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Third Year Revision Thread

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    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 \Phi_n be the n'th cyclotomic polynomial. Recall that if K is a field of characteristic prime to n, then the roots of \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 \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 \Phi_n(x) in Z iff \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 \Phi_n in K are precisely the primitive n'th roots of unity in K' part. Why is this important?
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    If \Phi_n(x) \equiv 0 \pmod p for some x then \mathbb{F}_p contains a primitive nth root of unity \xi_n, say. Then \{ 1, \xi_n, \dots, \xi_n^{n-1} \} is a subgroup of \mathbb{F}_p^{\times} of order n. Apply Lagrange.

    Edit: Removed comment about the converse; what I wrote was a pile of trout.
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    (Original post by nuodai)
    If p=an+1 then p=an+1 and 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?
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    (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+1 then n divides \left| \mathbb{F}_p^{\times} \right|. This is more useful than it looks because for this direction you don't need a root of \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 p\, |\, n, say n=bp then (X^n-1) \equiv (X^b - 1)^p \pmod p so you don't have any primitive nth roots.
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    (Original post by nuodai)
    I wrote a load of rubbish, ignore that.

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

    n divides  |\mathbb{F}_p^{\times}|. We thus want an element of \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).
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    (Original post by around)
    Ok.

    n divides  |\mathbb{F}_p^{\times}|. We thus want an element of \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.)
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    (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
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    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 X^4 - 2X^2 - 25. This has roots X = \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
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    (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 \alpha = \sqrt{1 + \sqrt{26}} and \beta = \sqrt{1 - \sqrt{26}}. Then we wish to find [\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, so the only question is whether [\mathbb{Q}(\alpha,\beta):\mathbb  {Q}(\alpha)] = 1 or 2, i.e. whether \beta \in \mathbb{Q}(\alpha). This happens if and only if \alpha \beta \in \mathbb{Q}(\alpha), but \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.
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    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.
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    (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 S \subseteq P. Let \bar S = S \cup \left\{ \bigvee \{ x, y \}\, :\, x,y \in S \right \}, which is directed by definition. Argue that \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?
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    (Original post by nuodai)
    Let S \subseteq P. Let \bar S = S \cup \left\{ \bigvee \{ x, y \}\, :\, x,y \in S \right \}, which is directed by definition. Argue that \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)
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    (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 \bar S to be the closure of S under the operation of successively appending unions of pairs of members.
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    So glad I dropped logic for the exam
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    (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.
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    More logic:

    Let V be a real vector space, suppose V has a basis of cardinality \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 \aleph_1 by taking \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 \mathbb{R}^{\aleph_1^{\omega}} which simplifies to 2^{\aleph_1}.
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    Well, the set of all finite subsets of \aleph_1 is a strict subset of {\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 \sum_{n=0}^{\infty} {\aleph_1}^n finite subsets of \aleph_1, but {\aleph_1}^n = \aleph_1 for all finite n, so there are at most \aleph_1 \times \aleph_0 = \aleph_1 finite subsets of \aleph_1.
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    (Original post by Zhen Lin)
    Well, the set of all finite subsets of \aleph_1 is a strict subset of {\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 \sum_{n=0}^{\infty} {\aleph_1}^n finite subsets of \aleph_1, but {\aleph_1}^n = \aleph_1 for all finite n, so there are at most \aleph_1 \times \aleph_0 = \aleph_1 finite subsets of \aleph_1.
    Ah, thanks. I was trying to work along these lines, but I was distracted by thoughts of R^omega.
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    So the Rep theory exam questions aren't as nice as I expected (/wanted) them to be...
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    (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 p^aq^b 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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