.    I am aware of proofs that use the compactness of $[0,1]$; however I have been told that a simpler proof exists using only the isomorphism theorems for rings along with basic facts about ideals. I have tried on my own to prove it this way but still have no success; is it possible?  (My attempts go something like: since $M$ is maximal, the quotient ring $R/M$ is a field. Then it would be nice to use the fact that $R/M$ has no zero divisors, but I can't quite make it work. Alternatively, consider $M \cap M_c$.)
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.
68376
Let $A$ be a ring and let $J$ be a right-sided ideal of $A$. We call the set $I_{A}(J)=\lbrace a \in A \mid aJ\subset J\rbrace$ the idealizer of $J$.  Show that $I_{A}(J)$ is the largest subring of that $A$ containing $J$ as an ideal.
95617
Question: Let $V \subset M(n,n,\mathbb{R})$ be the set of all symmetric, real $(n \times n)$ matrices, that is $a_{ij} = a_{ji}$ for all $i,j$. Show that $V$ is a subspace of $M(n,n,\mathbb{R})$ and calculate dim$(V)$.  My attempt so far: First part: To show that $V$ is a subspace I need to show:  (a) $ 0 \in V$ and   (b) $\forall A,B \in V:   (i) A + B \in V   (ii) \lambda A \in V$   For (a) I would say: Let $a_{ij} \in 0$(this should represent a zero matrix, is that how to write it?)   $a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V$   For (b) I am actually confused since I would first think: both a $(2 \times 2)$ matrix and a $(3 \times 3)$ matrix belong to $V$ but addition of matrices of different size is undefined $\Rightarrow$ $V$ is not closed under addition $\Rightarrow$ $V$ is not a subspace of $M(n,n,\mathbb{R})$... what am I missing here? (To start I don't really understand the notation $M(n,n,\mathbb{R})$... what exactly does the $\mathbb{R}$ represent there?).  Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use "$\in$" to denote "is an element of matrix"?  (i)Let $a_{ij},a_{ji} \in A$ and $b_{ij}, b_{ji} \in B$. Let $A,B \in V$   $\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}$   $A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})$   $\Rightarrow C \in V$   (ii) Let $A\in V, \lambda \in \mathbb{R}$. Let $a_{ij},a_{ji} \in A$.  $\Rightarrow a_{ij} = a_{ji}$   $\lambda \cdot A = A&#39;$ with $\lambda a_{ij} = \lambda a_{ji} \Rightarrow A&#39; \in V$   Second part: I feel that I understand the answer... For an $(n \times n)$ matrix, the diagonal length $ = n$ and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total$(n^{2})$, half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim$(V) = n + \frac{n^{2}-n}{2}$.   Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable?
28416
Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$?     deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field.  In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the  $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies  $$M \text{adj}(M) = \det(M).$$  If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus:     If $M$ is a counterexample, then $\text{adj}(M) = 0$.   When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case.   Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as $$M^k q(M) = 0$$  where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term.  Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence     If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent.    This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that $$q(M) = M - \lambda$$  where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case.   Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so     We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$.
63514
Let $R$ be a commutative ring with $1$ and let $J$ be a proper ideal of $R$ such that $R/J \cong R^n$ as $R$-modules where $n$ is some natural number. Does this imply that $J$ is the trivial ideal?   Basically I am trying to prove/disprove that if $J$ is a proper ideal of $R$ and $R/J$ is free then $J=0$ and above is my work.
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
25570
I want to prove that the kernel of the evaluation map $s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i$ where $a = (a_1,\dots, a_n) \in \mathbb{C}^n$ is the ideal generated by $\{x_1 - a_1, \dots, x_n - a_n\}$.    The proof in the book first shows it easily for the case $a = 0$, and says that by substitutions of variables $x_i' = x_i - a_i$, you can show it for the rest of the cases.  I know that $f \in \ker s_a$ iff $f(a) = 0$.  And $\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1 - a_1, \dots, x_n-a_n], x_i \mapsto x_i - a_i$ is a surjective ring homomorphism.  Please give a hint and not the full answer.  Thanks.