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Let $K \subset L \subset M$. Let $f(x)$ be the minimal polynomial for $\alpha$ over $M$. Moreover, suppose $f(x)\in L[x]$, then is the minimal polynomial for $\alpha$ over $L$ also $f(x)$?

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I don't think you need to mention $K$, here. –  Dylan Moreland Oct 12 '11 at 23:48
    
Yes : Denote $f_L$ the minimal polynomial over $L$. Because $f(\alpha) = 0$ and $f\in L[X]$, then $f_L$ divides $f$. But because $f_L(\alpha) = 0$ and $f_L \in M[X]$, then $f_L$ divides $f$. –  Joel Cohen Oct 12 '11 at 23:55

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up vote 5 down vote accepted

Yes. Let's write $g(x)$ for the minimal polynomial of $\alpha$ over $L$. Since $f(x) \in L[x]$ has $\alpha$ as a root, we have $g \mid f$. But $L[x] \subset M[x]$, so by the same reasoning over $M$ we have $g \mid f$. Since both polynomials are monic, we have equality.

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