mathimatikoi.org

Let \(\displaystyle{R}\) an associative ring ,where \(\displaystyle{1}\) is the unit. Prove that if \(\displaystyle{1-xy}\) has a multiplicative inverse then \(\displaystyle{1-yx}\) has a multiplicative inverse, where \(\displaystyle{x,y\in R}\).

We see that for each \(\displaystyle{g,h \in D}\) and \(\displaystyle{f\in S}\) ,

\(\displaystyle{\left (g-h \right )\circ f=g\circ f-h\circ f=f\circ g-f\circ h=f\circ \left ( g-h \right )}\)

\(\displaystyle{\left ( g\circ h \right )\circ f=g\circ \left ( h\circ f \right )=\left ( g\circ f \right ...

Let \(\displaystyle{R}\) an associative ring with unit and order \(\displaystyle{p^{2}}\) , where \(\displaystyle{p}\) is a prime number.

Prove that \(\displaystyle{R}\) is commutative.

Let \(\displaystyle{R}\) be an associative division ring. Prove that \(\displaystyle{Z\left ( R \right )}\) is a field, where \(\displaystyle{Z\left ( R \right )=\left \{ z\in R:rz=zr , \forall r\in R \right \}}\) and that \(\displaystyle{R}\) can't be a vector space over \(\displaystyle{Z\left ( R ...

Thank you Demetres Christofides for your nice and short solution. Although ,the part you write that \(\displaystyle{R}\) has characteristic
\(\displaystyle{2}\) wasn't so clear to me.Maybe it's a matter of symbolism.Anyway i explained it like this , if \(\displaystyle{x=0}\) then \(\displaystyle ...

Prove that the groups \(\displaystyle{\left ( R,+ \right )}\) , \(\displaystyle{\left ( R-\left \{ 0 \right \},\cdot \right )}\) are never isomorphic, where \(\displaystyle{\left ( R,+,\cdot \right )}\) is a field.

Suppose that \(\displaystyle{\left ( 1 \right )}\) is true. Let \(\displaystyle{R\neq \left\{0\right\}}\) a finite division ring. It is obvious that

\(\displaystyle{0^{1}=0}\). Now let \(\displaystyle{x\in R-\left\{0\right\}}\). If \(\displaystyle{{\forall} k,l \in \mathbb{N} , k\neq l, x^{k ...

Let \(\displaystyle{r_{1},r_{2}\in R : r_{1}r_{2}= 0}\) ,

if \(\displaystyle{r_{1}\neq 0}\) then there exists unique \(x_{1} : r_{1}= r_{1}x_{1}r_{1}\)

since \(\displaystyle{r_{1}\left (x_{1}+r_{2} \right )r_{1}=r_{1}x_{1}r_{1}+\left (r_{1}r_{2} \right )r_{1}=r_{1}+0=r_{1}}\) and \(\displaystyle ...

Let \(\displaystyle{A=\left ( a_{ij} \right )}\) an \(\displaystyle{n\times n}\) matrix.

By applying the changes referred to it's determinant we are taking the following determinant

\(\left | B \right |=\begin{vmatrix}
a_{11}+a_{1n} &a_{12}+a_{11} &\cdots &a_{1n}+a_{1\left ( n-1 \right )} \\
a ...