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4

Certainly not. For fixed $x_1, \ldots, x_n$, the linear map $F^n\to F$, $(c_1, \ldots, c_n)\mapsto \sum c_ix_i$ has an at least $(n-1)$-dimensional kernel, whereas the space of vectors with $c_1=\ldots =c_n$ is one-dimensional. For $n>2$ we have $n-1> 1$.


3

It is indeed correct. Nevertheless, you have not used the condtion $(x,y)=1$. In fact, you do not need it, as $$ \mathbb Q(2^{1/x},2^{1/y})\subset\mathbb Q(2^{1/xy}), $$ even in the case when $x$ and $y$ are not relatively prime.


2

By definition $4.4^{-1}\equiv1$ mod $5$ so $4.4^{-1}+4^{-1}\equiv1+4^{-1}$ mod $5$. Also $4.4^{-1}+4^{-1}\equiv\left(4+1\right)4^{-1}\equiv0$ mod $5$ Proved is now $1+4^{-1}\equiv0$ mod $5$ or equivalently $4^{-1}\equiv-1$ mod $5$ or equivalently $4^{-1}\equiv4$ mod $5$ To 'find' $4^{-1}$ in situations like this just go on search for the unique element ...


2

Think about $\mathbb F_p$ as being the field generated by $1$ in $F$. That means that $$\mathbb F_p = \{ 1_F, 1_F+1_F,..., 1_F+1_F+...+1_F \}$$ Then $\mathbb F_p \subset F$ and $u \in F$. Now when you construct $\mathbb F_p(u)$ you construct it as a subfield of $F$. If the underlying set is the same, it follows that the subfield contains all elements.


2

Hints: Every field of characteristic zero has the rationals as a prime field, meaning: any field of characteristic contains an isomorphic copy of the rationals. This already solves (A) . In a positive characteristic $\;p>0\;,\;\;p\;$ a prime, the prime field is $\;\Bbb F_p:=\Bbb Z/p\Bbb Z\;$


2

No: with either $F=\mathbb{Q}$ or $\mathbb Z/7\mathbb Z$, $$ 1+2+(-3)=0,\qquad 5\cdot1+(-1)\cdot 2+1\cdot(-3)=0,\qquad 5\neq -1\neq 1 $$


2

In general I'd like to understand the case of number fields with exactly 1 complex place, and at least 1 real place. I'm able to produce lots of examples, but I don't get why they are not considered normal. One of the several possible equivalent definitions of "Galois," for a number field $K$, is that every embedding $K \to \overline{\mathbb{Q}}$ has ...


2

Yes, there are bijections between $\mathbb{R}$ and $\mathbb{R}\times \mathbb{R}$, and thus we can take any bijection $f:\mathbb{R}\rightarrow \mathbb{R}\times \mathbb{R}$ and declare it a field isomorphism. That is, it forms a field with addition $$ (a,b)+(c,d)=f(f^{-1}(a,b)+f^{-1}(c,d))$$ where the additive identity is $f(0)$, and multiplication $$ ...


2

There are many ways to do it while preserving addition as it is. Note that if $F$ is any field of characteristics $0$ and cardinality $2^{\aleph_0}$ then as a vector space over $\Bbb Q$, $F$ and $\Bbb R$ and $\Bbb{R\times R}$ are isomorphic. So we can transport the multiplicative structure from $F$ to $\Bbb R$ or $\Bbb{R\times R}$ while preserving addition. ...


1

First, let me point out that it doesn't make sense to say that there exists $\sigma \in G(E/F)$ such that $\sigma \alpha = \alpha'$, because the root $\alpha$ lives in $\overline{E}$, not necessarily in $E$. Here is how I suggest you prove this: write $f(x) = \prod_{i=1}^n e_i(x)$ for the factorization of $f$ in $E[x]$. Let $G$ act on the set $\{e_i\}$. ...


