Fraleigh's proof of Kronecker's Theorem I was reading Fraleigh's abstract algebra textbook on field extension and he gave a proof of Kronecker's Theorem, but there are several spots that I don't quite understand so any help would be very appreciated.
One question before I start: If $ F $ is a field, then is it true that $ F $ is a subfield of $ F[x] $ since I can view elements in $ F $ as constant polynomials in $ F[x]? $
Theorem: Let $ F $ be a field and $ f(x) \in F[x] $ be a nonconstant polynomial. Then there exists an extension field $ E $ of $ F $ and an $ \alpha \in E $ such that $ f(\alpha) = 0 . $
Proof: We know that $ f(x) $ can be factored into products of irreducible polynomials in $ F[x] $ so it is sufficient to prove the theorem when $ f(x) $ is irreducible. We define the map $$ \psi: F \to F[x]/\langle f(x) \rangle $$ by $ \psi(a) = a + \langle f(x) \rangle $ for $ a \in F, $ which is a one-to-one map. We defined addition and multiplication in $ F[x]/\langle f(x) \rangle $ by choosing any representatives, so we may choose $ a \in (a + \langle f(x) \rangle). $ Thus $ \psi $ is a homomorphism that maps $ F $ one-to-one onto a subfield of $ F[x]/\langle f(x) \rangle. $ We identify $ F $ with $ \{a + \langle f(x) \rangle \mid a \in F \} $ by means of this map $ \psi. $ Thus we shall view $ E = F[x]/\langle f(x) \rangle $ as an extension field of $ F. $ 
Can someone explain what the bold paragraph is saying? In particular, I don't understand what he meant when he said We defined addition and multiplication in $ F[x]/\langle f(x) \rangle $ by choosing any representatives, so we may choose $ a \in (a + \langle f(x) \rangle). $  The way I understand it is that since any element $ a \in F $ can be written as $ a = a + 0.f(x), $ so $ F $ is a subfield of $ F[x]/\langle f(x) \rangle, $ but I don't think it's the correct interpretation.
Now we show that $ E $ contains a zero of $ f(x). $ Let $ \alpha = x + \langle f(x) \rangle \in E. $ Consider the evaluation homomorphism $ \phi_{\alpha}: F[x] \to E, $ then we have: 
$$ \phi_{\alpha}(f(x)) = \phi_{\alpha}(a_{0} + a_{1}x + \dots + a_{n}x^n) = a_{0} + a_{1}(x + \langle f(x) \rangle) + \dots + a_{n}(x + \langle f(x) 
\rangle)^n \in E $$  where $ a_{i} \in F. $ 
Hence $ \mathbf{f(\alpha) = (a_{0} + a_{1}x + \dots + a_{n}x^n) + \langle f(x) \rangle = f(x) + \langle f(x) \rangle = \langle f(x) \rangle = 0 \in E}. $
Regarding the last sentence, $ f(x) + \langle f(x) \rangle = \langle f(x) \rangle $ because $ f(x) \in \langle f(x) \rangle, $ but why then is $ \langle f(x) \rangle = 0? $
 A: Remember how you defined sum and multiplication in factor rings (Section 26). Also as $a=a+0.f(x)$ then $a \in (a+⟨f(x)⟩)$, and you can choose $a$ as representative. For, example $x$ is representative of $x+⟨f(x)⟩$.
So, you are almost right . Since $\psi$ is a monomorphism (why?) , $F$ and $\psi(F)$ are isomorphic  and $\psi(F)$ is a subfield of $F[x]/\langle f(x) \rangle$.Then you can see $F$ as a subfield.
In the second part , remember that in $E$: the factor ring, the new zero is $⟨f(x)⟩$ , because in the way you defined the sum in $E$, $⟨f(x)⟩$ is the neutral element.
A: For your first question, the answer is: yes, it is true that $F$ is a subfield of $F[x]$ since you can view elements in $F$ as constant polynomials in $F[x]$.
Inside $F[x]/\langle f(x)\rangle$, we have for its elements: 
$$a+\langle f(x)\rangle=a+g(x)f(x)=a$$
for any polynomial $g(x)\in F[x]$. In other words, any polynomial multiple of $f(x)$ is by definition zero inside $F[x]/\langle f(x)\rangle$. 
This is equivalent to saying that, inside $F[x]$ i.e. when viewed as polynomials in $F[x]$, $a+g(x)f(x)\equiv a$ modulo $f(x)$, (that is: modulo the equivalence relation on $F[x]$ imposed by the ideal generated by $f(x)$). 
So we say that $a$ is a representative of $a+\langle f(x)\rangle$ in $F[x]/\langle f(x)\rangle$. Thus, the author means that the field operations (addition and multiplication) are identified through $a$, which is viewed either as an element of $F$ or as (..a representative of..) an element of $F[x]/\langle f(x)\rangle$. This is why 
$$
\psi: F \rightarrow F[x]/\langle f(x)\rangle \\ 
a\mapsto \psi(a)=a+\langle f(x)\rangle=a
$$ 
is a monomorphism embedding $F$ inside $E=F[x]/\langle f(x)\rangle$. 
Regarding the last sentence, as has already been mentioned above, the ideal $\langle f(x)\rangle$ of $F(x)$ generated by $f(x)$, is identically zero inside $F[x]/\langle f(x)\rangle$. 
