1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ But how can I put $a$ in $f(x) $ because $f(x) $ is over $F$ but $a$ is in $E$? $\endgroup$ Jul 24, 2021 at 9:59
  • 1
    $\begingroup$ 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, $a \in \phi(F[x])$. $\endgroup$ Jul 24, 2021 at 10:01
  • $\begingroup$ 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 $a$. $\endgroup$ Jul 24, 2021 at 10:06
  • $\begingroup$ Thanks @Teresa Lisbon $\endgroup$ Jul 24, 2021 at 10:15
  • $\begingroup$ You are welcome! If you wish you can self-answer and call me, since I feel a little uncomfortable writing such a short answer. $\endgroup$ Jul 24, 2021 at 10:19

1 Answer 1

1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .