$\phi(a)=a$ and $\phi(x)=\alpha$ (solution attempt) Let $L$ extension of field $K$ and $\alpha \in L$.
$\phi : K[x]\rightarrow L $ for
$\phi(a)=a$ ,$\forall a \in K$ and
$\phi(x)=\alpha.$
a)  Prove $\phi$ is homomorphism of ring
Hi, a few hours ago I asked for help to interpret the text of the question, now I come here to bring up doubts about an attempt to demonstrate item a).
From what I understand this function is doing two things simultaneously to $\phi(a)$ e $\phi(x)$.
Homomorphism:
(+) There are 3 possibilities:
(1)
$\phi(a+b)$, $a,b \in K$ $\rightarrow$ $\exists c \in K \rightarrow c=a+b$
$\phi(c)=c=a+b=\phi(a)+\phi(b)$
(2)
$\phi(a+x)$, $a \in K x \notin K$. Let $z=a+x$ we will have 2 possibility:
$z \in K$ or $z \notin K$.
For $z \in K$, $\phi(z)=z=a+x=\phi(a)+x$.
For $z \notin K$ , $\phi(z)=\alpha$.
Both possibilities go wrong because what we want $\phi(a+x)=\phi(a)+\phi(x)=a+\alpha$
(3)
$\phi(x+y)$, $x,y \notin K$ Applying the same method as above I arrive at. $\phi(z)$=x+y or $\phi(z)=\alpha$.
(.) There are 3 possibilities:
(4) $\phi(ab)$, $a,b \in K$ $\rightarrow \exists c \in K \rightarrow c=ab$,then $\phi(c)=c=ab=\phi(a)\phi(b)$
(5)
$\phi(ax),a \in K$ $x \notin K$ , K is field $ax \in K$ $\rightarrow \exists c \in K \rightarrow c=ax$,then $\phi(c)=c=ax=\phi(a)x$.What is different from $\phi(ax)=\phi(a)\phi(x)=a\alpha$
(6)
$\phi(xy), x,y \notin K$. Let $z=xy \rightarrow \phi(z)$ have 2 possibilite : $z \in K$ or $z \notin K$.
For $z \in K$:
$\phi(z)=z=xy$
For $z \notin K$.
$\phi(z)=\alpha$
both are different from $\phi(xy)=\phi(x)\phi(y)=\alpha^2$
My doubts are what I'm leaving to consider or use? is there an easier proof than simply separating in by your chances?
 A: This is the evaluation homomorphism $\phi_\alpha$. You can find a short proof here: How to prove that the evaluation map is a ring homomorphism?
I am not sure you are really understanding the definition of a polynomial ring. It has no sense to write something like $x \notin K$ since $x$ in this context is just a formal symbol, it doesn't represent an element of the ring. What you have to show is that given two polynomials $f = a_n x^n + a_{n-1} x_{n-1} + \dots + a_0$, $g = b^m x^m + b_{m-1} x_{m-1} + \dots + b_0$ then $\phi(f) \phi(g) = \phi (fg)$ and $\phi(f) + \phi(g) = \phi (f+g)$.
As the link shows, for the product case it is sufficient to prove it for monomials. I suggest you reread your definition of a polynomial ring, so you get a feel for what they are in a more algebraic (rather than analytical) context.
About the evaluation homomorphism: As it name says, it is an "evaluation". You have probably studied polynomials "as functions" in analysis, for example $f(x) = x^2$, you give it a real number and it gives another real number.
In algebra, polynomials are defined "formally": not as functions but as a sum of (finite) terms of the form $a_n x^n$. The evaluation homomorphism $\phi_\alpha$ is equivalent to substituting every appearance of $x$ with the associated element (in this case $\alpha$) and calculating the result.
This turns out to be very useful, for example we can define the set of polynomials that have $\alpha$ as a root, as the kernel of the homomorphism $\phi_\alpha$. We can then apply the powerful theorems we have for homomorphisms to polynomial rings: for example, we know that the set of all polynomials that have $\alpha$ as root is an ideal (since the kernel of a ring homomorphism is always an ideal).
The importance of the evaluation homomorphism is that it connects both aspects of the polynomial: the fact that it can be viewed as a function with some special points (the root) or it can be viewed formally.
