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Let $K ⊂ L$ be an extension. Let $α ∈ L$ algebraic over $K$. Show that $α^2$ is algebraic over $K$.

Let $n$ be the degree of the minimal polynomial $p(x)$ of $α$ over $L$ and $m$ be the degree of the minimal polynomial $q(x)$ of $α^2$ over $L$.

Since $α^2 ∈ L(α)$, we have $L(α^2) ⊂ L(α)$, then $m\leq n$.

I have trouble understanding the last statement.

How do I know that $α^2 ∈ L(α)$ and why is $L(α^2) ⊂ L(α)$

Sumanta
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magnus
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    If $F$ is a field and $\alpha\in F$, then $\alpha^2\in F$ – J. W. Tanner Sep 08 '20 at 15:17
  • So α^2 ∈ L(α) because of the closure property of a field. Since α * α has to be in L. Got it thanks :). Any clue why L(α^2) ⊂ L(α)? – magnus Sep 08 '20 at 15:33
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    I think that follows from $L\subset L(\alpha)$ and $\alpha^2\subset L(\alpha)$ – J. W. Tanner Sep 08 '20 at 15:39
  • Thanks for the help, I understand it now :) – magnus Sep 08 '20 at 15:44
  • You can also directly find a polynomial whose zero is $\alpha^2$ by computing the determinant of the matrix of multiplication by $\alpha^2$. See https://math.stackexchange.com/questions/1130843/constructive-proof-that-algebraic-numbers-form-a-field . – tomasz Sep 08 '20 at 17:43

1 Answers1

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Since

$\alpha \in L \tag 1$

is algebraic over $K$,

$[K(\alpha): K] < \infty, \tag 2$

and since

$\alpha^2 \in K(\alpha), \tag 3$

$K(\alpha^2) \subset K(\alpha), \tag 4$

and thus

$[K(\alpha^2): K] \le [K(\alpha): K] < \infty, \tag 5$

which in turn implies $\alpha^2$ is algebraic over $K$.

Note Bene, 8 September 2020 10:20 AM PST: The above demonstration is easily extended to show that $\alpha^n$ is algebraic over $K$ for all $n \in \Bbb N$; one merely need observe that (3) may be replaced with

$\alpha^n \in K(\alpha), \tag 6$

whence

$K(\alpha^n) \subset K(\alpha), \tag 7$

from which

$[K(\alpha^n): K] \le [K(\alpha): K] < \infty, \tag 8$

that is, $\alpha^n$ is algebraic over $K$. End of Note.

Robert Lewis
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