Third Year Revision Thread
Maths and statistics discussion, revision, exam and homework help.
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Third Year Revision ThreadI'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?
Last edited by around; 08-05-2012 at 19:56. -
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Re: Third Year Revision Threadso it's clear what? sorry if i'm being obtuse.
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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Re: Third Year Revision ThreadI wrote a load of rubbish, ignore that.(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.
Yes. Except I think you mean (char(k), n) = 1, and not what you wrote.(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; 08-05-2012 at 20:18. -
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Re: Third Year Revision ThreadOk.(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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Re: Third Year Revision ThreadIt does when the group is cyclic. (Disclaimer: haven't really looked at the question.)(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).
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Re: Third Year Revision Thread(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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Re: Third Year Revision ThreadFinding 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; 09-05-2012 at 17:55. -
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Re: Third Year Revision ThreadHmm. (C_2)^3 doesn't appear as a subgroup of S_4.(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 ![[\mathbb{Q}(\alpha,\beta):\mathbb {Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb {Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]](http://www.thestudentroom.co.uk/latexrender/pictures/0c/0c78c4361bb87d603018df39927b8fb1.png "[\mathbb{Q}(\alpha,\beta):\mathbb {Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb {Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]")
. Then ![[\mathbb{Q}(\alpha):\mathbb{Q}] = 4](http://www.thestudentroom.co.uk/latexrender/pictures/b4/b452d2fc93515b0da29be385a2d2debe.png "[\mathbb{Q}(\alpha):\mathbb{Q}] = 4")
, so the only question is whether ![[\mathbb{Q}(\alpha,\beta):\mathbb {Q}(\alpha)]](http://www.thestudentroom.co.uk/latexrender/pictures/7c/7cd67e799ea4c1e06e6bf219a1fe2e28.png "[\mathbb{Q}(\alpha,\beta):\mathbb {Q}(\alpha)]")
= 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; 09-05-2012 at 22:21. -
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Re: Third Year Revision ThreadA 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 directed-complete 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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Re: Third Year Revision ThreadLet(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 directed-complete 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.
. Let 
, which is directed by definition. Argue that 
.
Edit/mini-hijack: Did you work out what the contradiction is in the proof of the division algorithm for ordinals?Last edited by nuodai; 23-05-2012 at 21:13. -
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Re: Third Year Revision ThreadI 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 ...(Original post by nuodai)
Let. Let , which is directed by definition. Argue that .
Edit/mini-hijack: 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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Re: Third Year Revision ThreadActually we're both wrong. You want(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 ...
to be the closure of 
under the operation of successively appending unions of pairs of members.
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Re: Third Year Revision ThreadBut it's so good! And it's not like PTJ's setting the ques...oh crap.(Original post by Daniel Freedman)
So glad I dropped logic for the exam -
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Re: Third Year Revision ThreadMore 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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Re: Third Year Revision ThreadWell, 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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Re: Third Year Revision ThreadAh, thanks. I was trying to work along these lines, but I was distracted by thoughts of R^omega.(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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Re: Third Year Revision ThreadThey are ridiculous.(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
