# A theoretical idea for proof of a theorem of field extension

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.

• But how can I put $a$ in $f(x)$ because $f(x)$ is over $F$ but $a$ is in $E$? Jul 24, 2021 at 9:59
• Ah, sorry. I'll clarify. @Math-Learner Consider the polynomial $f(x)=x$. This belongs to $F[x]$ rather obviously. Now, what is $\phi(f(x))$? Well, it's equal to $f(a)$, which by definition is the quantity in $F(a)$ obtained by substituting $a$ in place of $x$ in the polynomial $f(x)$. Now $f(x)=x$, so $\phi(f(x)) = f(a)=a$. Therefore, $a \in \phi(F[x])$. Jul 24, 2021 at 10:01
• The map is well defined because the way $F[a]$ is defined , one can show that for any $f(x) \in F[x]$ we have $f(a) \in F[a]$. If you want even more detailing, we should perhaps consider an example of a polynomial and an $a$. Jul 24, 2021 at 10:06
• Thanks @Teresa Lisbon Jul 24, 2021 at 10:15
• You are welcome! If you wish you can self-answer and call me, since I feel a little uncomfortable writing such a short answer. Jul 24, 2021 at 10:19

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 $$F(a)=F[a]$$.