When I was reading the following theorem from Joseph Gallian's Contemporary abstract algebra I got stucked.
Theorem
Let $F$ be a field and let $p(x) \in F[x] $ be irreducible over $F$. If $a$ is a zero of $p(x) $ in some extension $E$ of $F$, then $F(a) $ is isomorphic to $F[x] /\langle p(x) \rangle$.
Now the proof considers the function $\phi$ from $F[x] $ to $F(a) $ given by $\phi (f(x)) =f(a) $. Then after some steps it states that $\color{blue} {\phi (F[x])} $ $\color{blue} {\text{contains both } } $ $\color {blue} {F \text {and}\ a} $. I am unable to understand the cause of the blue line. If we take $f(x) \in F[x] $ as $f(x) =c$ where $c \in F$ then $\phi (f(x)) =f(a)=c $. Hence the image contains $F$. But I wonder how the image contains $a$. Please help me to understand. Thanks in advance.
Let $B$ be another ring and $\psi:F\to B$ a ring-homomorphism. Conser $\Psi:F[x]\to B$, s.t. $\Psi\vert_F = \psi$ then $\Psi$ is already determined by its image $\Psi(x)$, this is a variation of the so called universal property of polynomial fields (Further reading). In your case we just consider $\psi=\operatorname{id}_F$ and choose $\Psi:F[x]\to F[a]$ s.t. $\Psi\vert_F=\psi$ and $\Psi(x)=a$. Now $\operatorname{im} \Psi\subset F[a]$ is a subring which contains both $F$ and $a$, because $\Psi(x)=a$ and that the image of $\Psi$ is indeed a subring should be clear/can easily be checked. The kernel of this map is exactly $\langle p(x)\rangle$, because it is clear that $\ker \Psi \supset \langle p(x)\rangle$, now since $p(x)$ is irreducible it is prime (since $F[x]$ is UFD) and since it is $\neq 0$ the ideal is $\langle p(x)\rangle$ is maximal (because $F[x]$ is PID), hence it follows $\ker \Psi=\langle p(x)\rangle$ the homomorphism theorem establishes the isomorphism. Because now $F[x]/\langle p(x)\rangle\cong F[a]$ is a field we have $F(a)=F[a]$.