As the title says I have to show that for $R$ a commutative ring with neutral element, $ R/ I=\{x+I\mid x\in R\}$ is a ring if $I$ is an ideal and it isn't if $I$ is only a subring. It's the first time I am hearing about Rings so it's not that easy for me. I am proving the Ring conditions and the first one is that it has to be an abelian group. Let's start with the associative property: We take $x,y,z \in R/I$: We have to show that $(x+I+y+I)+z+I=x+I+(y+I+z+I)$ For example on the left hand side I guess that I have to do $((x+y)+I)+z+I$ and then $(x+y+z)+I$ and same with the other side but what gives  me the right to write $(x+I+y+I)=((x+y)+I)$, why can I take $x$ and $y$ together and then the same logic with $z$ and also (this might be very trivial) why is the $I$ only counted once and not 3 times. And also, what does the Ideal give me in this specific case that the subring does not? Thanks in advance for the help.
124226
From Wikipedia:     Given a ring R and a two-sided ideal I in R, we may define an   equivalence relation ~ on R as follows:  a ~ b if and only if a − b is in I.       Using the ideal properties, it is not difficult to check that ~ is a   congruence relation. In case a ~ b, we say that a and b are congruent   modulo I. The equivalence class of the element a in R is given by  [a] = a + I := { a + r : r in I }.       This equivalence class is also sometimes written as a mod I and called   the "residue class of a modulo I".      The set of all such equivalence classes is denoted by R/I; it becomes   a ring, the factor ring or quotient ring of R modulo I, if one defines  (a + I) + (b + I) = (a + b) + I; (a + I)(b + I) = (a b) + I.    Suppose that we no longer restrict $I$ to be an ideal, but just any set that is a subset of $R$ and is not an ideal of $R$. Then would it be impossible to satisfy  (a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = (a b) + I all the time if $I$ is not an ideal?
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.
28057
Let $R$ be a ring of the real functions with "+" and "x" and let $I_1:=\{f(x)\in R\mid f(1)=0\}$, $I_1$ is a maximal ideal?   I think yes, but don't know how to prove it.
104072
Question: Suppose that $R$ is a commutative ring without zero-divisors.   Show that the characteristic of $R$ is either $0$ or prime.   I have established that every element in a commutative ring $R$ without zero divisors have the same additive order $n$.   Now, if no such additive order n exists, then the characteristic of $R$ is $0$.  Obviously, if a finite additive order exists, Char of $R$ is finite.  How do I show that Char of $R$ is prime? It probably involves lagrange's theorem and the order of the element in $R$.  Hint is appreciated.  Thanks in advance.
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?)
28637
Let $R$ be a commutative ring with 1, we define  $$N(R):=\{ a\in R \mid  \exists k\in \mathbb{N}:a^k=0\}$$  and  $$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$  Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?  I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.
42228
Let $R$ be a ring and for $x,y\!\in\!R$, define the commutator as $[x,y]:=xy-yx$. An $r\!\in\!R$ is idempotent iff $r^2=r$.  How can one prove, that if every commutator is idempotent, then the whole ring is commutative, i.e. all commutators are zero?
70983
How to go with proving this : Let $R$ be a ring and let $I$ be an Ideal of $R$.  Prove that the factor ring $R/I$ is commutative iff $rs-sr \in I$ $\forall r$ and $s$ in $R$  any hint how to go with this...
81640
Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain.   My proof:  Take $r,s,a \in R$ with $ar = as$, and $p \in P$. Then  \begin{align*} par &amp;= pas\\ a p r &amp;= a p s &amp;&amp; \text{R is commutative} \end{align*} and since $P$ is an ideal of $R$, then $pr, ps \in P$. Since $P$ is an integral domain, then $a p r = a p s \implies p r = p s$, so $r = s$.  I'm not using the fact that $P$ is a prime ideal which leads me to believe that I messed up a step.
47800
Let $R$ be a commutative ring such that $R$ has no nontrivial ideal. Then show that either R is a field or R is a finite ring with prime number of elements and $ab = 0$ for all $a,b \in R$.  I am facing difficulty in proving the above!!
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
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.
77699
Let $A$ be a commutative ring. I am trying to show that if $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor then there exists $a$ in $A-\{0\}$ such that $af=0$ in $A[x_1,x_2,\ldots, x_r]$.    What I have tried so far as following. I am using induction on $r$ (not fixing the ring). The base case is for $r=1$ which I did assuming the minimal degree of $g$ for which $fg=0$ holds. So my induction assumption is for $r \geq 2$ whenever $f$ is in $A[x_1,x_2,\ldots x_r]$, $n&lt;r$ is a zero divisor there is $a$ in $A-\{0\}$ such that $af=0$. So for the final step I take $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor. Now since $A[x_1,x_2,\ldots, x_r]=A[x_1][x_2,x_3,\ldots, x_r]$ by induction hypothesis there is $g$ in $A[x_1]-\{0\}$ such that $fg=0$.      I cannot proceed further. Am I correct so far? Please help me. Thank you.
64919
Let $R$ a ring, and let $$\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}$$ with addition and multiplication as defined for polynomials.  We have that $R[[x]]$ is a ring containing $R[x]$ as a subring.  How to prove that if $a_0\in R$ is a unit, then  $\displaystyle\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$?
29107
Let R be a commutative ring with identity. Define a new operation $\circ$ on $R$ by for any $a,b \in R$  $$a \circ b=a+b-ab$$  a) Prove that $\circ$ is associative  b) Prove that R is a field iff the set $\{r \in R: r \ne 1\}$ is an abelian group with respect to the operation $\circ$  Starting with a) we must prove   $a \circ ( b \circ c)= (a \circ b) \circ c$  Left side: $a \circ ( b \circ c)=a \circ (b+c-bc)=a+b+c-bc-a(b+c-bc)=a+b+c-bc-ab-ac+abc$  Right side: $(a \circ b) \circ c =(a+b-ab) a \circ c=a+b-ab+c-(a+b-ab)c=a+b-ab+c-ac-bc+abc$ and we have thus proved that $\circ$ is associative  As for part b), I am a little confused on what is being asked exactly. Is it asking to go through ALL the properties of a field, closure, associativity, commutativity, identity, inverse, distributive, showing that all hold except for when $r \ne 1$?
119458
Let  $R$  be  a  ring  such  that  the  only  right  ideals  of  $R$  are  $(0)$  and  $R$.  Prove  that  either  $R$  is  a  division  ring  or  that  $R$  is  a  ring  with  a  prime  number  of  elements  in  which  $ab=  0$  for  all  $a,b\in R$.    I  don't  want  the  proof.  I  am  stuck  at  one  point.  Why  does $R$  have  to  be  finite  here?
60928
Let $R$ be a commutative ring. Show that for any $a,b \in R$ nilpotent that $a+b$ is also nilpotent in $R$.  We know $a^n = 0$ for some n and $b^m = 0 $ for some m, so consider $(a+b)^{m+n} = (a+b)(a+b)...(a+b)$  $n+m$ times.  I don't see how $R$ being commutative helps me here.