# Tag Info

3

The reason here is because the operation $[a,b]=ab-ba$ is called the commutator, so saying that $a$ and $b$ commute to $c$ is saying that their commutator is equal to $c$. This is in much the same was as you might say "$a$ and $b$ multiply to $c$" or "$a$ and $b$ add to $c$". The reason that this operation is called the commutator is that $[a,b]=0$ if ...

3

An infinite direct sum of nonzero modules is never finitely generated. Since $R$, as a left module over itself, is finitely generated, the result follows. However, in this case it's even easier: you can write $1=\sum_{\lambda\in\Lambda}x_\lambda$, with $x_\lambda\in I_\lambda$ (I changed slightly the notation, with $\Lambda$ as index set) and all but a ...

3

The trick (used by Goldie) is to show that if $aR\cap bR=\{0\}$, then the elements of $\{a^ib\mid i\in \Bbb N\}$ are $R$-linearly independent on the right. If $\sum_{i=0}^n a^ibr_i=0$, then $-br_0=\sum_{i=1}^n a^ibr_i$, but $a$ factors to the left side of this last expression. Therefore both sides are zero, so $r_0=0$, and $\sum_{i=1}^n a^{i-1}br_i=0$. ...

2

Saying that an operation $\star$ on a set $S$ is commutative amounts to sayig $a \star b = b\star a$ for all $a, b \in S$. This is typically phrased as saying that all elements in $S$ commute. If we didn't have to worry about commutativity on the level of individual elements, it would be silly to even describe an operation as commutative! So yes, in math ...

2

All central simple algebras over a global field are cyclic algebras. In Milne's class field theory notes he treats the case of number fields in section 2 of chapter VIII as a consequence of a corollary on the existence of a cyclic extension of a number field with finitely many specified local degrees. That same existence result on cyclic extensions is ...

1

Being able to extend an isometry to an isometry is rare. I don't think any nontrivial Banach space has this property. Here is a proof for $\ell^1$ and $\ell^\infty$, the two spaces of main interest to you. In both cases $e_1,e_2,\dots$ is the standard set of $0-1$ vectors, e.g. $e_2=(0,1,0,0,\dots)$. The case of $\ell^1$ Let $A=\{0,e_1,e_2\}$. Define ...

1

Hint: For singly generated ring structures $C_2\times C_2$, express them as a (non-unital) subrings of quotients of $\mathbb F_2[X]$. And for non-singly generated ring, express them as (not necessarily unital) subrings of $M_2(\mathbb F_2)$.

1

It seems you use the definition of associated primes given by Lam in Lectures on Modules and Rings, page 86. If your question refers to commutative rings, then on the same page Lam proves that for commutative rings the above definition of associated primes coincides with the usual definition (as annihilators of an element), and then the theory says that your ...

Only top voted, non community-wiki answers of a minimum length are eligible