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Doug B
science forum beginner
Joined: 03 Feb 2005
Posts: 27
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 Posted: Wed Jul 12, 2006 1:18 am Post subject:
Another Galois theory problem
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Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension. There, x^p-a splits, so pick r with
r^p=a. Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a, so
that r is a multiple root of x^p-a, contradiction. But this is not
rigorous, since in fields of characteristic >0, just because two
polynomials agree everywhere does not imply they are the same
polynomials. So even though the polynomials x^p-a and (x-r)^p agree
everywhere, I don't know how to show they are the same polynomials, if
indeed they even are. Also, I am implicitly assuming here that x^p-a
is irreducible, otherwise even if I could show it has multiple roots in
a splitting field, that would prove nothing. But I have no idea how to
show that x^p-a is even irreducible in the original field. And all
this is just one direction of the statement, the other direction I have
even less of a clue how to proceed... |
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Achava Nakhash, the Lovin
science forum beginner
Joined: 14 Sep 2005
Posts: 35
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| Posted: Wed Jul 12, 2006 5:16 am Post subject:
Re: Another Galois theory problem
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Doug B wrote:
| Quote: | Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension. There, x^p-a splits, so pick r with
r^p=a. Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a, so
that r is a multiple root of x^p-a, contradiction. But this is not
rigorous, since in fields of characteristic >0, just because two
polynomials agree everywhere does not imply they are the same
polynomials. So even though the polynomials x^p-a and (x-r)^p agree
everywhere, I don't know how to show they are the same polynomials, if
indeed they even are. Also, I am implicitly assuming here that x^p-a
is irreducible, otherwise even if I could show it has multiple roots in
a splitting field, that would prove nothing. But I have no idea how to
show that x^p-a is even irreducible in the original field. And all
this is just one direction of the statement, the other direction I have
even less of a clue how to proceed...
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I'm just talking off the top of my head here, and I am too sleepy to
really think. That being said, (x-r)^p = x^p - r^p is clear by basic
number theory, since expanding the left-hand side with the binomial
theorem and applying the trivial observation that all of the interior
binomial coefficients are divisible by p which is 0 in any field of
characteristic p, you see the truth of this theorem.
This looks like a homework problem, but you have shown your effort at
least. Have you seen that elements of an extension field being
separable is related to the derivative (formal, of course) of their
minimal polynomial? This statement helps a lot in what you are doing.
Hope this helps,
Achava |
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Arturo Magidin
science forum Guru
Joined: 25 Mar 2005
Posts: 1838
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| Posted: Wed Jul 12, 2006 5:31 am Post subject:
Re: Another Galois theory problem
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Doug B wrote:
| Quote: | Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension.
|
That last statement makes no sense whatsoever. A splitting field of a
(set of) polynomials over F is ->necessarily<- an algebraic extension!
It is generated by adjoining roots of polynomials, and therefore it is
generated over F by algebraic elements; hence the extension must be
algebraic. So why are you saying that you are "enlarging it if
necessary" to make it algebraic? Methinks you are missing some
very important and basic facts about field extensions!
| Quote: | There, x^p-a splits, so pick r with
r^p=a. Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a,
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This follows trivially from the fact that the characteristic is p. The
coefficient of x^i in the expansion of (x-r)^p is (p choose i), which
is a multiple of p for all i, 0<i<p.
| Quote: | so
that r is a multiple root of x^p-a, contradiction.
|
Indeed. See below.
| Quote: | But this is not
rigorous, since in fields of characteristic >0, just because two
polynomials agree everywhere does not imply they are the same
polynomials.
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Just multiply out (x-r)^p. Don't know why you are trying to evaluate
anything.
(By the way: it ->is<- true for polynomials of degree less than
char(F), though that is neither here nor there in this situation)
| Quote: | So even though the polynomials x^p-a and (x-r)^p agree
everywhere, I don't know how to show they are the same polynomials, if
indeed they even are.
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Yes,. they are. Just write out (x-r)^p and remember the characteristic
is p.
| Quote: | Also, I am implicitly assuming here that x^p-a
is irreducible, otherwise even if I could show it has multiple roots in
a splitting field, that would prove nothing.
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Think minimal polynomial: the minimal polynomial of r over F must
divide x^p-a. Since x^p-a is equal to (x-r)^p, then the minimal
polynomial of r over F must be of the form (x-r)^k for some k, 0<k<=p.
