Suppose that $f(x)$ and $g(x)$ are irreducible over $F$ and that $\gcd(~\deg g(x),\deg f(x)~)=1$. If $a$ is a zero of $f(x)$ in some extension of $F$, show that $g(x)$ is irreducible over $F(a)$

Attempt: Let $b$ denote a zero of $g(x)$ in some extension of $F$. We have :

$[F(a,b):F(a)][F(a):F]=[F(a,b):F] = [F(a,b):F(b)][F(b):F] ~~~~..... (1)$

$\deg f(x) = [F(a):F ] ~~~~...... (2)$

$\deg g(x) = [F(b):F ] ~~~~......(3)$

Since, $\deg g(x)$ and $ \deg f(x)$ are prime to each other

$\implies [F(b):F]$ divides $[F(a,b):F(a)][F(a):F] \implies [F(b):F]$ divides $[F(a,b):F(a)]$

How do I move ahead?

Let $a,b \in \mathbb Q~~|~~b \neq 0$. Show that $\mathbb Q(\sqrt a) = \mathbb Q(\sqrt b)$ if and only if there exists some $c \in \mathbb Q$ such that $a = bc^2$

Attempt: Case $1$: When $ a = {p_1}^2/{q_1}^2$ where $p_1,q_1 \in \mathbb Z,\gcd(p_1,q_1)=1$

Then : $\mathbb Q(\sqrt a) = \mathbb Q(p_1/q_1) =\mathbb Q = \mathbb Q(\sqrt b)$ only if $\sqrt b = \dfrac {p_2}{q_2}$ or $b=\dfrac {{p_2}^2}{{q_2}^2} $

In such a case : $\dfrac {{p_1}^2}{{q_1}^2}= \dfrac {{p_2}^2}{{q_2}^2} {(\dfrac {q_2 p_1}{p_2q_1})}^2$. Hence $c = (\dfrac {q_2 p_1}{p_2q_1})$

Did I do this correctly? How do i move ahead?

Thank you for your help..


You have $[F(b):F]\mid [F(a,b):F(a)]$. Now bound it from the other direction,

$$[F(a,b):F(a)]\le [F(b):F].$$

Note this is true for arbitrary $F,a,b$. Hint: consider $b$'s minpoly over $F(a)$ and $F$; what's a relationship between them. Once the indices are squeezed, they must be equal. Now what?

For your second question, you did Case 2 wrong. The number $\sqrt{a}$ can never be transcendental over the rationals, since it's a zero of $x^2-a$! First do the $a=bc^2\Rightarrow(\circ)$ part (easy) and then you can focus on the other direction. What kinds of things can you deduce from $\Bbb Q(\sqrt{a})=\Bbb Q(\sqrt{b})$? Right off the bat you should get a couple of containments, which means you can write things certain ways..

  • $\begingroup$ $b$'s minimal polynomial over $F$ is $g(x)$. if the minimal polynomial of $b$ over $F(a)$ is $g_1(x)$, then : $\deg g_1(x) \leq \deg g(x) \implies [F(a,b):F(a)] \leq [F(b):F] \implies [F(a,b):F(a)]= [F(b):F]$ . Is this correct? $\endgroup$ – MathMan Aug 14 '14 at 8:07
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    $\begingroup$ Good so far. Just to make sure you understand: why does $[F(a,b):F(a)]=[F(b):F]$ imply that $g(x)$ is irreducible over $F(a)$? Next, every element of $\Bbb Q(\sqrt{b})$ can be written as (fill in the blank). Same idea goes for the other field. Proceed from there. $\endgroup$ – whacka Aug 14 '14 at 8:14
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    $\begingroup$ No. Answer my question first: every element of $\Bbb Q(\sqrt{b})$ can be written in the form (fill in the blank). $\endgroup$ – whacka Aug 14 '14 at 8:16
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    $\begingroup$ Let $h$ be the minpoly of $b$ over $F(a)$. Since $g(b)=0$, we know $h(x)\mid g(x)$ in the ring $F(a)[x]$. But since $\deg h=\deg g$, we must have $h(x)=g(x)$, so $g$ - being a minimal polynomial over $F(a)$ - is irreducible over $F(a)$. | Next, you have $\sqrt{a}=p+q\sqrt{b}$ for some $p,q\in\Bbb Q$. Square both sides and solve for $\sqrt{b}$. What do you get? $\endgroup$ – whacka Aug 14 '14 at 8:24
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    $\begingroup$ No need to square that again. If we start with the assumption that $\sqrt{a}$ (and hence $\sqrt{b}$) is not rational, then that equation is a contradiction. This means we couldn't have divided by $2pq$, which means one of $p,q$ must be $0$. Clearly it must be $p$ that's zero, not $q$. Thus $\sqrt{a}=q\sqrt{b}$. Now square both sides. $\endgroup$ – whacka Aug 14 '14 at 8:55

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