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Third Year Revision Thread

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 Φn\Phi_n be the n'th cyclotomic polynomial. Recall that if K is a field of characteristic prime to n, then the roots of Φn\Phi_n 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 Φn(x)\Phi_n(x) in Z, for some x in Z iff p = an + 1 for some integer a.

So far, I've noticed that p divides Φn(x)\Phi_n(x) in Z iff Φn\Phi_n 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 Φn\Phi_n in K are precisely the primitive n'th roots of unity in K' part. Why is this important?
(edited 11 years ago)

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Reply 1
If Φn(x)0(modp)\Phi_n(x) \equiv 0 \pmod p for some xx then Fp\mathbb{F}_p contains a primitive nnth root of unity ξn\xi_n, say. Then {1,ξn,,ξnn1}\{ 1, \xi_n, \dots, \xi_n^{n-1} \} is a subgroup of Fp×\mathbb{F}_p^{\times} of order nn. Apply Lagrange.

Edit: Removed comment about the converse; what I wrote was a pile of trout.
(edited 11 years ago)
Reply 2
Original post by nuodai
If p=an+1p=an+1 then p=an+1p=an+1 and p1(modn)p \equiv 1 \pmod n, so it's clear.


so 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?
Reply 3
Original post by around
so it's clear what? sorry if i'm being obtuse.
I wrote a load of rubbish, ignore that.

If p=an+1p=an+1 then nn divides Fp×\left| \mathbb{F}_p^{\times} \right|. This is more useful than it looks because for this direction you don't need a root of Φn mod p\Phi_n\ \text{mod}\ p 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?

Yes. Except I think you mean (char(k), n) = 1, and not what you wrote.

Conversely if pnp\, |\, n, say n=bpn=bp then (Xn1)(Xb1)p(modp)(X^n-1) \equiv (X^b - 1)^p \pmod p so you don't have any primitive nnth roots.
(edited 11 years ago)
Reply 4
Original post by nuodai
I wrote a load of rubbish, ignore that.

If p=an+1p=an+1 then nn divides Fp×\left| \mathbb{F}_p^{\times} \right|. This is more useful than it looks because for this direction you don't need a root of Φn mod p\Phi_n\ \text{mod}\ p to be primitive.


Ok.

nn divides Fp× |\mathbb{F}_p^{\times}|. We thus want an element of Fp×\mathbb{F}_p^{\times} 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.

nn divides Fp× |\mathbb{F}_p^{\times}|. We thus want an element of Fp×\mathbb{F}_p^{\times} 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).


It does when the group is cyclic. (Disclaimer: haven't really looked at the question.)
Reply 6
Original post by Glutamic Acid
It does when the group is cyclic. (Disclaimer: haven't really looked at the question.)


:facepalm:

herping and derping all day long
Reply 7
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 X42X225X^4 - 2X^2 - 25. This has roots X=±1±26X = \pm \sqrt{1 \pm \sqrt{26}}, 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
(edited 11 years ago)
Original post by around

edit 2: it's a degree 8 extension and the galois group is (C_2)^3


Hmm. (C_2)^3 doesn't appear as a subgroup of S_4.

Let α=1+26\alpha = \sqrt{1 + \sqrt{26}} and β=126\beta = \sqrt{1 - \sqrt{26}}. Then we wish to find [Q(α,β):Q]=[Q(α,β):Q(α)][Q(α):Q][\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]. Then [Q(α):Q]=4[\mathbb{Q}(\alpha):\mathbb{Q}] = 4, so the only question is whether [Q(α,β):Q(α)][\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] = 1 or 2, i.e. whether βQ(α)\beta \in \mathbb{Q}(\alpha). This happens if and only if αβQ(α)\alpha \beta \in \mathbb{Q}(\alpha), but αβ=25∉Q(α)\alpha \beta = \sqrt{-25} \not\in \mathbb{Q}(\alpha).

So it's a degree 8 extension, and the only transitive subgroup of S_4 of order 8 is D_8.
(edited 11 years ago)
Reply 9
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.
Reply 10
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 SPS \subseteq P. Let Sˉ=S{{x,y}:x,yS}\bar S = S \cup \left\{ \bigvee \{ x, y \}\, :\, x,y \in S \right \}, which is directed by definition. Argue that Sˉ=S\bigvee \bar S = \bigvee S.

Edit/mini-hijack: Did you work out what the contradiction is in the proof of the division algorithm for ordinals?
(edited 11 years ago)
Reply 11
Original post by nuodai
Let SPS \subseteq P. Let Sˉ=S{{x,y}:x,yS}\bar S = S \cup \left\{ \bigvee \{ x, y \}\, :\, x,y \in S \right \}, which is directed by definition. Argue that Sˉ=S\bigvee \bar S = \bigvee S.

Edit/mini-hijack: Did you work out what the contradiction is in the proof of the division algorithm for ordinals?


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 ...

(also, nope. i'll ask about it in my supervision)
Reply 12
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 ...


Actually we're both wrong. You want Sˉ\bar S to be the closure of SS under the operation of successively appending unions of pairs of members.
So glad I dropped logic for the exam
Reply 14
Original post by Daniel Freedman
So glad I dropped logic for the exam


But it's so good! And it's not like PTJ's setting the ques...oh crap.
Reply 15
More logic:

Let V be a real vector space, suppose V has a basis of cardinality 1\aleph_1. 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 1\aleph_1 by taking 1ω\aleph_1^{\omega}, 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 R1ω\mathbb{R}^{\aleph_1^{\omega}} which simplifies to 212^{\aleph_1}.
Reply 16
Well, the set of all finite subsets of 1\aleph_1 is a strict subset of 10{\aleph_1}^{\aleph_0} (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 n=01n\sum_{n=0}^{\infty} {\aleph_1}^n finite subsets of 1\aleph_1, but 1n=1{\aleph_1}^n = \aleph_1 for all finite n, so there are at most 1×0=1\aleph_1 \times \aleph_0 = \aleph_1 finite subsets of 1\aleph_1.
Reply 17
Original post by Zhen Lin
Well, the set of all finite subsets of 1\aleph_1 is a strict subset of 10{\aleph_1}^{\aleph_0} (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 n=01n\sum_{n=0}^{\infty} {\aleph_1}^n finite subsets of 1\aleph_1, but 1n=1{\aleph_1}^n = \aleph_1 for all finite n, so there are at most 1×0=1\aleph_1 \times \aleph_0 = \aleph_1 finite subsets of 1\aleph_1.


Ah, thanks. I was trying to work along these lines, but I was distracted by thoughts of R^omega.
So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...
Reply 19
Original post by Daniel Freedman
So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...


They are ridiculous.

"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 paqbp^aq^b theorem. You should quote clearly any results you use."

And that's not the whole thing

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