Let $R$ be a ring. Set $M=\{ x\in R|\forall y\in R\Rightarrow 1-xy\;\; \text{unitary in}\; R\}$ and $J\left( R\right) =\cap \{ I|I\unlhd R, I \text{is a maximal ideal}\}$.      How to show that $M\subseteq J\left( R\right)$?
38498
I have in my notes that the Jacobson radical of a ring $R$ is:  $J(R) = \cap${$I$ | $I$ primitive ideal of $R$} $= \cap$ {$Ann_R M$ | $M$ simple $R$-module}.  I have now seen elsewhere that $J(R) = \{x ∈ R: xy-1 ∈ R^\times \text{ for all } y ∈R\}$  I was just wondering would anyone be able to provide an explanation as to how to get from one definition to the other? Thanks.
95198
Let $R$ be a commutative ring with identity. Assume that $R$ has exactly three distinct ideals: $\{0\},I, R.$      1) Show that if $a \in R-I$, then $a$ is a unit in $R$.      2) Let $a,b\ne0$ in $I$. Show that $ab=0$.
53351
Prove: Let $R$ a Artinian commutative ring with $1 \neq 0$. If $I$ is prime, then $I$ is maximal.     I got stuck on this. I understand that every ideal is generated by finitely many elements. Here is my approach:  It's enough to prove the following implication: $$ R/I \ \text{is a domain} \quad \Rightarrow \quad R/I \ \text{is a field}$$ Then only finding an inverse is left. So let $x \in R$ in arbitrary. We have to find an inverse element of $x+I \in R/I$, that as an element y so that $x \cdot y \in I+1$. We denote $I = (x_1, x_2, \cdots, x_n)$.  Now I have to find y so that $xy -1 = r_1 x_1+ r_2 x_2 + \cdots + r_n x_n$. This is where I no more knew what to do.   Your advise will be appreciated.
77094
Let $R =\{f:\{1,2,3,4,\cdots ,10\}\to \mathbb{Z}_2 \}$ be the set of  all $\mathbb{Z}_2$ -valued functions on the set $\{1,2,3......10\}$  of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?     1) $R$ has a unique maximal ideal.      2) Every prime ideal of $R$ is also maximal.      3 ) The number of proper ideals of $R$ is $511$.      4 ) Every element of $R$ is idempotent.   R is finite commutative ring so  every prime ideal  is maximlal ideal ,so I think option  2 is true   .But for the other options   I am  helpless  .
86956
If $I$ and $J$ are ideals in a ring $R$ with 1 which are  co-maximal, i.e $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n$ in $\mathbb{N}$  Work done:     Should I proceed using    By taking $$K=I^m+J^m \neq R$$  Let $$S=\{I,J \in R \mid I^m+J^n \neq R \}$$ be the required partially order set ordered by inclusion then  $$I+J \subseteq I^n+J \subseteq I^n+J^m \subseteq \ldots $$  Then, I think this chain doesn't have an upper bound...as $I^m+J^n \neq R$ for all $ m,n \in \mathbb N$. Am I right??  Thanks for any help!!
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?
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 $R/m^k$?
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?
19973
Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?
44835
If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$?  That is, is $I$ necessarily closed under multiplication by elements of $R$?  Surely $I$ is closed under multiplication by elements of $J$ (since $I$ is an ideal of $J$).  The "obvious" approach to proof fails so quickly that it must be false, that "an ideal of an ideal is not necessarily an ideal of the big ring".  Please provide a counterexample.  (Or a proof?)
54581
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. Let $s$ be the maximum number of linearly independent elements of $M$, while $t$ is the minimum number of a system of generators of $M$. Show that $s\leq t$.  It is immediate in the case of vector spaces, but my first attempt (explicit computation) didn't work here due to the fact that there are coefficients (elements of $R$) without an inverse. I tried in so much other ways spending a lot of time, and I should be very glad if you cold give me a precise, complete and rigorous proof. Thank you in advance.
27785
So I basically have to prove what is on the title.      Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.   One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.   I am having troubles proving the other inclusion. Any comments?  Thanks
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..
76124
Let $R = \mathbb{Z}_{(2)}$ be the localization of $\mathbb{Z}$ at the prime ideal generated by $2$ in $\mathbb{Z}$. Then prove that the ideal generated by $(2x-1)$ is maximal in $R[x]$.
36400
I'm making some exercises to prepare for my ring theory exam:     Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal   of $ℤ[i]$.   I know that $a+2l=b$ with $l\in ℤ$ (or should I say $l \in ℤ[i]$)? Therefore I can write $a+(a+2l)i=a+ai+2li$. So I don't have $0+i$ in my ideal. For even real parts I have even complex parts, and for odd real parts I have odd complex parts. But I don't see what I can conclude now.
40037
I have to prove that if $I\subseteq R$ is an ideal, and $1\in I$, then $I=R\,$. So I know $I\subseteq R$ is an ideal if $a,b\in I$ implies $a+b\in I$, and if $a\in I$, $r\in R$, then $ra\in I$. I'm finding it hard to put this into words. Since $1\in I$ and $I\subseteq R$ is an ideal, then let $a=1$ and $r=1$ so $1\cdot 1\in R\,?$
22268
Let $I$ ba a two-sided ideal of a ring $R$ and $M$ be an $R-$ left module. Prove that if $M$ is an $R-$ projective left module then $M/IM$ is an $R/I-$ projective left module.  It is easy to see that $M/IM$ has the structure of $R/I-$module. I need some ideas to prove that $M/IM$ is an $R/I-$ projective left module.  Thank you very much.