Timeline for Degree of prime power of an element [closed]
Current License: CC BY-SA 4.0
15 events
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| Jan 28, 2020 at 16:41 | history | edited | YCor | CC BY-SA 4.0 |
removed/replaced deprecated tag, improved title
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| Jun 21, 2016 at 6:50 | review | Reopen votes | |||
| Jun 21, 2016 at 14:05 | |||||
| Oct 13, 2015 at 6:36 | comment | added | eric | No! If you adjoin one cube root of 2 to the rationals to get $F$, and then let alpha be another one, then $F(\alpha)/F$ has degree 2 and is generated by something whose cube is in the ground field. The question is as subtle as the OP suggests and should not be on hold. As I say, I once found all this out whilst marking an exam where both the students and examiner had underestimated this question. I would answer if it was unonholded. Kummer theory does not apply when the $q$ th roots of unity are not in the ground field. Do I remember seeing @kconrad going on about this here a few years ago? | |
| Oct 12, 2015 at 22:26 | comment | added | Gerry Myerson | @eric, I wonder whether the only counterexamples are those coming from $q$th roots of unity. | |
| S Oct 12, 2015 at 20:27 | history | suggested | kjetil b halvorsen | CC BY-SA 3.0 |
fixed typo in title
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| Oct 12, 2015 at 19:58 | review | Suggested edits | |||
| S Oct 12, 2015 at 20:27 | |||||
| Oct 12, 2015 at 19:33 | comment | added | eric | @Joe Silverman: the problem is that the primitive qth roots of unity are not in the ground field. This question is more delicate than it seems and I don't know why it's on hold. A collegue of mine asked a question closely related to this in a Galois Theory final exam and several students handed in solutions like yours ;-) but in fact $\alpha^q$ can be in $F$! (as I found out when I was marking the question...). Just let $\alpha$ be a primitive $q$th root of unity and let $F$ be an appropriate field with $p$ a divisor of $q-1$. | |
| Oct 12, 2015 at 12:43 | comment | added | user9072 | Please do not post to Mathematics and this site at the same time, especially not without mentioning it. math.stackexchange.com/questions/1476194/… | |
| Oct 12, 2015 at 12:05 | history | closed |
Andrés E. Caicedo Felipe Voloch Derek Holt Marco Golla Gerry Myerson |
Not suitable for this site | |
| Oct 12, 2015 at 11:31 | review | Close votes | |||
| Oct 12, 2015 at 12:06 | |||||
| Oct 12, 2015 at 11:30 | comment | added | user81389 | Well, I simply thought that It is not an easy problem. I guess that you are giving a counterexample. As you mentioned, $ f(X)$ is a product of $p$ terms of the form $X-xi \alpha$. Hence one needs to show that some product of $p$ terms of the form $X-xi \alpha$ is irreducible over $F$. How can you show it? | |
| Oct 12, 2015 at 11:20 | comment | added | Joe Silverman | What makes you say it's not an exercise problem (or do you mean that you didn't happen to find it in a textbook)? You have $F\subseteq F(\alpha^q)\subseteq F(\alpha)$ with the degree of $F(\alpha)/F$ being the prime $p$. So what you're asking is whether it is possible for $\alpha^q$ to be in $F$. Suppose it is. Let $f(X)\in F[X]$ be the irreducible poly of $\alpha$ over $F$. Note that $g(X)=X^q-\alpha^q\in F[X]$ has $\alpha$ as a root, so $f(X)\mid g(X)$. Hence $f(X)$ is a product of terms of the form $X-\zeta\alpha$ with the $\zeta$ being primitive $q$'th roots of unity. Does that help? | |
| Oct 12, 2015 at 11:14 | history | edited | user81389 | CC BY-SA 3.0 |
deleted 1 character in body
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| Oct 12, 2015 at 11:03 | review | First posts | |||
| Oct 12, 2015 at 11:20 | |||||
| Oct 12, 2015 at 11:01 | history | asked | user81389 | CC BY-SA 3.0 |