Show that if $E$ is an extension of $F$ and $[E:F] = n$ and $p(x)$ is irreducible over $F$ and $\gcd(n, \deg p(x)) = 1$ $\implies $ $[E(a) : E] \le [F(a) : F] = \deg f(x)$ where $a$ is a zero of $p(x)$ in some extension of $F$.

I know that for any $a,b$ in an extension $E$, $[F(a,b):F(b)] \le [F(a):B]$, but I can't see how this would relate to showing the inequality.

  • 1
    $\begingroup$ It looks like $p$ turned into $f$ somewhere. And what is $a$? $\endgroup$ – carmichael561 Sep 26 '16 at 16:36
  • $\begingroup$ Notice the proper way to set $\deg p(x)$ in MathJax. $\qquad$ $\endgroup$ – Michael Hardy Sep 26 '16 at 16:39
  • $\begingroup$ Edited for clarity. $\endgroup$ – Oliver G Sep 26 '16 at 16:40

I don't know why you need all those extra conditions, but if $F \subset E \subset L$ is a field extension, $p\in F[x]$ is irreducible and $a$ is a root of $p$ in $L$, then you want to show that $$ [E(a):E]\leq [F(a):F] = \deg(p(x)) $$ For this, just note that $a$ is algebraic over $E$, since it is algebraic over $F$. Hence, it has a minimal polynomial $q(x) \in E[x]$. Since $p(x) \in E[x]$ and $p(a) = 0$, it follows that $$ q(x) \mid p(x) \text{ in } E[x] \qquad (\dagger) $$ In particular, $$ [E(a):E] = \deg(q(x)) \leq \deg(p(x)) = [F(a):F] $$

Now suppose $n:= [E:F]$ and $\deg(p(x))$ are relatively prime, then you get something stronger, viz. $$ q(x) = p(x) $$ To prove this, you need a corollary of the tower law, that states that

If $F\subset E_1$ and $F\subset E_2$ are two finite field extensions such that $[E_1:F]$ and $[E_2:F]$ are relatively prime, then $$ [E_1E_2:F] = [E_1:F][E_2:F] $$

In your case, take $E_1 = E$ and $E_2 = F(a)$, then note that $E_1E_2 = E(a)$. Hence, you get that $$ \deg(q(x))n = [E(a):E][E:F] = [E(a):F] = [E:F][F(a):F] = n\deg(p(x)) $$ and so $\deg(q(x)) = \deg(p(x))$ which, together with $(\dagger)$ proves that $q(x) = p(x)$ (since they are both monic).


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