Let $K/F$ be a finite extensions (i guess it has to be separable) and let $\alpha \in K$. Let $L$ be a Galois extension of $F$ containing K and let $H \leq Gal(L/F)$ be the subgroup corresponding to K.
Define Norm of $\alpha\in K$ to be $\large \Pi \sigma(\alpha)$ where the product is taken over all the embeddings of K into an algebraic closure of F.
Define Trace of $\alpha\in K$ to be $\large \Sigma \sigma(\alpha)$ where the product is taken over all the embeddings of K into an algebraic closure of F.
First Question is to prove that $\large \Pi \sigma(\alpha) \in F$ and $\large \Sigma \sigma(\alpha) \in F$
What i suppose is that $K/F$ has to be galois, I mean if it is galois, it would be easy to see
$\tau(\large \Pi \sigma(\alpha))=\large \Pi \tau\sigma(\alpha)=\large \Pi \sigma(\alpha)$ (for a fixed $\tau$, if $\sigma$ varies over all elements of galois group, $\tau\sigma$ varies over all elements of the group).i.e., $\large \Pi \sigma(\alpha)$ is fixed by $\tau$.
This holds for all $\tau\in Gal(K/F)$ thus, $\large \Pi \sigma(\alpha)$ has to be in $F$ as only elements which are fixed by all elements of $Gal(K/F)$ are elements of $F$. Thus, $\large \Pi \sigma(\alpha)$ has to be in $F$.
For similar reasons, one can see that $\large \Sigma \sigma(\alpha) \in F$.
So, If it is Galois, I am able to see, But, result is true in general.
I am unable to convince myself that $\large \Pi \sigma(\alpha) \in F$ and $\large \Sigma \sigma(\alpha) \in F$ for any finite Extension $K/F$
To be frank, I did not understand what does "all embeddings of $K$ into algebraic closures of $F$" mean. 🙁
Second Question is to prove $N_{K/F}(ab)=N_{K/F}(a)N_{K/F}(b)$ and $Tr_{K/F}(ab)=Tr_{K/F}(a)+Tr_{K/F}(b)$. This can be seen easily as $\sigma$ are homomorphisms.
Another Question is :
Let $m_{\alpha}(x)=x^d+a_d x^{d-1}+….+a_1 x+a_0\in F[x]$ be minimal polynomial for $\alpha\in K$ over $F$. Let $n=[K:F]$. Prove that $d$ divides $n$, that there are $d$ distinct Galois conjugates of $\alpha$ (does he mean $K/F$ is Galois??) which are all repeated $n$ times in the product and sum above and conclude that $N_{K/F}(\alpha)=(-1)^na_0^{n/d}$ and $Tr_{K/F}(\alpha)=-\frac{n}{d}a_{d-1}$.
For this, as $\alpha\in K$ and $n=[K:F]$, the collection $1,\alpha,\alpha^2,…..,\alpha^{n-1},\alpha^n$ is linearly dependent, so, i have $b_0+b_1\alpha+b_2\alpha^2+…..+b_n\alpha^n=0, b_i\in F$.
So, I have a polynomial $g(x)=b_0+b_1 x+b_2 x^2+…..+b_n x^n$ for which $\alpha$ is a root, by minimality of $m_{\alpha}(x)$ we see that $m_{\alpha}(x)$ divides $g(x)$ and so, $d$ (power of x in $m_{\alpha}(x)$) divides $n$ (power of x in $g(x)$)
Assuming $K/F$ is cyclic, i can some how see that there are $d$ distinct Galois conjugates of $\alpha$ (I may not be able to explain it clearly now, but i am quite sure about it).
But in general i am unable to see that.
Any help would be appreciated.
Best Answer
First, suppose that $L/K$ is separable. Fix an algebraic closure $\overline{K}$ for $K$, and let $M$ be a Galois closure for $L/K$ in $\overline{K}$.
EDIT: If the idea of Galois closure isn't known to you, think about it as follows. Since your extension is separable, it is primitive, say $L=K(\alpha)$. Note then that necessarily $K(\alpha')/K$ is separable for every root $\alpha'$ of the minimal polynomial $m_\alpha$ of $\alpha$ over $K$. In the algebraic closure $\overline{K}$, all the roots $\alpha'$ exist. Then, the splitting field $F$ of $m_\alpha$ is just $K(\{\alpha':m_\alpha(\alpha')=0\})$. Since each $\alpha'$ is separable the full extension $F$ is separable. What this basically shows is that every finite separable extension of $K$ is contained in a finite Galois extension of $K$ (in fact, if you fix an algebraic extension, this is the minimal Galois extension containing $K$). So, when I say "take a Galois closure" I just mean "think about a Galois extension containing our extension". This is a common technique, for it allows you to deduce results about separable extensions by first proving it for Galois extensions, and then deducing it for certain subextensions.
$\text{ }$ $\text{ }$
Note that for any embedding $\tau\in\text{Hom}_K(L,\overline{K})$ one necessarily has that $\tau(L)\subseteq M$. Thus, for every $\sigma\in\text{Gal}(M/K)$ and every $\tau\in\text{Hom}_K(L,\overline{K})$ one has that $\sigma\circ\tau\in\text{Hom}_K(L,\overline{K})$. Moreover, since $\sigma\circ\tau=\sigma\circ\tau'$ implies that $\tau=\tau'$, you can see, in particular, that the mapping $\tau\mapsto \sigma\circ\tau$ is an injection $\text{Hom}_K(L,\overline{K})$ to itself, and so necessarily a bijection. From this you can see that for every $\sigma\in\text{Gal}(M/K)$ and $x\in L$ one has that
$$\begin{aligned}\sigma\left(\text{Tr}_{L/K}(x)\right) &= \sigma\left(\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\right)\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}(\sigma\circ\tau)(x)=\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\\ &= \text{Tr}_{L/K}(x)\end{aligned}$$
Since $M/K$ is Galois, this implies that $\text{Tr}_{L/K}(x)\in K$.
Do something similar for the norm map.
For your second question. Use the fact that $F\subseteq F(\alpha)\subseteq K$ and the fact that degree is multiplicative in towers. To deduce the other result, consider the polynomial
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(\alpha))$$
Convince yourself that for each distinct Galois conjugate $\beta$ of $\alpha$, that $\sigma(\alpha)=\beta$ precisely $\frac{n}{d}$ times. Thus,
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(x))=m_\alpha(T)^{\frac{n}{d}}$$
Compare the constant term and the coefficient of $T^{n-1}$ on both sides to get both of your results.