 $p(u) = 0$ gives a linear dependence relation between them. The same reasoning applies to $1,v,\dots, v^{n-1}$.

Why do we define $\tilde{\sigma}$ in that way? Why does it satisfy the properties of the theorem?

Recall that we wanted a map $\tilde{\sigma} : K[u]\to K'[v]$ such that $\tilde{\sigma}(a) = \sigma(a)$ for all $a\in K$ and such that $\tilde{\sigma}(u) = v$. Now every element of $K[u]$ is a polynomial in $u$ with coefficients in $K$, so a map satisfying those properties is $\tilde{\sigma}(\sum_{i = 0}^m a_i u^i) := \sum_{i = 0}^m \sigma(a_i) v^i$ (do you see why $\tilde{\sigma}(a) = \sigma(a)$ for all $a\in K$ and why $\tilde{\sigma}(u) = v$?).
It remains to verify that $\tilde{\sigma}$ is an isomorphism. First, we will show injectivity. If $\sum_{i = 1}^m a_i u^i$ is a nonzero element of $K[u]$, then we may assume that $a_m\neq 0$ and that $m\leq n-1$ (why?), so that $\tilde{\sigma}(\sum_{i = 0}^m a_i u^i) = \sum_{i = 0}^m\sigma(a_i) v^i$. But then $\sigma(a_m)\neq 0$, and since $1,v,\dots, v^m$ are linearly independent, $\sum_{i = 0}^m\sigma(a_i) v^i\neq 0$. Hence, $\tilde{\sigma}$ is injective.
To see that $\tilde{\sigma}$ is surjective, choose any element $\sum_{i = 0}^{n-1} a'_i v^i\in K'[v]$. It is easy to verify that $\sum_{i = 0}^{n-1} \sigma^{-1}(a'_i) u^i$ maps to $\sum_{i = 0}^{n-1} a'_i v^i$ under $\tilde{\sigma}$, so that $\tilde{\sigma}$ is an isomorphism (I'll
$ p(u) = 0 $ gives a linear dependence relation between them. The like reasoning enforce to $ 1,v,\dots, v^{n-1}$. 

 Why do we define $ \tilde{\sigma}$ in that direction? Why does it meet the properties of the theorem? 

 Recall that we wanted a function $ \tilde{\sigma }: K[u]\to K'[v]$ such that $ \tilde{\sigma}(a) = \sigma(a)$ for all $ a\in K$ and such that $ \tilde{\sigma}(u) = v$. immediately every element of $ K[u]$ is a polynomial in $ u$ with coefficient in $ K$, so a function satisfying those properties is $ \tilde{\sigma}(\sum_{i = 0}^m a_i u^i): = \sum_{i = 0}^m \sigma(a_i) v^i$ (do you experience why $ \tilde{\sigma}(a) = \sigma(a)$ for all $ a\in K$ and why $ \tilde{\sigma}(u) = v$ ?). 
 It remains to verify that $ \tilde{\sigma}$ is an isomorphism. First, we will usher injectivity. If $ \sum_{i = 1}^m a_i u^i$ is a nonzero element of $ K[u]$, then we may assume that $ a_m\neq 0 $ and that $ m\leq n-1 $ (why ?), so that $ \tilde{\sigma}(\sum_{i = 0}^m a_i u^i) = \sum_{i = 0}^m\sigma(a_i) v^i$. But then $ \sigma(a_m)\neq 0 $, and since $ 1,v,\dots, v^m$ are linearly independent, $ \sum_{i = 0}^m\sigma(a_i) v^i\neq 0$. Hence, $ \tilde{\sigma}$ is injective. 
 To see that $ \tilde{\sigma}$ is surjective, choose any component $ \sum_{i = 0}^{n-1 } a'_i v^i\in K'[v]$. It is easy to verify that $ \sum_{i = 0}^{n-1 } \sigma^{-1}(a'_i) u^i$ maps to $ \sum_{i = 0}^{n-1 } a'_i v^i$ under $ \tilde{\sigma}$, so that $ \tilde{\sigma}$ is an isomorphism (I'll
 $p(u) = 0$ gives a linear dependtnce relation betcwen thxm. The aame rearoning applies to $1,v,\dots, v^{n-1}$.

Whb do we dtyine $\tilde{\sigma}$ in tfat way? Wjy does ut setisfy the propecfies of the tgcorem?

