Is there a Ring $R$ with $(R,+) \cong (R^\times,\cdot)$? If $R$ is finite, clearly only the trivial ring does it (for cardinality reasons). But what about infinite rings? Are there even fields as example?
121528
Let $R$ be a commutative ring with unity, and let $R^{\times}$ be the group of units of $R$. Then is it true that $(R,+)$ and $(R^{\times},\ \cdot)$ are not isomorphic as groups ?    I know that the statement is true in general for fields. And it is trivially true for any finite ring (as $|R^{\times}| \le |R|-1&lt;|R|$, so they are not even bijective).  I can show that the groups are not isomorphic whenever $\operatorname{char} R \ne 2$ , but I am unable to deal with $\operatorname{char} R=2$ case ... Please help. Thanks in advance.
113834
Does there exist a totally ordered ring with zero divisors? I can't think of an example right now.
61499
Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?   Every element of $S$ is a unit in $S^{-1}R$. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$.
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?
76530
I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?
65578
Prove that in general the set of the right zero divisors is different from the set of left zero divisors in ring $R$.   The ring is arbitrary. Does exist general proof for it ?  Attention that the problem is comparing two sets. Thank you.
35475
$\DeclareMathOperator{\ass}{Ass}$     Let $R$ be a Noetherian ring and $S$ be a multiplicative set in $R$. Then show that   $\ass_R(S^{-1}M)=\ass_R(M)\cap \{P\in \text{Spec}(R):P\cap S=\emptyset\}$   Attempt: Let $P\in \ass_R(S^{-1}M)$. We argue that $S\cap P=\emptyset$. Say $P=\text{Ann}(m/s)$ for some $m/s\neq 0$ in $S^{-1}M$. If $r\in P\cap S$, then from $r\cdot (m/s)=0$ we get $(r/1)\cdot (m/s)=0$. But $r/1$ is invertible in $S^{-1}R$ and thus we have $m/s=0$, a contradiction. So we must have $S\cap P=\emptyset$.  The main difficulty I am having is in showing that $P\in \ass_R(M)$. I suspect that $P=\text{Ann}(m)$. Clearly, if $r\cdot m=0$, then $r\cdot (m/1)=0$ and therefore $r\in P$. So $\text{Ann}(m)\subseteq P$. On the other hand, if $r\in P$, then $(r\cdot m)/s=0$, giving an element $u\in S$ such that $(ur)\cdot m=0$. But I am getting nowhere with this.
2693
For curiosity: can a ring of positive characteristic ever have infinite number of distinct elements? (For example, in $\mathbb{Z}/7\mathbb{Z}$, there are really only seven elements.) We know that any field/ring of characterisitc zero must have infinite elements, but I am not sure what happens above.
56222
Show that a ring $ R $ with exactly $ n $ zero divisors (different from $0$) has cardinality atmost $ (n+1)^2 $.    I have shown that annihilator of any element among the zero divisor is a subset of the zero divisor which proves it is finite. now i think i have to show that it's coset space is finite.please help me.
123834
Problem is this:      Suppose for any element $r$ in a ring $R$ with unity $1$, $r$ or $1-r$ is invertible. Then show that non-invertible elements form an ideal of $R$.   It is crystal clear that $-r$ is non-invertible if $r$ is. So I tried to show the sum of two non-invertible elements is non-invertible, and the product of non-invertible elements with any ring element is non-invertible but I'm stuck.  Any help will be appreciated.
66738
Let $R$ be a ring with unity (not necessarily commutative) and $I$ an ideal of $R$.  Suppose that for every element $a \in I$ there exists an element $c\in I$ such that $ac=a$. Note that $c$ is related to $a$. Now we have the following question:  Can we say that  for every element  $a\in I$ there exists an idempotent element $c\in I$ such that $ac=a$?    Of course we have many examples such that the answer is true for them but in general  we don't know.
70782
Let D be the ring of integers of a number field K. Let I be a non-zero ideal of D. Why D/I is finite?  Would you help me please? Thank you in advance.
72316
I know it does not make sense that if a subring of a ring $R$ is commutative, then $R$ is also commutative. (For example, the set consisting of the matrices whose all entries except (1,1)-entry are zero, is a subring of ring of 2x2 real matrices.)  I also considered the case of a ring containing a subring with identity, but I had no ideas. Maybe it seems that it does not make sense, either.  Who give me some examples supporting my guess?
68822
Question: If $R$ is a finite boolean ring, then show $R \cong \mathbb Z_2 \times \mathbb Z_2 \times \cdots\times \mathbb Z_2$.   I know that   $\mathrm{char}(R) =2$  $R$ has $2^k$ elements for some $k$  (this is because, if $p\neq 2$ is a prime with $p\mid|R|$, then there exists a non-zero $x \in R$ s.t. $px=0$ and $p $ is odd $\implies x=0$, contradiction).   Does this suffice to answer my question?
26257
How do I exactly find the zero divisors and units of a ring in the form of $\mathbb{Z}_n \times \mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $\mathbb{Z}_6 \times \mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $\mathbb{Z}_n \times \mathbb{Z}_m$. Also, the units of $\mathbb{Z}_6 \times \mathbb{Z}_2$ would be elements that send some element $s$ in $\mathbb{Z}_n \times \mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.
42912
It's well known that if $K$ is a finitely generated extension of some field $E$, then any intermediate field $F$, $E\subseteq F\subseteq K$, is also finitely generated over $E$.  I'm curious, does the same hold for rings?      Say $S$ is a ring, and $R\supset S$ is a finitely generated extension of $S$. If $T$ is any intermediate ring, is it necessarily true that $T$ is finitely generated over $S$ as a ring?   Is it as simple as saying that for any $t\in T$, $T$ can be generated by the generators of $R$ over $S$? I feel unsure about this statement, since it's not clear to me that the generators of $R$ over $S$ need be in $T$.  If not, what is an example what shows otherwise? Thanks.
114155
In which kinds of rings $R$ does the following hold:  $$(a)\cdot (b) = (ab) \; ?$$  With $a, b\in R$, $(a)$ denoting the (two-sided) ideal generated by $a$ and the multiplication of ideals $I, J\subset R$ defined as  $$ I\cdot J = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I, y_i \in J \biggr\}\, .$$  It seems to me that it only holds for commutative rings with $1$. Is that right?  Ok, I'm trying a proof:     Let $R$ be commutative with $1\in R$. Then $(a)\cdot (b) = (a b)$ for any $a, b\in R$.   First let $I_a := \{ra : r\in R\}$, we're going to show that  $$(a) = I_a\, .$$ $I_a$ is obviously an ideal. Since $1 \in R$ we have $1 \cdot a \in I_a$, so $(a) \subset I_a$. On the other hand, any $x \in I_a$ can be written as $x = ra$ and so must be an element of $(a)$. This proves that $(a) = I_a$.  Then $$(a)\cdot (b) = I_a \cdot I_b = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I_a, y_i \in I_b \biggr\}  = \biggl\{\sum_{i=1}^n (r_i a)  (s_i b) : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b \sum_{i=1}^n r_i s_i : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b r : r \in R \biggr\} = I_{ab} = (ab)\, .$$  We obviously had to use that $R$ is commutative. In the last step we also used that every $r \in R$ can be written as $r=\sum_{i=1}^n r_i s_i$. This is because $1 \in R$, so with $n=1$ we have $r = 1\cdot r$.