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5

A Euclidean Domain is usually defined to be an integrals domain in which there is a division algorithm. Whatever definition you had should be equivalent to this. It turns out that every Euclidean Domain is a Primitive Ideal Domain, and every Primitive Ideal Domain is a Unique Factorization Domain, but the other direction for both statements is false. EDIT: ...


4

A finite-dimensional division algebra over $\mathbb{C}$ is a division algebra $D$ over $\mathbb{C}$ which is finite-dimensional as a complex vector space. The way you'll use this hypothesis is the following: every $d \in D$ has the property that the elements $\{ 1, d, d^2, \dots \}$ must be linearly dependent, since there are infinitely many of them, so it ...


4

You seemed to be new to abstract algebra and you just want to feel the subject, I mean, to see how the results work and why they are how they are instead of just looking at proofs. Maybe my recommendation could sound a little controversial but here it goes: 1) First you need some motivation and a book which explains each thing in easy terms and easy words. ...


3

It is as simple as it looks. We first have $p(0)=0$. Consider $\alpha\in k,\, \alpha\neq 0$. It is a polynomial of degree $0$. Perform the Euclidean division of $\alpha$ by $f$ to get $\alpha=0\cdot f+\alpha$. This means $\forall \alpha\in k,\,p(\alpha)=\alpha$ and therefore $$\begin{align}p(f(x))&=p(\sum_0^n\alpha_i ...


3

I also had a lot of trouble with this when I first saw it. Here is the explanation that made sense to me with the details written out. If $k$ is a field, and $p(X)$ is an irreducible polynomial in the ring $k[X]$, then the ideal $\mathfrak p$ generated by $p(X)$ is a maximal ideal (because, for example, $k[X]$ is a principal ideal domain). Thus the ...


3

This is not a definition, it is a property of free objects. Every object surjecting onto a free object satisfies this too, and these do not have to be free. (Consider the group $S_n \times \mathbb{Z}$ for instance.) I advise you to read any text / book on universal algebra. There, you can find concise definitions and constructions of free objects.


2

We suppose that $H\subset Z(G)$, then the cardinal of $G$ is inferior to $4$. The problem is trivial. Suppose that $H$ is not contained in $Z(G)$, then $C_G(H)$ contains $Z(G)$ and $H$, this implies that the cardinal of $Z(G)$ is 2, otherwise the cardinal of $C_G(H)\geq 8$. Let $G'=G/Z(G)$ and $H'$ the projection of $H$ to $G'$ by $p:G\rightarrow G/Z(G)$. ...


2

Let us start with the basics. Let $f: E\rightarrow E$ be a field isomorphism with $\mathbb{Q}\subset E$ being a subfield. Then we have $$ f(x+y)=f(x)+f(y); f(xy)=f(x)f(y) $$ It is now easy to see that $$ f(nx)=nf(x), f(1)=1 $$ Therefore we have $$ f(p*\frac{1}{p})=p*f(\frac{1}{p})=1\rightarrow f(\frac{1}{p})=\frac{1}{p} $$ and thus $f$ must leave ...


2

An element of $S_3$ is a map from $\{1,2,3\}$ to itself. If we are given $\tau\in M$ then a priori this is a map from $\{1,2,3,4\}$ to itself. Of course we can always restrict such a map to a subset, thus obtaining a mpa $\{1,2,3\}\to\{1,2,3,4\}$. But by the given condition that $\tau(4)=4$ (and hence $\tau(x)\ne 4$ for $x\ne 4$) we see that we can consider ...


2

Given $\sigma\in\ S_3$ define $f(\sigma)$ to be the permutation in $S_4$ fixing $4$ and agreeing with $\sigma$ on $1$, $2$ and $3$.


2

I want to include "Abstract Algebra" by I. N. Herstein. Admittedly, I chose this gem based on its reviews, but the author offers a unique writing style, extremely lucid explanations and tries to motivate topics using historical developments in the field, which helps immensely to clarify seemingly unintuitive topics. I understand there is another well known ...


