Tag Info

Hot answers tagged

38

The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the "extent" to which two elements fail to commute. That is, given elements $a,b$, we wish to "compare" $ab$ and $ba$. In the context of a group, we only have one operation: "multiplication". One ...


29

Here is a more abstract and unified approach to commutators. I could phrase this whole answer in terms of category theory (and this is probably the best way to think of it), but I won't in order to make the answer more accessible. Let $G$ be a group. We want to make it commutative. What does that mean? We want to find a commutative group $A$ and a ...


15

$\mathbb{Z}\times\mathbb{Z}$. There are only $(-1,1),(-1,-1),(1,1),(1,-1)$ as invertible elements but infinitely many zero divisors $(a,0)$ for every $a$.


6

Another perspective on this comes from considering the relationship between a Lie group and its corresponding Lie algebra. Details are at On the relationship between the commutators of a Lie group and its Lie algebra, but the upshot is that the ring commutator $[A,B]=AB-BA$ can be regarded as an infinitesimal version of the group commutator ...


6

I assume you mean $(-1)^{2} = 1$ for this answer. Furthermore, I assume you mean rings, not groups as in the tag, because I assume the square is the multiplicative operation, yet the minus sign refers to additive inverse. These things should be clarified. The answer is yes. In general, we can show that $(-1)(x) = -x$ in the following way: $x + (-1)(x) = ...


5

The general criterion is that no number can be found with more than one valid, distinct factorization. This might sound like I'm merely rephrasing the question, but it's actually a reframing of the question. Plenty of numbers (infinitely many, to be precise) in $\mathbb{Z}[\sqrt{-5}]$ have more than one factorization. $6$ is just the easiest to find. To ...


4

Find $u_i \in A$ such that $$\alpha^{-n} + u_{n-1} \alpha^{1-n} + \dotsc + u_1 \alpha^{-1} + u_0 = 0.$$ Multiplying with $\alpha^n$ yields $$1+ (u_{n-1} + \dotsc + u_0 \alpha^{n-1})\alpha = 1 + u_{n-1} \alpha + \dotsc + u_1 \alpha^{n-1} + u_0 \alpha^n = 0.$$ Now you see that $$\alpha^{-1} = -(u_{n-1} + \dotsc + u_0 \alpha^{n-1}) \in A.$$


4

Consider $R=\mathbb Z$, $M=\mathbb Z$ and $N=\mathbb Z/2\mathbb Z$.


4

The two ideals are in fact equal. Let $ra + na \in (a)$ and $sb + mb \in (b)$ where $r, s \in R$ and $n, m \in \mathbb{Z}$, then $$(ra + na)(sb + mb) = rsab + mrab + nsab + nmab = (\underbrace{rs + mr + ns}_{\in\ R})ab + (\underbrace{nm}_{\in\ \mathbb{Z}})ab \in (ab).$$ As $(a)(b)$ consists of finite sums of elements of the form $(ra + na)(sb + mb)$ and ...


4

Note, if $R$ has a unit, then the condition $xa = x$ for all $x \in R$ implies that $a$ is the unit. To see this, let $x = 1_R$ be the unit of $R$, then $a = 1_Ra = 1_R$. An example of a ring $R$ and a non-unit element $a \in R$ with $xa = x$ for all $x \in R$ is $$R = \left\{\begin{bmatrix}r & 0\\ s & 0\end{bmatrix} \mid r, s \in ...


3

The power set of $\: \{\hspace{-0.02 in}0,\hspace{-0.05 in}1\hspace{-0.02 in},\hspace{-0.04 in}2,\hspace{-0.04 in}3,...\hspace{-0.04 in}\} \:$ is an example of such a ring.


3

If zero is considered to be a correct counterexample for this problem, then the problem is not a very good one, having a trivial answer for a pedantic reason. We can surmise this was not the intended effect, and that the author meant for something meaningful to happen. In all likelihood, the author probably meant for zero to count as a zero divisor. This is ...


3

A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 2 in PRIMES 2015 reading project problem set #1. (Ignore Section 0.1; it has nothing to do with this.) Unlike Martin's answer, this one directly gives a formula for $\dbinom{xy}{n}$ as a nonnegative linear combination of products ...


3

An example of a difference: $\mathbb{Z}/3\mathbb{Z}$ is projective (and free) as a $\mathbb{Z}/3\mathbb{Z}$-module but is not projective (nor free) as a $\mathbb{Z}$-module. However things like simplicity and indecomposability will remain the same, which is easy to prove since the action of $A/\mathfrak a$ is given by the action of $A.$ The only thing ...


