Im struggling to think of a ring that satisfies this. I thought maybe a subset of Mat$(2,\mathbb{R})$ with all elements in the General Linear group but I am not sure that would form a ring. Any suggestions?
116247
What are some examples of a non-commutative division ring other than quaternions?
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!
33999
Suppose $K \subset L$ is a finite field extension of degree $m$. Is it true that there exists some natural $n$ such that $L$ is isomorphic to a subfield of $M^{n\times n}(K)$, the ring of $n\times n$ matrices with entries in $K$? What is the smallest dimension (as a function of $m$) for which an isomorphism exists?  The example I had in mind was the 2-dimensional extension $\mathbb{R} \subset \mathbb{C}$ with $\mathbb{C}$ being isomorphic to $\{ \begin{pmatrix}a&amp;-b\\b&amp;a\end{pmatrix} : a, b \in \mathbb{R}\}$.
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. $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq...$)
2133
So as seen on this question , we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, because the right hand side does not have an element of order $nm$.  But take a more general setting. Let $R$ be a commutative ring, and let $A,B$ be ideals in $R$. $A$ and $B$ are said to be comaximal if $A+B=R$. If $A, B$ are comaximal, then we have: $R/AB \cong R/A \times R/B$. In this setup, is the converse true?      If we have $A, B$ ideals in $R$ such that $R/AB \cong R/A \times R/B$, do we always have that $A$ and $B$ are comaximal?
51212
Let $(A,+,\cdot)$ be a unity ring with the property that if $x \in A$ and $x^2=0$ then $x=0$. Consider the set $M=\{a\in A | a^3=a\}$. Prove that:      a) $2a\in Z(A)$, $\forall a\in M$, where $Z(A)$ denotes the centre of the ring $A$;   b) $ab=ba$, $\forall a,b\in M$.   My attempts revolved around the fact that an idempotent element in a reduced ring is central. So, since for $a\in M$ we have that $(a^2)^2=a^2$, it follows that $a^2\in Z(A)$, $\forall a\in M$. The next thing I wanted to use in order to solve a) was that $Z(A)$ is a subring of $A$, so if I had proved that $(a+1)^2 \in Z(A)$, $\forall a\in M$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true. Another idea that I had was to prove that $M$ is a subring of $A$. Of course, this didn't work out because I cannot even prove that $M$ is closed under addition. Again, I don't know if this is true and it most likely isn't. As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $x^3=x$ for any $x$ in that ring is commutative, but since $(M,+,\cdot)$ is almost definitely not a ring, this doesn't help. EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.
40447
Give an example of a finite, non-commutative ring, which does not have a unity.    I can't think of any thing which fits this question. I was thinking $M_2(\mathbb{R})$ but it has the identity. Any help is appreciated.
57225
Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements in the ring $R=k[X_1,X_2,X_3,X_4]$? Can it be generated with three elements? (Here $k$ is a field.)    Thanks for any help.
90773
I'm doing some sheets for my Abstract Algebra class and I can't seem to remember the group defined as $\mathbb{R}^{\times}$. It's obviously some variation of $\mathbb{R}$ but I'm away from college on reading week so can't ask my tutor. If someone could clear up the confusion I'd be grateful.
39294
Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.  Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$ S^{-1} R \otimes M \cong S^{-1} M$$  Now I'm slightly confused.      Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)
52999
Let $A$ be a commutative ring, $M$ an $A$-module and $N_1, N_2$ two submodules of $M$.  If we have $M/N_1 \cong M/N_2$, does this imply $N_1 \cong N_2$?  This seems so trivial, but I just don't see a proof... Thanks!
37204
As stated in the question, I am looking for a Noetherian ring over which the Euclidean algorithm is not valid.  I am trying to construct non-trivial examples of Noetherian rings.  Thank you.
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?
1594
Let $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?  I know how to do this with $GL_2(\mathbb{R})$ and $SL_2(\mathbb{R})$ but with $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.
71157
I have the following problem:  Is it possible to construct ring homomorphism from $M_2(\mathbb Z)\to \mathbb Z$, or in other words, a homomorphism from ring of all $2\times2$ matrices over the integers into integers?  I tried determinant, trace and mapping that maps matrix to it's element in the position $(1,1)$ but none of that obviously works, which led me to believe there might not be such a homomorphism. Determinant doesn't work because it is obviously not a linear map.  Trace doesn't work because it doesn't respect multiplication.  And the mapping to the position $(1,1)$ also doesn't work because $$\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}\begin{pmatrix}1&amp;1\\1&amp;1\end{pmatrix}=\begin{pmatrix}2&amp;2\\2&amp;2\end{pmatrix}.$$
50207
Let $R$ be a ring with unity. How can I prove that group of invertible elements of $R$ is never of order $5$?   My teacher told me and my colleagues that problem is very hard to solve. I would be glad if someone can provide me even a small hint because, at this point, I have no clue how to attack the problem.
45349
I want to know if the set $\{(1, 1, 1), (3, 2, 1), (1, 1, 0), (1, 0, 0)\}$ spans $\mathbb{R}^3$. I know that if it spans $\mathbb{R}^3$, then for any $x, y, z, \in \mathbb{R}$, there exist $c_1, c_2, c_3, c_4$ such that $(x, y, z) = c_1(1, 1, 1) + c_2(3, 2, 1) + c_3(1, 1, 0) + c_4(1, 0, 0)$.  I've looked around the internet, but all the answers I found involve setting up a matrix and finding the determinant, and I can't do that here because my matrix isn't square. What am I missing here?