2

Show first $A(B\cap C)\subseteq AB\cap C$. Since $A\subseteq C$ we have $A(B\cap C)\subseteq C(B\cap C)\subseteq CC\subseteq C$ and because of $B\cap C\subseteq B$ we have $A(B\cap C)\subseteq AB$. So together we get $A(B\cap C)\subseteq AB\cap C$. Show now $A(B\cap C)\supseteq AB\cap C$. Let be $x\in AB\cap C$. So $x\in C$ and there is $a\in A\subseteq ...


2

If $x\in C$ and $x=ab$ with $a\in A\subseteq C$ and $b\in B$ then $$b=a^{-1}x\in C$$ So $b\in B\cap C$ and $x=ab\in A(B\cap C)$, as was to be shown. Converse: If $x=az$ with $a\in A\subseteq C$ and $z\in B\cap C$ then $x\in AB$ (since $a\in A$ and $z\in B$) and $x\in C$ (since $a\in C$ and $z\in C$). So $x\in AB\cap C$


2

Let $L$ be a Lie algebra and $I\subset T(L)$ be the smallest ideal containing $\{xy-yx-[x,y]|x,y\in L\}$. Then by definition, $U(L)=T(L)/I$. Now every element of $L$ can be seen as an element of $U(L)$ via the map $$f:L\to U(L)$$ defined as the composite of the inclusion $i:L\to T(L)$ and the projection $\pi:T(L)\to U(L)$. The claim is that ...


2

Let $A(x)=(x+1)(x+2)\cdots (x+p-1)$ and let $B(x)=x^{p-1}-1$. Note that every $x$ from $1$ to $p-1$ is a root of $A(x)-B(x)$, and $A(x)-B(x)$ has degree $\lt p-1$. For the roots of $A(x)$ in the $p$-element field are $-1$ to $-(p-1)$, which is $1$ to $p-1$ in reverse order. And the fact that every non-zero element of the field $\Bbb{F}_p$ is a root of $B(x)$ ...


2

Here is a theorem you may not be aware of: Theorem: Let $H$ be a subgroup of a group $G$. Then the following are equivalent $H$ is normal in G. Every left coset of $H$ is also a right coset of $H$. In particular, in your second point, by saying that $aH = Ha'$, you have implicitly assumed that $H$ is normal. To see why the theorem is ...


1

An element $u \in \mathbb{Z}_n^{\times}$ is a generator if and only if $\varphi(n)$ is the smallest positive integer $k$ such that $u^k = 1$. To verify this condition it suffices to verify that $u^{\frac{\varphi(n)}{p}} \neq 1$ for all prime divisors $p$ of $\varphi(n)$, and you can do this using binary exponentiation. The fewer prime divisors $\varphi(n)$ ...


1

The operation is associative, but you are confusing identity and idempotent. One has $oo = o$ and $ee = e$ and hence both $o$ and $e$ are idempotent. However $oe = eo = e$ and $oo = o$. Thus $o$ is an identity. The resulting structure is a monoid.


1

If you consider a field of two elements, $0,1$ with $0\neq 1$, we have the addition structure $$0+0=0$$ $$ 0+1=1$$ $$1+0=1$$ $$1+1=0$$ the multiplication structure differs for fields; we don't consider the inclusion of $0$ in the structure because there are no zero divisors. And again, we have $$0\cdot 0=0$$ $$0\cdot 1=0$$ $$1\cdot 0=0$$ $$1\cdot 1=1$$ ...


1

First notice that even times odd is even, not odd, and so $e$ is not a multiplicative identity. [EDIT] This matches what you at the moment have as the "2nd structure". The most common way to see this operation is as the multiplication in the field $\mathbb{Z}_{2} = \mathbb{Z}/2\mathbb{Z}$ ("even" $=0$, "odd" $=1$); if you work with addition of even/odd, you ...


