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Hi guys! :-)
I'm new to the forum. Currently doing a course on Galois Theory - it's a fascinating subject but our lecturer is a bit....erm....under-par! :-s
He's only doing abstract theory & NO examples, but our exam is 100% applications! So I was wondeing if you guys could possibly help me please?
Here are a couple of questions from the last few years exam - if you have any idea how to solve ANY of them I would be EXTREMELY greatful! :-) Please could I ask you to be explicit with you methods, use of theorems, etc - these will LITERALLY be the first worked examples I've ever seen!
Thanks everyone! :-)
Questions:
1) Let the complex number A be a root of X^7+6X^2+2. Find the degree of Q(A):Q.
2) Let the complex number Zeta be a 256th root of unity. Find the degree of Q(Zeta:Q) & the minimal polynomial for Zeta over Q.
3) Let the complex number Zeta be a 12th root of unity. Find the Galois Group Gal(Q(Zeta):Q) & all subfields of Q(Zeta).
4) Find the minimal polynomial of Sqrt(6)+Sqrt(3) over (a) Q(Sqrt(2)) & (b) Q(Sqrt(3)).
Thanks so much in advance! :-) x
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[/align]Last edited by JaneFairfax (2009-03-05 12:49:45)
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1) Let the complex number A be a root of X^7+6X^2+2. Find the degree of Q(A):Q.
This question is trivial so long as you know the following:
1. If m is irreducible with root a, then m is the minimal polynomial of a.
2. Eisenstein's Criterion
3. Gauss's lemma
If you don't have all of those, or if you don't see how to piece them together, let me know.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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I made a mistake with 4(b) earlier; the following is the correct solution.
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Wow - thanks for the quick replies guys!! :-D Much appreciated!
JaneFairfax - I don't quite follow your method for 2) - Why do you look at the gcd's of n compared with 128? And why do you stop at 64? And why go up in multiples of 2?
Like I said these are LITERALLY my first worked examples, so any further explanation of the method used would be EXTREMELY appreciated! :-)
Many thanks again. x
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PS - is there any other way to show the polynomial you arrive at in 4(a) is irreducible over Q(Sqrt(2)) other than to show that Sqrt(6)+Sqrt(3) is not an element of Q(Sqrt(2)) & hence the polynomial has no roots in Q(Sqrt(2)) ?
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JaneFairfax - I don't quite follow your method for 2) - Why do you look at the gcd's of n compared with 128? And why do you stop at 64? And why go up in multiples of 2?
Since
for any , this must mean thatHence
.Therefore if
, and for any in other words, is a primitive 128th root of −1.If
, and for any i.e., is a primitive 64th root of −1.Similarly
⇒ is a primitive 32nd root of −1. And so on and so forth.Last edited by JaneFairfax (2009-03-06 11:28:11)
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PPS - Ricky - I think I've got the answer now (7 yes?) - thankyou! :-) But do you need Gauss' Lemma? Or doesn't the irreducibility of the polynomial over Q come straight from the Eisenstein Criterion? (with p = 2)
Many thanks. :-)
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PS - is there any other way to show the polynomial you arrive at in 4(a) is irreducible over Q(Sqrt(2)) other than to show that Sqrt(6)+Sqrt(3) is not an element of Q(Sqrt(2)) & hence the polynomial has no roots in Q(Sqrt(2)) ?
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Thank you LadyFairfax - but I don't understand the meaning or significance of your 2nd & 3rd lines about the primitive root of unity. How do you know this gcd for k holds? And why does that lead us to conclude line 3?
Thank you very much again. :-) x
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Well, you can't know it's irreducible unless it has no roots in the set & therefore we need to know that the obvious root ( Sqrt(6)+Sqrt(3) ) is not in the set? (unless we are just going to assume that to be the case) :-)
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See. How in the world can you write
in the form
for some rational numbers
and ? How?Offline
PPS - Ricky - I think I've got the answer now (7 yes?) - thankyou! :-) But do you need Gauss' Lemma? Or doesn't the irreducibility of the polynomial over Q come straight from the Eisenstein Criterion? (with p = 2)
Many thanks. :-)
Eisenstein allows you to conclude it is irreducible over Z, Gauss's Lemma tells you that monic polynomials that are irreducible over Z are also irreducible over Q.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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But in order to prove that JaneFairfax you would have to assume that you CAN write it in that form and then show this leads to a contradiction. (a very easy contradiction - you get that 1+Sqrt(2) = a/Sqrt(3) + [Sqrt(2)]b/Sqrt(3) for rational a &b, which is obviously impossible.) I know it's obvious, but if yuou want to be thorough.... :-)
Could you please clarify your stuff about gcds in part (2)? :-) Thank you. x
Ricky - thanks. :-)
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Could you please clarify your stuff about gcds in part (2)? :-) Thank you. x
I already did! Didnt you see my post above?
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Oh yes sorry, my mistake! :-s lol! I must have missed it. Thank you very very much! :-D x
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