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10

The empty set cannot be a torsion group nor can it be a torsion free group because the empty set cannot be a group. A group must have an identity element.


9

You probably are confusing the empty set (which is not a group) with the trivial group, having only the identity as element. This group is indeed torsion (any element has finite order) and torsionfree (any non identity element has infinite order). In this way the statements every subgroup of a torsionfree group are torsionfree and every quotient ...


8

One way to see this is that any finite extension of $\mathbb{Q}$ will be countable (being a finite dimensional vector space over $\mathbb{Q}$. Since $\mathbb{R}$ is not countable, it cannot be a finite extension of $\mathbb{Q}$.


6

Perhaps the distinction is between "the function $f$ is constant on its domain" (which is true for the polynomial written here, with domain the finite field -- the values is everywhere 1) and "a polynomial in which the coefficients of $x, x^2, \ldots$ are all zero", which one might reasonably call a "constant polynomial". I believe that the meaning intended ...


5

Another way: since $\pi$ is trascendental, no polynomial of $\Bbb Q[X]$ has $\Bbb R$ as its descomposition field.


3

Hint. What can you say about the degree of the minimal polynomial of a non-separable element?


3

The most common name is the length of the group element, which is defined to be the minimum length of words that define the element. Of course it depends on the choice of finite generating set, but changing $X$ will decrease or increase the length by at most a constant factor.


3

Hints in steps: The map $\;\varphi: I\times I\to A\;,\;\;\varphi(r,s):=rs\;$ is bilinear, so we get a unique homomorphism $\;\phi: I\otimes I\to A\;$ . Assume $\; 2\otimes 2+ x\otimes x= r\otimes s\;$ . Prove that then $\;4 + x^2=rs\;$ . Reach a contradiction by analyzing the (two) differentpossibilites of $\;r,s\;$: or $\;\deg r=0\;$ or ...


2

By Sylow's theorems, the group must have a normal $7$-subgroup and a normal $13$-subgroup, and so it is the direct product of these two subgroups, and is cyclic. Then the problem then becomes a special case of this problem. The upshot is that the only possible size of $G/Z(G)$, if it is cyclic, is $1$.


2

I think the implication is true: Write $f=\prod\limits_{i=1}^nf_i^{r_i}$ for the irreducible factors $f_i$ of $f$ and $f+1=\prod\limits_{j=1}^m\tilde{f}_j^{s_j}$, respectively. Then, as in the comment above, one has that $\prod\limits_{i=1}^nf_i^{r_i-1}$ and $\prod\limits_{j=1}^m\tilde{f}_j^{s_j-1}$ divide the derivation $f'=(f+1)'$ and thus, as $f$ and ...


2

This follows from the Mason--Stothers theorem, which is the polynomial analogue of the $ABC$ conjecture. We need to be careful about stating the theorem in a way that works in characteristic $p$, as follows. Theorem (Mason, Stothers): If $k$ is a field and $a(x)$, $b(x)$, and $c(x)$ are nonzero polynomials in $k[x]$ such that (i) $a(x) + b(x) = c(x)$, ...


2

By construction, $KG$ has $K$ dimension $|G|$, so yes, it is finite dimensional. The group algebra $KG$ is defined by forming the free $K$-vector space using the elements of $G$ as a basis, and then enforcing the "obvious" multiplication rules to make it into an algebra. Thus if $G$ is finite, $KG$ is finite dimensional.


2

The symmetric power $S^p(V)$ of a vector space $V$ is defined as the quotient $V^{\otimes p} / (v_1 \otimes \dotsc \otimes vp = v_{\sigma(1)} \otimes \dotsc \otimes v_{\sigma(p)} : v_i \in V, \sigma \in \Sigma_p)$. The natural isomorphism $V^{\otimes p} \otimes V^{\otimes q} \to V^{\otimes p+q}$ extends to a linear map $S^p(V) \otimes S^q(V) \to S^{p+q}(V)$. ...


2

It is a standard fact that irreducible representations of finite abelian groups are $1$-dimensional. Every finite-dimensional (you forgot this assumption!) representation is a direct sum of irreducible representations. The claim follows.


1

This is actually a direct consequence of Schurs lemma. I recommend you to study Schurs lemma and see if you can deduce the result.


1

A number $p>1$ is prime if and only if the only solutions of the equation $x^2\equiv 1\ (\ mod\ p\ )$ are {$-1,1$} If only a weak pseudoprimetest is made (check, if $a^{p-1}\equiv 1\ (\ mod\ p\ )$) and $p$ passes this test, but the rabin-test fails for this base, then a nontrivial solution of $x^2\equiv 1\ (\ mod\ p\ )$ can be derived (showing that ...


1

Consider $D=\Bbb R$, $n=2$ and the set of matrices $$\begin{pmatrix}0&a\\0&a'\end{pmatrix}$$ This is a left ideal, and it is proper. You can probably generalize this yourself.