If k>1, then this gives you the inseparabilit, and if k=1, that tells
you that a already had a p-th root in F, which contradicts your choice
of a.
| Quote: | But I have no idea how to
show that x^p-a is even irreducible in the original field.
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You don't have to (although it will be, since it will turn out that for
an irreducible polynomial to be inseparable, it must be of the form
g(x^p) for some polynomial g).
| Quote: | And all
this is just one direction of the statement, the other direction I have
even less of a clue how to proceed...
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If every algebraic extension is separable, let a in F and consider the
polynomial x^p-a. Let r be a root, and consider F(r). Now argue as
above about what the minimal polynomial of r must look like, and use
the fact that the extension is separable to deduce that r is already in
F.
Arturo Magidin. sans .sig |
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Timothy Murphy
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 275
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| Posted: Wed Jul 12, 2006 1:40 pm Post subject:
Re: Another Galois theory problem
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Doug B wrote:
| Quote: | Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension.
|
Aren't you being too clever?
If every extension is separable, then the extension
adding the pth root c of a is separable
But the minimal polynomial of c divides x^p - a = (x-c)^p.
It's not x^p - a, since this is not separable.
It's easy now to deduce that c is in the ground field.
On the other hand, a non-separable polynomial is of the form f(x^p).
If every element has a pth root you can express this as g(x)^p.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
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Jim Heckman
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 121
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| Posted: Fri Jul 14, 2006 1:23 am Post subject:
Re: Another Galois theory problem
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On 11-Jul-2006, "Doug B" <doug_protocols@yahoo.com>
wrote in message <1152667122.492475.163730@h48g2000cwc.googlegroups.com>:
| Quote: | Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
I don't really know how to proceed either direction. I've had the most
luck proving if it is perfect, it has the pth root property. I do this
by supposing it does not. Then there's an a such that x^p-a has no
roots. I take a splitting field and, enlarging it if necessary, assume
it is an algebraic extension. There, x^p-a splits, so pick r with
r^p=a.
|
You don't need a splitting field. Let irr(F,r) be the irreducible
polynomial of r over F. In F(r), x^p-a = x^p-r^p = (x-r)^p, so
irr(F,r) divides this and so irr(F,r) = (x-r)^n for some n <= p.
If n > 1 then r is separable over F which contradicts that F is
perfect, so n = 1 and r is in F.
| Quote: | Then what I *WANT* to write is (x-r)^p = x^p-r^p = x^p - a, so
that r is a multiple root of x^p-a, contradiction. But this is not
rigorous, since in fields of characteristic >0, just because two
polynomials agree everywhere does not imply they are the same
polynomials. So even though the polynomials x^p-a and (x-r)^p agree
everywhere, I don't know how to show they are the same polynomials, if
indeed they even are. Also, I am implicitly assuming here that x^p-a
is irreducible, otherwise even if I could show it has multiple roots in
a splitting field, that would prove nothing. But I have no idea how to
show that x^p-a is even irreducible in the original field.
|
In fact it isn't, but that doesn't matter. See above.
| Quote: | And all
this is just one direction of the statement, the other direction I have
even less of a clue how to proceed...
|
Assume there's an r inseparable over F, with f(x) = irr(F,r). Then
f'(r) = 0 since r is inseparable, so f(x) must divide f'(x), so
f'(x) = 0 since the degree of f'(x) is less than that of f(x). So
f(x) is in F[x^p], so f(x) = a_n*x^{pn} + a _{n-1}*x^{p(n-1)}... +
a_0 and since each a_i has a pth root in F, say b_i^p = a_i, then
f(x) = (b_n*x^n)^p + (b_{n-1}*x^{n-1})^p + ... + b_0 = (b_n*x^n +
b_{n-1}*x^{n-1} + ... b_0)^p, contradicting that f(x) is
irreducible over F.
--
Jim Heckman |
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ggl@jmilne.org
science forum beginner
Joined: 23 Oct 2005
Posts: 8
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| Posted: Fri Jul 14, 2006 2:19 am Post subject:
Re: Another Galois theory problem
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Doug B wrote:
| Quote: | Hi, I am trying to show that a field of characteristic p>0 is perfect
iff it has the property that every element has a pth root (the
definition of perfect field I have is, "F is perfect if every algebraic
extension is separable").
See http://www.jmilne.org/math/CourseNotes/math594f.html page 23 of the |
notes. |
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