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4

Let $a$ and $b$ be any two distinct elements of $F$. Then we have $F(a)=F(b)=F$ but the minimal polynomial of $a$ over $F$ is $x-a$, whereas the minimal polynomial for $b$ over $F$ is $x-b$, which are different. This is just a trivial example of the general behavior. Let $F$ be any infinite field, and let $a$ be anything that's algebraic over $F$ ...


3

Suppose $n$ is not prime and let $n=p_1^{k_1}\cdots p_s^{k_s}$ be the canonical factorization of $n$. Then $p_1^{k_1}\cdots p_i^{k_i}X$ and $p_{i+1}^{k_{i+2}}\cdots p_s^{k_s}X$ are nonzero elements of $\mathbb{Z}/n\mathbb{Z}[X]$ whose product is $0$ so this is not a domain and therefore not a UFD. Conversely, if $n$ is prime, then ...


3

1) The definition is right. 2) No. That not the same. 3) That's wrong. Consider $x(x^2+1)$ over the reals, which is not algebraically closed.


3

No, this relies heavily on the fact that $n=1$. Here is a general proof. Suppose $P\leq G$ is a Sylow $2$-subgroup. Then there is a homomorphism $\pi:G\to S_{m}$ induced by the action of $G$ on the set of left cosets of $P$. If $G$ is simple then $\pi$ must be injective because it is not trivial. This means that $$2^nm\leq m!$$ so $$2^n\leq (m-1)!$$ But ...


3

Indeed, the set of commutators need not be a subgroup (take a look at this thread). But that's not a problem, because the commutator subgroup of a group $G$ is defined to be the subgroup generated by the set of commutators of $G$ (Wikipedia link).


2

$\newcommand{\FF}{\mathbb{F}}$ Given $a=\sum_{k} a_k T^k \in \FF_p((T))$ and $b = \sum_{r} b_r T^{-r}\in \FF_p((1/T))$, there's a unique element $c=\sum_{n}c_k T^k \in \FF_p[T,1/T]$ such that $a_k - c_k =0$ for $k < 0$ and $b_r - c_{-r}= 0$ for $r \leq 0$. This makes it possible to identify the quotient with $\prod_{n\in \mathbb{Z}} \FF_p$ by sending ...


1

$\DeclareMathOperator{Aut}{\operatorname{Aut}}$ $\Aut(\Bbb C)$ contains the conjugation mapping $z^*$ and the multiplication by $i$ mapping as automorphisms. Now notice that $(iz)^* = -iz^* \neq iz^*$.


1

It is explained in the Wikipedia article on group actions. Such an action can have additional properties. From Wikipedia: "In addition to continuous actions of topological groups on topological spaces, one also often considers smooth actions of Lie groups on smooth manifolds, regular actions of algebraic groups on algebraic varieties, and actions of group ...


1

Let $G$ consists of upper triangular $2\times2$ matrices with ones on the diagonal. As $G$ is 1-dimensional the trivial subgroup $M$ is the only Zariski closed subgroup. Yet any additive subgroup of the base field gives rise to an intermediate subgroup (assume characteristic zero to avoid the possibility of finite subgroups that would surely be also Zariski ...


1

Yes. Since $1\in \mathfrak{I}$, if $f \in \mathcal{O}(\mathbb{C})$, $f = f1 \in \mathfrak{I}$.


1

There are only two such equivalence relations. In one, all three elements are equivalent. In the other, 1~2 (as required) but 3 is not equivalent to any element (other than itself).



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