I'm sorry for asking such a simple question but I'm having a hard time thinking about all possible ideals of $M_2(Z)$. So let's consider an arbitrary matrix $\begin{pmatrix} a &amp; b \\ c &amp; d\end{pmatrix}$ where $a,b,c,d \in Z$. How can I start constructing ideals?
4257
What are the left and right ideals of matrix ring? How about the two sided ideals?
108762
I'm trying to track down an example of a ring in which there exists an infinite chain of ideals under inclusion. (i.e. $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq...$)
39872
What are the proper ideals, prime ideals, maximal ideals of $\mathbb{Z}[i]$, the ring of Gaussian integers.    Check whether $(1+i)$ is prime or maximal ideal.     Can someone help me please. I have no idea how to crack this problem. Thanks for your time.
108734
I can prove that the ideal $(4, 2x, x^2)$ in $\mathbb Z[x]$ is not principal. But I failed to prove that this cannot be generated by two elements.  It's really difficult for me. Would you give me a hint.
48295
Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$.   I tried proving that the only elements outside the ideal are the invertibles of $K$ (I should still prove that this implies maximality, but it shouldn't be too difficult).  Is there a better strategy, or another stategy?
123656
We are given ($a_1,a_2,....,a_m$) and ($b_1, b_2,....,b_n$) sequences with non-negative integers.   Decide whether it's possible and if it is construct a matrix $\Re^{m x n}$ of ones("1") and zeros("0") where the number of ones("1") in row $i$ is $\forall i $ : $a_i$ and in column $j$ is $\forall j$ : $b_j$   Any hint is greatly appreciated.
97142
Please give me a hint for this problem.  Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then $$ f  \   \colon (R/I) \otimes_{R} M \to  M/IM$$  defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.   And show that if $L$ is a left ideal, then  $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
32992
I'm self studying Steven Roman's Advanced Linear Algebra, and this is problem 10 of Chapter 8.      Let $A,B\in M_2(\mathbb{C})$, $A^2=B^3=I$, $ABA=B^{-1}$, but $A\neq I$ and $B\neq I$. If $C\in M_2(\mathbb{C})$ commutes with $A$ and $B$, then $C=rI$ for some $r\in\mathbb{C}$.   Is there a way to solve this without writing out arbitrary matrices and attempting to solve a huge system of equations? The only thing I observe is that $A=A^{-1}$, so $B\sim B^{-1}$, so $B$ and $B^{-1}$ have the same characteristic polynomial. I'm stymied trying to show $C$ is diagonal, let alone a multiple of $I$. Thanks for any ideas.  I should add that I know that the center $Z(M_n(\mathbb{C}))$ consists of scalar multiples of $I$, but I don't see any reason to assume or prove $C$ commutes with everything.
110838
Let A be a commutative ring with unity ring, M be a maximal ideal and $b\notin M$ . Show that there exists $q \in M, s\in A$ such that for every $a \in A$ we have $a=sba+qa$. Show that this is true only when $M$ is a maximal ideal.   Suppose if we take $A=2\mathbb Z$ and $M=4\mathbb Z$, $a=6.$ Then by the given statement $\exists$ $s\in A$, $q\in M$ s.t. $6=6(2s+q)$ $\implies$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$  Please someone give some hints. how can solve this.  Thank you..
43419
Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring $A$ (not necessarily commutative), the matrix ring $M_n(A)$ is simple if and only if $A$ is simple.   How to prove?
60832
It well known that if $R$ is a commutative ring then every ideal of $M_n(R)$ is $M_n (I)$ where $I$ is an ideal of $R$. My question is: Suppose that $I$ and $J$ are ideals of $M_n(R)$, i.e., $I=$M_n(A)$ and $J=$M_n(B)$. When is the statement $M_n(AB)=M_n(A)M_n(B)$ correct?
95042
I am looking to find the left, right and two sided ideals of the ring R = $\begin{bmatrix} \mathbb{Q} &amp; \mathbb{Q}\\ 0 &amp; 0 \end{bmatrix}$. It is in finding the left ideals that I am stuck. Finding the Right Ideals To find the right ideals I considered a nonzero element $\begin{bmatrix} a &amp; b\\ 0 &amp; 0 \end{bmatrix}$ of an arbitrary right ideal $I_{r}$. I  then considered two cases. Case 1: $a\neq0$ We have $\begin{bmatrix} a &amp; b\\ 0 &amp; 0 \end{bmatrix}\begin{bmatrix} a^{-1} &amp; 0\\ 0 &amp; 0 \end{bmatrix}=\begin{bmatrix} 1 &amp; 0\\ 0 &amp; 0 \end{bmatrix}\in I_{r}$. Hence for any $\begin{bmatrix} c &amp; d\\ 0 &amp; 0 \end{bmatrix} \in R$ we have: $\begin{bmatrix} c &amp; d\\ 0 &amp; 0 \end{bmatrix}=\begin{bmatrix} c &amp; 0\\ 0 &amp; 0 \end{bmatrix}\begin{bmatrix} 1 &amp; 0\\ 0 &amp; 0 \end{bmatrix}+\begin{bmatrix} 1 &amp; 0\\ 0 &amp; 0 \end{bmatrix}\begin{bmatrix} 0 &amp; d\\ 0 &amp; 