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6

Let $R$ be a ring with unity, take $r\in R \setminus \{0\}$, and suppose that $(r) = R$ as ideals. In the commutative case, this means that $r$ is invertible. Why? Because every element of $(r)$ is of the form $ar$, so we have $ar=1$ for some $a\in R$. But in the non-commutative case, if we are looking at two-sided ideals, the elements of $(r)$ include ...


3

You can proceed as usual by starting with $F\stackrel{f}\to A\to 0$ and $H\stackrel{h}\to C\to 0$, where $F$ and $H$ are free of finite rank. Then show that there is an exact sequence $G=F\oplus H\stackrel{g}\to B\to 0$. Now consider $F'=\ker f$ and so on. You have a short exact sequence $0\to F'\to G'\to H'\to 0$. Now use the result for finitely generated ...


2

Such a map can exist, but not for every noncommutative ring. For example, no such map can exist over a division ring. This is because the annihilator of the left tensor product module is a nonzero left ideal, hence the left tensor product is always trivial in that case. For an example where the map does exist, let $T$ be the tensor algebra over a field, and ...


2

Take $A=\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle=\mathbb R[x,y]$. Then the ideal $I=\langle y,x-1\rangle\subset A$ is not principal but $I^2$ is principal . Hence $I\oplus I\cong A\oplus I^2\cong A\oplus A$ as $A$-modules.


2

Hints. If $S$ is a simple left $R$-module, then $S^n$ is a simple left $M_n(R)$-module. (Note that any $M_n(R)$-submodule of $S^n$ has the form $S_1^n$ for some $R$-submodule $S_1$ of $S$.) $\operatorname{ann}_{M_n(R)}S^n=M_n(\operatorname{ann}_RS)$. The map $X\to X^n$ induces a bijective correspondence between the isomorphism classes of $R$-modules, ...


2

No. There exists a ring with $a,b$ such that $ab=1$ and $ba\neq 1$. Even though $1$ is a unit, neither $a$ nor $b$ are units. Looks like I had trouble finding this before when I was writing this solution: Examples of such rings with $a$ and $b$ like that are scattered throughout the site, but a little hard to search for. I found one here at this related ...


2

The problem is that you are generalizing the real Gaussian integral to the Grassmannian case rather than the complex Gaussian integral $\int dz d\bar{z} e^{-(z, Az)}$, where $(z, A z) = \sum_{i,j} z_i A_{ij} \bar{z}_j$ and $d z d\bar{z} = dz_1 d\bar{z}_1 \ldots dz_n d\bar{z}_n$. You can see that the determinant will not arise by looking at the ...


2

Yes. If $N$ were projective, then the short exact sequence $$0\to e_iJ \to S_i \to N \to 0$$ of right $R$-modules would split, so $e_iJ$ would be a right $R$-module direct summand of $R$, and therefore of the form $fR$ for some idempotent $f$. But the Jacobson radical contains no nonzero idempotents.


1

Here is an alternative way that runs along the lines of Jeremy's answer for a while: If $S_i/e_iJ$ were projective so that $0\to e_iJ \to S_i \to S_i/e_iJ \to 0$ splits in Mod-$R$, $e_iJ$ would be an $R$-summand of $S_i$, hence an $R$-summand of $R$. But $e_iJ \subseteq J$, so this is saying a submodule of $J$ is a summand of $R_R$. This is a problem since ...


1

Since $210=2\cdot 3\cdot5\cdot 7$, the additive group must be the cyclic group $A=\mathbb Z/210\mathbb Z$ (classification of finite abelian groups). To distinguish our multiplication from the usual multiplication on $A$, we write $\odot$. Note that $\bar 0=0+210\mathbb Z\in A$ is additive neutral (and $0\odot x=0$ for all $x\in A$), but $\bar 1=1+210\mathbb ...


1

Consider $A := \begin{pmatrix} \ast & \ast \\ 0 & \ast\end{pmatrix}\subset\text{Mat}_{2\times 2}({\mathbb k})$ and put $e_1 := \tiny\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $e_2 := \tiny\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$ and $\alpha := \tiny\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$. Then, with respect to the ...


1

For a counterexample take the first Weyl algebra $\mathbf C[p,q]$ where $pq-qp=1$, which is a subalgebra of ${\rm End}_{\mathbf C}\mathbf C[X]$. Here $p$ acts on polynomials by $p f(X)=f'(X)$ and $q$ acts by $qf(X)=Xf(X)$. It should be checked $q,p$ are not invertible, yet $pq-qp=1$ means $(p)=(q)=1$. The point is that in noncommutative cases, even principal ...



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