Xerall that we wakted a map $\dilde{\sigma} : K[u]\do K'[r]$ such that $\tilde{\sigma}(a) = \sigma(a)$ for wll $a\in K$ and such that $\tilcq{\sigja}(u) = v$. Now every element of $K[u]$ is z polynmmial in $u$ wiyh coefficients in $K$, so a lap datisfying those pgoperties iw $\tijee{\sigma}(\sum_{i = 0}^m a_i u^i) := \sum_{i = 0}^m \sigmz(a_i) v^i$ (do you see why $\tilde{\sigmx}(a) = \sngma(a)$ for aol $a\ij K$ and why $\vilde{\spgma}(u) = v$?).
It remains to verify that $\tilde{\sigka}$ ms ab isomorphism. First, wx will show injectivyty. If $\suk_{i = 1}^m a_i u^i$ is a nobzwro enemett ow $K[u]$, thtn xe jay asdumx that $a_m\nes 0$ and that $m\leq n-1$ (why?), so that $\tyode{\sigma}(\sum_{i = 0}^m a_i t^i) = \sum_{i = 0}^m\sigma(a_i) v^i$. But then $\sigma(a_m)\neq 0$, ahd since $1,v,\dots, v^m$ are lunearly independent, $\som_{i = 0}^m\sigmw(a_i) v^i\neq 0$. Hence, $\tilde{\sigma}$ is injective.
To see thdt $\timae{\snnix}$ id surjective, choose any element $\sum_{i = 0}^{n-1} a'_i v^i\yh L'[v]$. It is easy to verify thay $\duk_{y = 0}^{n-1} \sigma^{-1}(a'_i) o^i$ maps to $\sum_{i = 0}^{n-1} a'_i v^i$ undfr $\tildg{\sigma}$, so that $\uilde{\xigma}$ is an isomorphism (I'll
$p(u) = 0$ gives a linear dependence them. same reasoning to $1,v,\dots, v^{n-1}$. in way? Why does satisfy the properties the theorem? Recall that we wanted map $\tilde{\sigma} : K[u]\to K'[v]$ such that $\tilde{\sigma}(a) = \sigma(a)$ for all $a\in and such that $\tilde{\sigma}(u) = v$. Now every element of $K[u]$ is a in with in so a map satisfying those properties is $\tilde{\sigma}(\sum_{i = 0}^m a_i u^i) := \sum_{i = 0}^m v^i$ (do you see why $\tilde{\sigma}(a) = \sigma(a)$ all $a\in K$ and $\tilde{\sigma}(u) = v$?). It remains verify $\tilde{\sigma}$ is isomorphism. we show injectivity. If = 1}^m a_i u^i$ is a nonzero element of $K[u]$, then we may assume that $a_m\neq 0$ that $m\leq so that = a_i = \sum_{i = But then $\sigma(a_m)\neq 0$, and since linearly independent, $\sum_{i = 0}^m\sigma(a_i) v^i\neq 0$. Hence, is injective. see that $\tilde{\sigma}$ is surjective, choose element $\sum_{i = 0}^{n-1} a'_i v^i\in K'[v]$. It easy to verify that $\sum_{i = 0}^{n-1} \sigma^{-1}(a'_i) u^i$ maps to $\sum_{i = 0}^{n-1} a'_i $\tilde{\sigma}$, so that $\tilde{\sigma}$ an isomorphism (I'll
 $p(u) = 0$ gives a linear dependence rElation betWeen tHem. the SaMe reAsonIng applies to $1,v,\dOTs, v^{n-1}$.

why do we define $\tilde{\sigmA}$ in thAt WAy? WhY DoEs it sAtisfy tHE pROPerTiEs Of tHe THeOrem?

REcaLl that wE wanted a maP $\tiLdE{\sigma} : K[u]\to K'[v]$ SUcH that $\tilde{\SigMa}(a) = \sigma(a)$ for All $A\in K$ anD sUch THat $\tiLde{\Sigma}(U) = v$. Now eVEry eleMent of $K[u]$ iS a POlynomIAl in $u$ wiTH CoEffiCients in $K$, so a map saTIsFYing those propeRties iS $\tILdE{\SIgmA}(\suM_{i = 0}^m a_i u^i) := \sum_{I = 0}^m \Sigma(A_I) v^i$ (do yoU SeE WHY $\tiLDe{\sigma}(a) = \sigma(A)$ for all $a\in K$ ANd wHy $\tildE{\sIgmA}(U) = v$?).
It reMains To VEriFy that $\tilde{\SigmA}$ is an isomOrphisM. first, we WIll show InjectIviTy. IF $\sum_{I = 1}^M a_I u^I$ is A nONzeRO eLemENt oF $K[u]$, then wE mAy AssumE thaT $A_M\NEq 0$ anD thAt $m\lEq n-1$ (whY?), so that $\tilde{\sIgmA}(\sum_{I = 0}^M a_i U^i) = \sum_{I = 0}^m\sigMa(a_i) V^i$. but thEn $\sigmA(a_m)\neQ 0$, aNd since $1,v,\dots, v^m$ aRe liNearly indEpeNdEnt, $\SuM_{i = 0}^m\siGMa(a_i) v^i\Neq 0$. henCe, $\tilde{\Sigma}$ is INjeCtIVE.
to See that $\tilde{\sigma}$ iS sURJeCtive, choOse any ELeMeNT $\sum_{i = 0}^{n-1} a'_i V^i\In K'[V]$. It iS EAsy to VeriFY tHat $\sum_{i = 0}^{n-1} \Sigma^{-1}(a'_I) U^i$ MaPs to $\sum_{I = 0}^{n-1} A'_i v^i$ unDeR $\tiLde{\Sigma}$, SO thaT $\tilde{\Sigma}$ is aN isomORphism (I'll
 $p(u) = 0$ gives a linear  dependenc e rel ati onbe twee n th em. The same r e ason ing applies to $1,v,\d ots,v^ { n-1} $ .
Whydo we d e fi n e  $\ ti ld e{\ si g ma }$ in  th at way?  Why doesitsa tisfy the pr o pe rties of t hetheorem?

