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7

In the ring of $2\times 2$ matrices, say with real entries, look at $$\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ and $$\begin{pmatrix}0&0\\1&0\end{pmatrix}$$


4

A counterexample occurring in nature (for some rather generous definition of "nature") comes from the fact that there are abelian groups $A$ that have direct sum decompositions both as a direct sum $B\oplus C$ of two indecomposable groups, and as the direct sum of infinitely many non-zero groups. If $R$ is the endomorphism ring of such a group $A$, and $e$ ...


2

For $r\in\Bbb C$ let $$I_r=\left\{\begin{bmatrix}a&br\\0&0\end{bmatrix}:a,b\in\Bbb Q\right\}\;.$$ If $$\begin{bmatrix}c&s\\0&t\end{bmatrix}\in R\quad\text{and}\quad\begin{bmatrix}a&br\\0&0\end{bmatrix}\in I_r\;,$$ then ...


2

Indeed, the conclusion $\alpha^{i}(I_N)=I_{N+i}$ does not follow. In fact, the claim of the theorem is incorrect in this generality. The theorem is true if $\alpha$ is an automorphism, and then the questionable implication also becomes true. To see that it is not necessarily true if $\alpha$ is not surjective, consider the following example. Let $R=k[x]$ be ...


1

Let $R$ and $S$ be arbitrary rings and $R$ be a subring of $S$. We know that $\mathrm{rad}S = J$. What can we say about $\mathrm{rad}R$? To answer this question in general is very difficult. But in special cases, for example, when $R$ is two-sided ideal, we can easily find $\mathrm{rad}R$. And this theorem gives the answer how we can do it.


1

The statement is false; the fields $\Bbb Q_p$ and $\Bbb C_p$ of $p$-adic rationals and complexes are (metrically) complete normed fields, which are of course not isomorphic to $\Bbb R,\Bbb C,\Bbb H$. The correct statement along these lines is known as the Frobenius theorem: Theorem (Frobenius 1877): Every finite-dimensional (associative) division ...



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