Suppose $F \subset K$ is a field extension and $\alpha \in K$. If $[F[\alpha]:F] = 5$, prove that $F[\alpha ^2] = F[\alpha]$.

Question

Suppose $F \subset K$ is a field extension and $\alpha \in K$. If $[F[\alpha]:F] = 5$, prove that $F[\alpha ^2] = F[\alpha]$.

F as a basis of $F[\alpha] = ${$1, \alpha, \alpha^2, \alpha^3, \alpha^4$}

$F[\alpha]$ = {$a_0 + a_1 \alpha + a_2 \alpha^2 + a_3 \alpha^3 + a_4 \alpha ^4 $}

$F[\alpha^2]$ = {$a_0 + a_1 \alpha^2 + a_2 \alpha^4 + a_3 \alpha^6 + a_4 \alpha ^8 $}

All elements of $F[\alpha ^2]$ can be written as elements of $F[\alpha]$, as $\alpha ^ 6 = \alpha * \alpha ^ 5 = \alpha$, and $\alpha^8 = \alpha^3$. Thus $F[\alpha ^2] \subset F[\alpha]$

Is this all valid so far? I am stuck proving $ F[\alpha] \subset F[\alpha^2]$

Answer 1

Since

$\alpha^2 \in F[\alpha], \tag 1$

we have

$F[\alpha^2] \subset F[\alpha]; \tag 2$

thus

$[F[\alpha]:F] = [F[\alpha]:F[\alpha^2]][F[\alpha^2]:F]; \tag 3$

since

$[F[\alpha]:F] = 5, \tag 4$

we find that

$[F[\alpha]:F[\alpha^2]][F[\alpha^2]:F] = 5; \tag 5$

thus

$[F[\alpha]:F[\alpha^2]] = 1 \tag 6$

or

$[F[\alpha]:F[\alpha^2]] = 5; \tag 7$

if (7) holds, then

$[F[\alpha^2]:F] = 1, \tag 8$

whence

$\alpha^2 \in F, \tag 9$

but then

$[F[\alpha]:F] = 2, \tag{10}$

since (9) implies $\alpha$ satisfies a polynomial in $F[x]$ of degree at most $2$; the degree cannot be $1$ since then $\alpha \in F$, contradicting the given (4); (10) however also contradicts (4); therefore (7) is false and so (6) holds, implying

$F[\alpha^2] = F[\alpha]. \tag{11}$

$OE\Delta$.