1

1) Show that $[LM : K] \le [L : K][M : K]$. 2) WLOG let $[L : K] = 2$. Show that $[LM : K]/[M : K]$ is either $1$ or $2$ (note: $M \subseteq LM$). 3) If $[LM : K] = [M : K]$, show that $L \subseteq M$. Deduce that $K = L \cap M = L$, contradicting $[L : K] = 2$.


1

If you set an arbitrary $F$-linear combination of the $\frac{1}{\alpha-\lambda_i}$'s equal to zero and multiply by the product of the $\alpha-\lambda_i$'s, then you get an $n-1$ degree polynomial in $\alpha$ equal to zero. Thus the polynomial is zero. (The product of the $\alpha-\lambda_i$'s is not zero because $\alpha$ is not in $F$.) edit: Let ...


1

For (a), I'm not quite sure whether your proof is correct, as you seem to have assume that $f$ is the identity in the proof. Basically, if you want $f$ to fix the rationals, then $f(2)=2$. But $f(\alpha)^3=f(\alpha^3)=2,$ so $f(\alpha)=\alpha$, which forces $f$ to be the identity on everything else by the properties of automorphisms. For b, note that the ...


1

In a) how do you make the assumption that $f(\alpha)=\alpha$? In b) you are asked to compute the Galois group of $L$ over $\mathbb{Q}$. Hint for both parts of the exercise: First show that any automorphism of the fields in question fixes $\mathbb{Q}$. Then try to find out where it could possibly map $\alpha$ and $u$; since $\alpha$ and $u$ satisfy a ...


1

$\;(B)\implies (C)\;$ . Let $\;a\in L\;$ and let $\; a=a_1,a_2,...,a_n\;$ be all its conjugates (i.e., all the roots of the minimal polynomial $\;m_a(x)\;$ of $\;a\;$ over $\;K\;$ ) We know that for any $\;1\le i\le n\;$ exists $\;1\le k(i)\le n\; $ .s.t $\;j(a_i)=a_{k(i)}\;$, and since $\;j\;$ is $\;1-1\;$ we get that ...


1

Do this by definition. $\mathbb{Q}(\sqrt{2})$ is by definition the intersection of all subfields that contain $\mathbb{Q}$ and $\sqrt{2}$. So, to prove that $\mathbb{Q}(\sqrt{2}) \subseteq L$, it is enough to establish two things: (1) $\mathbb{Q} \subseteq L$ and (2) $\sqrt{2} \in L$. Fact (2) is given to you for free. It remains to prove fact (1): ...


1

Field homomorphisms are always injective; there is no need to specify that $\phi$ is a monomorphism. If $\alpha \in E$ and $\alpha=\alpha_1,\dotsc,\alpha_n$ are the conjugates of $\alpha$ (i.e. the roots of the minimal polynomial of $\alpha$), then $\phi$ induces a map $\{\alpha_1,\dotsc,\alpha_n\} \to \{\alpha_1,\dotsc,\alpha_n\}$ (why?) which is in fact ...


1

I think I know you. I gave this homework in order that you work on it and to learn the course. Please do not post it on Internet before you hand it in. Sincerely, Your teacher


1

Let $\alpha$ be a root of $f(x)$ and let $m_\alpha(x)$ be the minimal polynomial of $\alpha$ over $K$. It's easy to conclude that $n=\deg\left(m_\alpha(x)\right)=[K(\alpha)\colon K]$. It holds that $K\preceq K(\alpha)\preceq F(\alpha)$ and $K\preceq F\preceq F(\alpha)$. Since all these extensions are finite it follows that $$\begin{align} ...


1

If I'm not mistaken in understanding of your question. Let's $G$ be an abelian group of order $p^n$. Does a subgroup of order $p^k$ exist? Yes, it exists: If $G$ is cyclic then subgroup of order $p^k$ is generated by element $g = p^{n-k}$. It is an implication of the fact that $g$ has order $p^k$. Every abelian group of order $p^n$ can be represented as a ...



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