1

Yes. The usual formula for the quadratic equation works for every field of characteristic $\neq 2$. Nothing special happens when you apply this to $\mathbb{C}$.


1

Let $P$ be a nonzero projective in the category. Consider its injective envelope $D$ in the category of abelian groups, which is a torsion divisible module. Thus $D$ is a direct sum of Prüfer groups and so there is a nonzero morphism $f\colon P\to G$ where $G$ is a Prüfer group, say $G=\mathbb{Z}(p^\infty)$, for some $p$. Let $G_0=\{0\}\subset G_1\subset ...


1

If char$\,K\mid n\;$, say $\;n=pr\;,\;\;p\;$ a prime, then $\;x^n-1=x^{pr}-1=(x^r-1)^p\;$ and this polynomial has at most $\;r<n\;$ different roots in $\;K^*\;$ and we're done. Otherwise, define $$W_n:=\left\{\;a\in K\;;\;w^n=1\;,\;\;n\in\Bbb N\;\right\}$$ It's easy to check $\;W_n\;$ is a subgroup of $\;K^*\;$ , and since it is finite it is then ...


1

Yes, it is true. Consider the assertion that $[x,y_1 \cdots y_n]$ can be expressed as a product of conjugates of $[x,y_n],\ldots,[x,y_1]$, in that order. For $n=2$, it can be verified that $[x,y_1 y_2] = [x,y_2][x,y_1]^{y_2}$. Fix $n$ and assume the assertion holds for smaller values of $n$. Then $[x,y_1 \cdots y_n] = [x,y_n] [x,y_1 \cdots y_{n-1}]^{y_n}$. ...


1

You almost done it. Consider $F=\ker\pi\oplus M$, $x\in\ker\pi$, $x=\sum r_ie_i$ with $r_i\in\mathfrak m$, and write $e_i=u_i+v_i$ with $u_i\in\ker\pi$ and $v_i\in M$. Then $x=\sum r_iu_i+\sum r_iv_i$, so $$x-\sum r_iu_i=\sum r_iv_i\in\ker\pi\cap M=\{0\},$$ hence $x=\sum r_iu_i\in\mathfrak m\ker\pi$.


1

There is no need to construct the field of fractions of $\mathbb{Z}[i]$ specifically, since the construction works "uniformly" for every integral domain. The field of fractions of $\mathbb{Z}[i]$ is $\mathbb{Q}[i] \subseteq \mathbb{C}$ because $\mathbb{Q}[i]$ is a field containing $\mathbb{Z}[i]$, and clearly the smallest one.


1

There are commutative domains (for example, any non-Noetherian valuation domain) which are not PIDs but for which every finitely generated torsion-free module is free.


1

One can obtain solutions by getting each of the individual summands to be zero: For $k\in\{1,\dots,n\}$, since $\mathbb R$ does not contain zero divisors, $$x_k(x_k+p)^2=0$$ holds iff $x_k=0$ or $x_k+p=0$. Hence, the set of solutions is a superset of $$\{0,-p\}^n=\{\,x\in\mathbb R^n\,\mid\,\forall k\in\{1,\dots,n\}.\;x_k\in\{0,-p\}\,\}\text.$$


1

$F \cong \ker(\pi) \oplus M$ implies $F/\mathfrak{m}F \cong \ker(\pi)/\mathfrak{m} \ker(\pi) \oplus M/\mathfrak{m} M$. Count dimensions.


1

Hint: If $A \neq \{0\}$ then $A$ must contain a positive number (why? inverses!) so there exists a smallest positive number in $A$, call this $a$. Define $\mathbb Z \to A$ by $n \mapsto na$ and try to show that this is surjective (use the remainder theorem) and injective (easy).


1

The ring $\mathbb Z/p^e\mathbb Z$ with $e\ge2$ is not a field because it has divisors of zero. Specifically, the product of $p+p^e\mathbb Z$ and $p^{e-1}+p^e\mathbb Z$ is zero. The additive group of $\mathbb F_{p^e}$ is isomorphic to $(\mathbb Z/p\mathbb Z)^e$, whereas that of $\mathbb Z/p^e\mathbb Z$ is cyclic.


1

Hint: An element '$a$'belongs to the Kernel of a mapping if it is mapped to identity. Note, $(0,0)$ is the identity of $\mathbb{Z}_3\times\mathbb{Z}_6$ $h(a)=(a\pmod3,2a\pmod6)=(0,0)$. Can you find all such $a$'s?


1

The isomorphism of Hom group is already very good, but if you want to you can attack it this way: If $f:M^m\to N^n$ was a nonzero homomorphism, then $f$ must be nonzero on one of the copies of $M$, and the restriction to this copy creates an injection of $M$ into $N^n$. I'll recycle the letter and call the image of this injection $M$ again. Since $N^n$ is ...



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