Re cal l that  w e w a nteda m ap $\ tilde{ \ sigma}  : K[u]\t oK '[v]$s uch tha t $\ tild e{\sigma}(a) = \s i gm a (a)$ for all $ a\in K $a nd s uch  th at $\tilde {\ sigma } (u) = v $ .N o w  ev e ry element of  $K[u]$ isa  po lynomi al  in $u$ wi th co ef f ici ents in $K$ , so  a map sa tisfyi n g those propert ies is  $\ til de{\ s ig ma }(\ su m _{i =0}^ m  a_ i u^i) : =\s um_{i  = 0 } ^ m \sig ma( a_i)  v^i$  (do you seewhy  $\t i lde {\sig ma}(a ) =\s igma( a)$ fo r all  $ a\in K$ and why  $\t ilde{\sig ma} (u ) =  v $?).I t rema ins  to  verify  that $ \ til de { \ s ig ma}$ is an isomorp hi s m .First, w e will sh ow injectiv it y.If $ \ s um_{i  = 1 } ^m  a_i u^i $ is a no nz ero ele me nt of$K [u] $,thenw e ma y assu me that$a_m\ n eq 0$ and that $m\leq n-1$ ( w hy ? ) ,s o th at$\tilde{\si gma} ( \sum _{i=  0 }^m a_i u ^i) =  \ s um _ {i = 0}^m\sigma(a_i )v^i$.But t hen $\sigma(a _m)\neq 0$ , a nd since  $1, v ,\ d ots, v^m$ arelinea rly indepe n dent, $\ sum_{ i = 0}^m \sigma(a_ i )  v^i\neq  0$ . H enc e,$ \ ti lde{\sigma}$i s  inj ec tive.
T o s ee that  $\ til de{ \si gm a}$ is su rjective ,ch oo se  an y ele m ent $\su m_ {i=0}^ {n-1} a'_i v ^i\in  K'[ v] $. Itis easy to v erif yth at $ \su m_ {i =0}^{ n -1}  \sigma ^{-1}(a'_ i)u ^i$ma ps  to $\s um_{i = 0}^{n -1 } a'_i v^i $und er $\t i l de{\sigm a}$, so that $\tilde{\s i gma}$ i s a n iso morp hism (I'l l
 $p(u)_= 0$_gives a linear dependence_relation between_them._The same_reasoning_applies to $1,v,\dots,_v^{n-1}$.

Why do we_define $\tilde{\sigma}$ in that_way? Why does_it_satisfy the properties of the theorem?

Recall that we wanted a map $\tilde{\sigma} : K[u]\to_K'[v]$_such that_$\tilde{\sigma}(a)_=_\sigma(a)$ for all $a\in K$_and such that $\tilde{\sigma}(u) =_v$. Now_every element of $K[u]$ is a polynomial in_$u$_with coefficients in_$K$, so a map satisfying those properties is $\tilde{\sigma}(\sum_{i_= 0}^m a_i u^i) := \sum_{i_= 0}^m \sigma(a_i)_v^i$_(do_you see why $\tilde{\sigma}(a)_= \sigma(a)$ for all $a\in K$_and why $\tilde{\sigma}(u) = v$?).
It remains_to verify that $\tilde{\sigma}$ is an isomorphism._First, we will show injectivity. If_$\sum_{i = 1}^m a_i u^i$_is a_nonzero element of $K[u]$, then_we may assume_that $a_m\neq_0$ and that_$m\leq n-1$ (why?), so that $\tilde{\sigma}(\sum_{i_= 0}^m a_i_u^i) = \sum_{i = 0}^m\sigma(a_i) v^i$._But_then $\sigma(a_m)\neq 0$,_and_since_$1,v,\dots, v^m$_are linearly independent,_$\sum_{i_= 0}^m\sigma(a_i)_v^i\neq_0$. Hence, $\tilde{\sigma}$ is injective.
To see_that_$\tilde{\sigma}$ is surjective, choose any element $\sum_{i_= 0}^{n-1} a'_i v^i\in_K'[v]$._It is easy to_verify that $\sum_{i = 0}^{n-1}_\sigma^{-1}(a'_i) u^i$ maps to $\sum_{i =_0}^{n-1} a'_i_v^i$ under_$\tilde{\sigma}$, so that $\tilde{\sigma}$ is an isomorphism (I'll