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5

Let $K(\alpha)/K$ be an algebraic extension of fields, and suppose $\alpha$ has minimal polynomial $f(x)$ over the base field $K$, with $\deg f(x)=n$ (with $n\ge2$). Let $A\in M_n(K)$ be the companion matrix associated to $f(x)\in K[x]$. Then the there is an isomorphic copy of $K(\alpha)$, call it $L$, sitting inside the matrix algebra $M_n(K)$, which the ...


4

A $k$-algebra structure on $A$ can simply be seen as a ring homomorphism $k\rightarrow Z(A)$ into the center of $A$. Thus if you have a $k$-algebra structure on $M_n(D)$, you have a morphism $k\rightarrow Z(M_n(D))$. But $Z(M_n(D))=Z(D)$ consists of matrices of the form $d I_n$ for $d\in Z(D)$. See here for instance for a proof of this fact. Thus, the $k$-...


4

If you don't care about mapping $1$ to $1$, then we can embed any field $K$ into the ring $A$ of $2\times 2$ matrices over $K$, by sending $a\in K$ to $\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix}$. This preserves addition and multiplication, and $K$ clearly this does not lie in the center of $A$. If you care about mapping $1$ to $1$, then we have to ...


3

Consider the ring $R = \left( \begin{smallmatrix} k & k \\ 0 & k \end{smallmatrix} \right)$ of upper triangular $2 \times 2$-matrices over some field $k$, then $J(R) = \left( \begin{smallmatrix} 0 & k \\ 0 & 0 \end{smallmatrix} \right)$ and for $e = \left( \begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right)$ we get $e J(R) e =...


2

As noted in Rostami's answer, you do get that $J(R)$ is nil, hence is nilpotent if it is finitely generated. (Here is an alternative proof, just for fun. Since idempotents are locally 0 or 1, these assumptions imply $J(R) = $nil$(R)$ holds locally, hence globally.) But here is the main point of my answer: an example of a quasilocal ring whose maximal ...


2

Let $a$ be an element of $R$. If $\left<a\right>=\left<a\right>^2$, then it is clear that the exists $e^2=e\in \left<a\right>$ such that $\left<a\right>=\left<e\right>$. Now, if $0\not= a\in J (R)$, then $\left<a\right>$ is nilpotent or idempotent. If $\left<a\right>$ is idempotent, by above argument there exists $e^...


1

I don't know about using the Zassenhaus formula, but I did obtain the following expression for the terms that are first-order in $d$: $$\sum_{n\geq1}\frac{c^{n-1}t^n}{n!}\sum\limits_{r,s\geq0,\ \ r+s=n-1}\hat{X}^r\hat{Y}\hat{X}^s. $$ Basically, expand $e^{t(c \hat{X} + d \hat{Y})}$ as a power series: $$e^{t(c \hat{X} + d \hat{Y})}=\sum_{n=0}^\infty \frac{t^...


1

This is indeed not true. As you observe, $T(V\oplus W)$ is the coproduct of $T(V)$ and $T(W)$ in the category of $K$-algebras, which is typically larger than $T(V)\otimes_K T(W)$. For a very simple example, take $V=W=K$. Then $T(V)$ and $T(W)$ are both polynomial rings in one variable over $K$; write $T(V)=K[x]$ and $T(W)=K[y]$. Then $T(V)\otimes_K T(W)\...



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