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?
Third Year Revision Thread
Announcements  Posted on  

Applying to Uni? Let Universities come to you. Click here to get your perfect place  20102014 


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? 
(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. 
(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). 
(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). 
(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 
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)^3 
(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. 
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. 
(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? 
(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) 
(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 ... 
So glad I dropped logic for the exam

(Original post by Daniel Freedman)
So glad I dropped logic for the exam 
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 . 
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 .

(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 . 
So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...

(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
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: