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Let $M$ be a normal extension of $F$. Suppose that $a_1, a_2$ are in $M$ and are the roots of the minimal polynomial of $a_1$ over $F$, and $b_1,b_2$ are the roots of minimal polynomial of $b_1$ over $F$. Determine whether or not there is an $F$-automorphism $Z$ from $M$ to $M$ with $Z(a_1)=a_2$, and $Z(b_1)=b_2$.

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Is this homework? –  Asaf Karagila Feb 7 '12 at 11:39
    
In the second sentence, are $a_1, a_2$ the roots of the minimal polynomial in the sense that the minimal polynomial of $a_1$ over $F$ has degree two and splits as $(X - a_1)(X - a_2)$ over $M$? Same question for the $b_i$. [Also, I fixed up your formatting and tried to find a more descriptive title; admittedly, the title I've given it is slightly awkward.] –  Dylan Moreland Feb 7 '12 at 13:54
    
Actually, you didn't demand separability so the news could be worse than that, I guess. –  Dylan Moreland Feb 7 '12 at 14:30
    
An anonymous user suggested an edit to change the hypotheses to include that these are the only roots. I rejected the edit, because it changes the question and its answer, so the original poster should do such an edit. Is that ok? –  Jyrki Lahtonen Feb 8 '12 at 8:15
    
@Jyrki, I rejected that edit, too, and for the same reason. This could be one of those cases where the anonymous editor is the original poster, so I suggested the person contact a moderator if that was the case. –  Gerry Myerson Feb 8 '12 at 23:19

1 Answer 1

As Dylan points out in the comments, it is not clear whether you want $a_1,a_2$ to be the only roots of the minimal polynomial of $a_1$; likewise, $b_1,b_2$. I'll point you toward a counterexample in the case where they are not the only roots. Let the first polynomial have distinct roots $a_1,a_2,a_3$ (and perhaps others), let $b_i=2a_i$ for all $i$; you won't find an automorphism that takes $a_1$ to $a_2$ and $b_1$ to $b_3$.

EDIT: in partial and elliptical response to awllower's comment, there's no automorphism taking $\sqrt2$ to $-\sqrt2$ and also taking $\root4\of2$ to $-\root4\of2$.

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Let us then assume the other case; what can we say about the circumstance? Is there such an automorphism? And if we only require one condition? Is there any difference whereby attached? –  awllower Feb 8 '12 at 8:58

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