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Let $x^n -a \in F[x]$ be an irreducible polynomial over $F$, and let $b \in K$ be its root, where $K$ is an extension field of F. If $m$ is a positive integer such that $m|n$, find the degree of the minimal polynomial of $b^m$ over $F$.

My solution:

$[F(b^m):F]=[F(b^m):F(b)][F(b):F] \Rightarrow n\le [F(b^m):F]$


$F(b^m)\subset F(b) \Rightarrow [F(b^m):F]\le [F(b):F]=n$




I didn't use the fact that $m|n$, where am I wrong? I need help how to solve this problem.


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So, one thing you got wrong is the inclusion of fields. You have $F \subset F(b^m) \subset F(b)$, and not $F \subset F(b) \subset F(b^m)$. – Rankeya Nov 1 '12 at 21:45
Your very first equation is not true in general. Think of $b=\sqrt 2$ and $m=2$. – Phira Nov 1 '12 at 21:45
First of all - $F(b^m)/F(b)$ is not a field extension - its the other way around! – Belgi Nov 1 '12 at 21:45
Also, as a hint, if $m| n$, then $n = mk$ for some integer $k$. Then, $b^n - a = 0 \Rightarrow (b^m)^k - a = 0$. Then $b^m$ satisfies a polynomial of degree lower than $n$. What is it? – Rankeya Nov 1 '12 at 21:47
Then why did you write $[F(b^m):F(b)]$? – Rankeya Nov 2 '12 at 6:09
up vote 4 down vote accepted

Hint Let $km = n$ then if $b^{mk} - a = 0$ then $(b^m)^k - a = 0$ so maybe $x^k-a$ is the minimal polynomial?

Hint' Show that if $x^k-a$ has a factor then so does $x^{mk} - a$.

Given a field extension $K/L$ then $K$ is a vector space with coefficients in $L$ of dimension $\left[K:L\right]$ which is called the degree of the field extension.

The vector space $F(b^m)$ is spanned by $F$-linear combinations of the basis vectors $\left\{1,b^m,b^{2m},\ldots,b^{(k-1)m}\right\}$ so $\left[F(b^m):F\right] = k$.

Furthermore $\left[F(b):F\right] = n$ and $\left[F(b):F(b^m)\right] = m$ (prove these, for the second one use that $b$ is the minimal polynomial of $z^m - b^m$ [why can we not just use $z-b$?] in $F(b^m)$) so by $mk = n$ we have the identity $\left[F(b):F(b^m)\right]\left[F(b^m):F\right] = \left[F(b):F\right]$.

Why is $F(b^m)$ spanned by $F$-linear combinations of $\{1,b^m,b^{2m},…,b^{(k−1)m}\}$?

$F(b^m)$ is the field generated by all well defined sums differences products and fractions of the elements $F \cup {b^m}$. So that means it includes $b^m, (b^m)^2, (b^m)^3, \ldots$ but since $b^m$ satisfies a polynomial every power of $b^m$ higher or equal to $k$ can be reduced by it to a linear combination of lower powers. Similarly $(b^m)^{-1} = a (b^m)^{k-1}$, of course the sum of linear combinations is again a linear combination so we have seen that $F$-linear combinations of $\{1,b^m,b^{2m},…,b^{(k−1)m}\}$ span $F(b^m)$. The fact it's an independent basis (i.e. cannot be made smaller) comes from the polynomial being minimal.

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Then [F(b^m):F]=k? – user42912 Nov 2 '12 at 23:14
@user42912, $c = c \cdot 1 + 0 \cdot b^m + 0 \cdot b^{2m} + \ldots + 0 \cdot b^{(k-1)m}$ I don't really understand what you're asking here. – sperners lemma Nov 4 '12 at 8:22
@user42912, I'm glad I could help. – sperners lemma Nov 4 '12 at 17:05
@user42912, if that identity holds then you have just written down a polynomial of degree smaller than the minimal one, contradiction! – sperners lemma Nov 4 '12 at 18:19
This last technique is really beautiful and elegant. Thank you again for the pacience. – user42912 Nov 4 '12 at 19:24

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