0 \end{bmatrix} \in R$. That is $I_{r}=R$. Case 2: $a=0$ We have $K=\{b\in\mathbb{Q}\mid\begin{bmatrix} 0 &amp; b\\ 0 &amp; 0 \end{bmatrix}\in I_{r}\}$ is an additive subgroup of $\mathbb{Q}$ and $I_{r}=\begin{bmatrix} 0 &amp; K\\ 0 &amp; 0 \end{bmatrix}$. Moreover one can verify that $\begin{bmatrix} 0 &amp; K\\ 0 &amp; 0 \end{bmatrix}$ is a right ideal of $R$. Hence in this case any ideal is of the form $\begin{bmatrix} 0 &amp; K\\ 0 &amp; 0 \end{bmatrix}$ where $K$ is an additive subgroup of $\mathbb{Q}$. Finding the Left Ideals I have tried a similar strategy as above. I considered a nonzero element $\begin{bmatrix} a' &amp; b'\\ 0 &amp; 0 \end{bmatrix}$ of an arbitrary left ideal $I_{l}$. I then attempted to consider two cases. Case 1: $b'=0$ We have $K'=\{a'\in\mathbb{Q}\mid\begin{bmatrix} a' &amp; 0\\ 0 &amp; 0 \end{bmatrix}\in I_{r}\}$ is an additive subgroup of $\mathbb{Q}$ and $I_{l}=\begin{bmatrix} K' &amp; 0\\ 0 &amp; 0 \end{bmatrix}$. Moreover one can verify that $\begin{bmatrix} K' &amp; 0\\ 0 &amp; 0 \end{bmatrix}$ is a leftideal of $R$. Hence in this case any left ideal is of the form $\begin{bmatrix} K' &amp; 0\\ 0 &amp; 0 \end{bmatrix}$ where $K'$ is an additive subgroup of $\mathbb{Q}$. It is for the second case that I am stuck, I have no idea how to proceed. Moreover when finding the ideals, I will simply check whether each left/right ideal I find is a right/left ideal. Overall is this the best way to have approached this question, it feels awfully long.
98754
Consider the subring $R\subset\Bbb Q$ , $R=\{\frac ab\ | \ a,b\in\Bbb Z, b\text{ odd}\}$  I am struggling with the following questions:  (1) Prove that the ideals of $R$ are $\{0\}$ and $2^nR$ for $n\ge 0$  (2) Prove that $R$ has 2 prime ideals and 1 maximal ideal  For (1) I can see that $2^nR$ is an ideal for $n\geq0$ but I'm not sure how to show all ideals other than $\{0\}$ are of this form  With question (2),  $2^nR\subsetneq 2^{n-1}R\subsetneq...\subsetneq 2R\subsetneq R$  so $2R$ is clearly the only maximal ideal and is therefore prime, but I am not sure where to go from here.  Any suggestions?
13664
Let $S$ be the ring of all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$?   Thanks in advance.
47255
Consider the matrix $$ M=\left(\begin{matrix} 0&amp;0&amp;0&amp;1\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ \end{matrix}\right). $$ Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n&gt;1$?    One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?
66721
Let $aR$ be a nonzero ideal in a PID $R$. Show that $R/aR$ is a ring with only finitely many ideals.  Honestly, I do not know how to start. Appreciate any tips.
91928
I am trying to mimic the interesting proof from  to give a generating set of the ideal correspond to the $d$-uple embedding: We fix the notation $\Delta = \big\{ (\nu_0, \dots, \nu_n) \in {\mathbb{N}}^{n+1}_0 ~:~ \nu_0 + \dots + \nu_n = d \big\}$. The polynomial map is $\theta : k[y_\nu : \nu \in \Delta] \rightarrow k[x_0, \dotsc, x_n]$ given by $$ \theta ( y_\nu ) := x^{\nu_0}_0 \cdots x^{\nu_n}_n $$ and we know that ${\mathrm{ker}}~\theta$ is the ideal of the $d$-uple embedding, i.e., $\rho_d({\mathbb{P}}^n) = Z({\mathrm{ker}}~\theta)$ (cf. Exercise I.2.12 Hartshorne). Now let $W \leq k[y_\nu : \nu \in \Delta]$ denote the ideal $$ W := \Big\langle y_{\tau_1} y_{\tau_2} - y_{\tau^{\prime}_1} y_{\tau^{\prime}_2} ~:~ \tau_1, \tau_2, \tau^{\prime}_1, \tau^{\prime}_2 \in \Delta, \tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2 \Big\rangle $$ Clearly, $W \subseteq {\mathrm{ker}}~\theta$. Using the lemma in the above link, we need to produce a $k$-subspace $T \subseteq k[y_\nu : \nu \in \Delta]$ with the conditions $$ T + W = k[y_\nu : \nu \in \Delta], \hspace{.2in} \theta \lvert_{T} = {\mathrm{injective}} $$ After a few thoughts it seems reasonable to think of the following subspace as a candidate for $T$: First define the equivalence relation on $\Delta \times \Delta$ given by $(\tau_1, \tau_2) \sim (\tau^{\prime}_1, \tau^{\prime}_2)$ if $\tau_1 + \tau_2 = \tau^{\prime}_1 + \tau^{\prime}_2$. Let $\Omega \subseteq \Delta \times \Delta$ be a subset with a unique representative from each class of $\sim$. Now let $T$ be the $k$-span of the monomials $y^{\alpha_1}_{\tau_1} \dotsc y^{\alpha_r}_{\tau_r}$ (with the distinct elements $\tau_1, \dotsc, \tau_r \in \Delta$ and $\alpha_1, \dotsc, \alpha_r \geq 1$) such that $(\tau_i, \tau_j) \in \Omega$ for each $i \neq j$. Clearly, $\theta$ is injective if restricted to $T$. I got lost in the step : Each monomial $M$ of $k[y_\nu : \nu \in \Delta]$ satisfy $M \equiv b$ mod $W$ for some $b \in T$. Did I do anything wrong?