3

Yes, there is. But first there is the matter of integral closure (I prefer to think of it as "completeness") to attend to. If $d \equiv 1 \pmod 4$, then you need $\frac{1 + \sqrt{d}}{2}$ instead of $d$. For example, with $d = 13$, $1 + \sqrt{13}$ is an algebraic integer, but so is $\frac{1 + \sqrt{13}}{2}$. For fun, try this in your calculator: ...


3

If $\mathfrak{a}$ is a modular right ideal, then every right ideal $\mathfrak{b}$ containing $\mathfrak{a}$ is modular as well, because $r-er\in\mathfrak{a}$ implies $r-er\in\mathfrak{b}$. Note also that $e\notin\mathfrak{a}$, if $\mathfrak{a}$ is a proper ideal. Applying Zorn's lemma is indeed the way to go. Consider the family of right ideals containing ...


2

You can relate the commutator for a Lie group $G$ to the (ring-type) commutator of its Lie algebra $\mathfrak{g}$. Say $g$ lies infinitesimally close to the group identity, $g = \exp(\epsilon\,X)$ for some $X \in \mathfrak{g}$. Likewise say $h = \exp(\epsilon\,Y)$ for some $Y \in \mathfrak{g}$. Then, $ghg^{-1}h^{-1}$ also lies infinitesimally near the ...


2

In both cases, commutators measure how far the object is from being commutative. A group is commutative when its operation is commutative. A ring is commutative when its multiplication is commutative. (Addition in a ring is always commutative.) Hence, the definition of commutator in each case reflects what it means to commute, as it should.


2

In the semigroup theory S is called a left ideal in the semigroup R under multiplication. See any semigroup theory book(J.M.Howe: Semigroup theory for example)


2

Take $\mathbb{Z}\times R$ for $R$ any nontrivial ring with only finitely many invertible elements.


2

$f(x+y) = f(x)+f(y)$ fails. Can you think of an example?


2

After inverting primes outside $\Sigma$, we can assume $A=A_\Sigma$. Hensel's lemma is being used as the following assertion: if $C$ is a finite ├ętale $A$-algebra, then any section $C \to B$ lifts uniquely to a section $C \to A$. Apply this with $C = A[x]/(x^n - a)$, for any $n$ prime to $\Sigma$ and the section $C\to B$ corresponding to the identity ...


1

$\operatorname {Frac}(\mathbb Z[[x]])\subsetneq \mathbb Q((x))$ because $\sum\frac {1}{n!}x^n\in \mathbb Q((x))\setminus \operatorname{Frac}(\mathbb Z[[x]])$


1

There are some problems with the problem itself. For starters $f:\mathbb{R}\rightarrow \mathbb{R}$ does not make sense with domain and codomain as stated. So we could either extend the codomain to $\mathbb{C}$, or restrict the domain to $\mathbb{R}_{\geq 0}$. But now what group are we talking about? Under what operation? For example, ...


1

For any ring $R$, and any ideal $J$, $R/J$, as an $R$-module, is cyclic, meaning it is generated by a single element, namely the coset of $1$, $1+J$. But certainly there are $R$-modules that aren't cyclic. An ideal is cyclic if and only if it is principal, so, excluding rings where every ideal is principal, you have many examples of ideals for which this ...


1

Let $K=\mathbb{Q}(\sqrt{-26})$, and $\sigma$ its non-trivial automorphism. We have ${\cal O}_K=\mathbb{Z}[\sqrt{-26}].$ If $I^3 = (1+\sqrt{-26})$, then $$N_{K/\mathbb{Q}}(I)^3=(I I^\sigma)^3 = I^3 (I^\sigma)^3 = (1+\sqrt{-26})(1-\sqrt{-26})=(27)=(3)^3.$$ Then $I I^\sigma=N_{K/\mathbb{Q}}(I)=(3)$. Since $3$ is prime, $I$ (and $I^\sigma$) must be prime ...


1

This is not an answer, but rather an extended comment. In your comments, you state that the identity element $s$ is unique. I'm not sure how to prove this from the axioms, but OK. Anyway, I'm not convinced you've looked at enough examples. Why don't you do $D_8$ and $Q_8$? If you take a group $G$, with $u_1$ the usual group multiplication, then your ...


1

This is well-studied under the heading of "cancellability," and Lam's crash course on the topic is very nice. Are there any simple conditions on $R$-modules $M,A$ and $B$... The readiest one is that if $R$ has stable range 1 and $M$ is finitely generated and projective, then it cancels from $M\oplus A\cong M\oplus B$. You can find this, for example, in ...


1

(answer from a discussion with @peter a g) Our objective is to form a homomorphism from $M''$ to $M$. Call $\{e_1,e_2,...,e_n\}$ a basis of the free module $M''$. As it is a basis, every element of $M''$ can be written as a linear combination of the basis elements in a unique way. At this point, we construct our homomorphism from $M''$ to $M$ by first ...



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