1

Maybe you have in mind that we can define $AB = \{ab \mid a \in A, b \in B\}$ for $A,B \subseteq R$. Let denote this product by $A \ast B$, just to distinguish it from the product of cosets defined by $(a+K)\cdot (b+K)=ab+K$ Then, if I understand well, you want to compare $(a+K)\cdot (b+K)$ and $(a+K)\ast (b+K)$. Since $K$ is a $2$-sided ideal of $R$: ...


1

Rotman's "Introduction to the Theory of Groups" is a great book. It is focused on groups only (unlike some books on abstract algebra that sometimes skim over the subject), and Rotman's style makes it very readable. He usually includes proofs to every claim, a good deal of examples, and useful exercises.


1

If $S$ is any commutative $R$-algebra and $M$ is any monoid, then $$R[M] \otimes_R S \cong S[M]$$ in $S\mathsf{-Alg}$. In fact, if $T$ is an $S$-algebra, then (denoting the evident forgetful functors $C \to D$ here by $(-)|_D$) $$\hom(R[M] \otimes_R S,T) \cong \hom(R[M],T|_{R\mathsf{-Alg}}) \cong \hom(M,T|_{R\mathsf{-Alg}}|_{\mathsf{Mon}})$$ $$ = ...


1

I am quite suprised nobody has mentioned this book so far. I suggest you try "Algebra" by Michael Artin. This book is suitable for a beginner to intermediate level course in abstract algebra, especially group theory and uses the linear group as one of its recurring examples, which I.N.Herstein's book doesnt


1

If $char(F)$ divides $|G|$, then you can show that $\sum_{g\in G}g$ is a nonzero central nilpotent element, so it is in the Jacobson radical. The Jacobson radical of a semisimple ring is zero of course.


1

Suppose that $|G|=0$ in $k$ but $kG$ is semisimple. Then the augmentation map $$\epsilon:kG\to k:g\mapsto 1$$ splits because it is surjective. Let $s:k\to kG$ be a splitting and let $s(1)=\sum_{g\in G}\lambda_g g$. It is a $kG$-module map, thus $$\sum_{g\in G}\lambda_g g=s(1)=s(h1)=hs(1)=\sum_{g\in G}\lambda_g hg$$ for every $h\in G$. But that implies that ...


1

If $p = \operatorname{char} k > 0$ divides $\# G$, then we can't construct the averaging projection $\frac{1}{\#G} \sum_g$. For a concrete example of what can go wrong, consider the representation of $G = \mathbb{Z}_p$ on $V = \mathbb{F}_p^2$ that sends a generator $g$ to \begin{align*} g \to \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}. ...


1

Suppose $f(x) = a_0 + a_1x + \cdots + a_nx^n \in F[x]$ for some field $F$. If $u = x + \langle f(x)\rangle$ (the coset of $x$ in $F[x]/\langle f(x)\rangle$), then: $f(u) = a_0 + a_1(x + \langle f(x)\rangle) + \cdots + a_n(x + \langle f(x)\rangle)^n$ ...now a natural question is how we can multiply an element of $F[x]/\langle f(x)\rangle$ by an element of ...


1

No, and in fact there's a counterexample that's a subgroup of $\mathbb{Q}^2$. See http://mathoverflow.net/questions/90586/are-these-abelian-groups-free for more details.


1

If $K$ is any field containing $\mathbb{Q}$ and $f:K\to K$ is any automorphism, then $f(q)=q$ for all $q\in\mathbb{Q}$. First, $f(0)=0$ and $f(1)=1$ by definition of a homomorphism. It follows that $f(2)=f(1)+f(1)=1+1=2$, $f(3)=f(2)+f(1)=2+1=3$, and so on, so $f(n)=n$ for all $n\in\mathbb{N}$. Since $f$ preserves additive inverses it follows that $f(n)=n$ ...



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