I'm going to have a lot of questions, and I'm going to ask them here.
2010 Paper 1 Galois Theory:
Let n be a +ve integer, let be the n'th cyclotomic polynomial. Recall that if K is a field of characteristic prime to n, then the roots of in K are precisely the primitive n'th roots of unity in K.
Using this fact, prove that if p is a prime number not dividing n, then p divides in Z, for some x in Z iff p = an + 1 for some integer a.
So far, I've noticed that p divides in Z iff mod p has a root  presumably here you're meant to use the hint given. But it doesn't guarantee the existence of roots, which is what you need, and I have no idea how to prove the converse direction.
I guess what I find hard is the 'roots of in K are precisely the primitive n'th roots of unity in K' part. Why is this important?
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 08052012 19:53
Last edited by around; 08052012 at 19:56. 
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 08052012 20:02
If for some then contains a primitive th root of unity , say. Then is a subgroup of of order . Apply Lagrange.
Edit: Removed comment about the converse; what I wrote was a pile of trout.Last edited by nuodai; 08052012 at 20:09. 
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 08052012 20:08
thanks for the help withe the second part btw. is it true that the roots of Phi_n are always primitive n'th roots, and that the content of the hint amounts to that if (char(k), p) = 1 then all the primitive n'th roots are roots of Phi_n? 
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 08052012 20:14
(Original post by around)
so it's clear what? sorry if i'm being obtuse.
If then divides . This is more useful than it looks because for this direction you don't need a root of to be primitive.
(Original post by around)
thanks for the help withe the second part btw. is it true that the roots of Phi_n are always primitive n'th roots, and that the content of the hint amounts to that if (char(k), p) = 1 then all the primitive n'th roots are roots of Phi_n?
Conversely if , say then so you don't have any primitive th roots.Last edited by nuodai; 08052012 at 20:18. 
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 08052012 20:28
(Original post by nuodai)
I wrote a load of rubbish, ignore that.
If then divides . This is more useful than it looks because for this direction you don't need a root of to be primitive.
divides . We thus want an element of such that x^n = 1. But we can't use Cauchy's theorem, because n isn't necessarily prime (and it's not necessarily true that n dividing the order of a group imples there is an element of order n). 
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 08052012 20:44
(Original post by around)
Ok.
divides . We thus want an element of such that x^n = 1. But we can't use Cauchy's theorem, because n isn't necessarily prime (and it's not necessarily true that n dividing the order of a group imples there is an element of order n).Post rating:1 
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 08052012 21:21
(Original post by Glutamic Acid)
It does when the group is cyclic. (Disclaimer: haven't really looked at the question.)
herping and derping all day long 
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 09052012 17:35
Finding Galois Groups:
I know of two ways of computing the Galois groups of polynomials: reduce mod p and hope you get lucky (i.e. if deg P prime, and we can find a 5 cycle and a transposition we know the Galois group has to be S_n) and explicitly calculate roots and field homomorphisms.
Take, for instance, the polynomial . This has roots , and is hence a degree 8 extension over Q. I'm guessing that the Galois group is Q_8, but to show this properly, would I have to find elements (that is, homomorphisms) that correspond to i, j and k?
edit: it's not a degree 8 extension. i'm not even sure what degree it is as an extension of Q...
edit 2: it's a degree 8 extension and the galois group is (C_2)^3Last edited by around; 09052012 at 17:55. 
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 09052012 21:48
(Original post by around)
edit 2: it's a degree 8 extension and the galois group is (C_2)^3
Let and . Then we wish to find . Then , so the only question is whether = 1 or 2, i.e. whether . This happens if and only if , but .
So it's a degree 8 extension, and the only transitive subgroup of S_4 of order 8 is D_8.Last edited by Glutamic Acid; 09052012 at 22:21.Post rating:1 
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 23052012 20:39
A directed subset of a poset P is some subset D such that every pair of elements of D have an upper bound in D. We say P is directedcomplete if every directed subset of P has a least upper bound.
Show that P is complete iff it is directed complete and every finite subset of P has a join.
I have no idea how you'd even begin to show the reverse implication. 
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 23052012 20:52
(Original post by around)
A directed subset of a poset P is some subset D such that every pair of elements of D have an upper bound in D. We say P is directedcomplete if every directed subset of P has a least upper bound.
Show that P is complete iff it is directed complete and every finite subset of P has a join.
I have no idea how you'd even begin to show the reverse implication.
Edit/minihijack: Did you work out what the contradiction is in the proof of the division algorithm for ordinals?Last edited by nuodai; 23052012 at 21:13. 
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 23052012 23:57
(Original post by nuodai)
Let . Let , which is directed by definition. Argue that .
Edit/minihijack: Did you work out what the contradiction is in the proof of the division algorithm for ordinals?
(also, nope. i'll ask about it in my supervision) 
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 24052012 00:02
(Original post by around)
I think you need to take a few more unions: you want S^{bar} to be S, union the sup of any pair, union the sup of any triple, union ... 
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 24052012 01:37
So glad I dropped logic for the exam

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 24052012 08:59
(Original post by Daniel Freedman)
So glad I dropped logic for the exam 
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 26052012 19:20
More logic:
Let V be a real vector space, suppose V has a basis of cardinality . What is the cardinality of V?
My answer:
Every element in V can be written as a finite linear combination of elements of the basis (this is the definition of basis). So we pick out all possible finite subsets of by taking , and then every vector in V is a function from one of these subsets to R.
So the cardinality of V is the cardinality of which simplifies to . 
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 26052012 19:56
Well, the set of all finite subsets of is a strict subset of (don't mix ordinal and cardinal notation!) so it's not immediately clear that this gives the correct answer. A slightly better approach would be to argue that there are at most finite subsets of , but for all finite n, so there are at most finite subsets of .
Post rating:1 
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 26052012 20:00
(Original post by Zhen Lin)
Well, the set of all finite subsets of is a strict subset of (don't mix ordinal and cardinal notation!) so it's not immediately clear that this gives the correct answer. A slightly better approach would be to argue that there are at most finite subsets of , but for all finite n, so there are at most finite subsets of . 
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 27052012 01:42
So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...

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 27052012 07:20
(Original post by Daniel Freedman)
So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...
"Show that the degree of a complex irreducible character of a ﬁnite group is a factor of the order of the group.
State and prove Burnside’s theorem. You should quote clearly any results you use."
And that's not the whole thing
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Updated: June 8, 2012
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