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  1. J

    Question on Galois Group

    Yes that makes sense that its a Galois extension, because we're told that each of the roots are distinct (and there are n of them) so it is a splitting field and has characteristic 0, thus it's Galois. Although your argument makes perfect sense, saying that |GalQQ(z)|=[Q(z):Q]=1 if the group is...
  2. J

    Question on Galois Group

    I am trying to show that if z, z2, z3, ..., zn=1 are n distinct roots of xn-1 in some extension field of Q (the rationals), then GalQQ(z) (the galois group of Q(z) over Q) is abelian. Would I be wrong to say that since the galois group we're talking about here only involves an extension field...
  3. J

    Finite Field Question

    Ahh you're right! I was being distracted; all I need is Fermat's Little Theorem. Thanks a lot.
  4. J

    Finite Field Question

    I am trying to prove that if c is a root of f(x) in Z_p then c^p is also a root. It seems very simple but I can't think how to approach it. Any insight on this would be greatly appreciated, and sorry for not using the latex but it seems to be acting up.
  5. J

    Simple Field Extension Question

    Oh wow yea, ok it's pretty clear. I think I was complicating things. Thanks.
  6. J

    Simple Field Extension Question

    I'm currently trying to prove that (for a field extension K of the field F) if u\in K and u^2 is algebraic over F then u is algebraic over K. I thought of trying to prove it as contrapositive but that got me nowhere--it seems so simple but I don't know what to use for this. Any help with this...
  7. J

    Solving questions

    confirm
  8. J

    Solving questions

    Are you having trouble with (q1)? For (q2) think of the fact that since a*a=e, for all a in G, a is it's own inverse.
  9. J

    N^2 Modular Arithmetic

    Try considering the two cases where n^2is either even or odd and how that relates to the congruence modulo 3.
  10. J

    Polynomials in Z modulus 9

    I'm trying to figure out how to prove that every polynomial in \mathbb{Z}_9 can be written as the product of two polynomials of positive degree (except for the constant polynomials [3] and [6]). This basically is just showing that the only possible irreducible polynomials in \mathbb{Z}_9 are the...
  11. J

    Simple question on disproving a group isomorphism

    I definitely agree, it's good to able to prove it in a number of ways. Thanks you guys for both suggestions.
  12. J

    Simple question on disproving a group isomorphism

    Oh ok I didn't think of showing Q isn't cyclic, that's probably the simplest way to do it now that I think of it. Thanks a bunch.
  13. J

    Simple question on disproving a group isomorphism

    It would still be in integers right? That's why I'm confused because I can't seem to show that that property disallows the isomorphism.
  14. J

    Simple question on disproving a group isomorphism

    I am trying to prove that the additive groups \mathbb{Z} and \mathbb{Q} are not isomorphic. I know it is not enough to show that there are maps such as, [tex]f:\mathbb{Q}\rightarrow \mathbb{Z}[/itex] where the input of the function, some f(x=\frac{a}{b}), will not be in the group of integers...
  15. J

    Proof on Order of Elements in a Group

    Oh ok, it just hit me it makes perfect sense. I wasn't thinking hard enough about the actual definition of order. Thank you both for your help.
  16. J

    Proof on Order of Elements in a Group

    I'm trying to figure out how to prove the following... If a, b \in G where G is a group, then the order of bab^{-1} equals the order of a. I'm rather stumped because the group is not necessarily abelian and it seems like it would have to be in order to directly show that you can rearrange...
  17. J

    Simple prime/GCD proof question

    ohh hah yes of course, thanks.
  18. J

    Simple prime/GCD proof question

    Hello, I'm working out of Hungerford's Abstract Algebra text and this proof has been bothering me because I think I know why it works and it's so simple but I can't figure out how you would show a rigorous proof of it... If a=p_1^{r_1}p_2^{r_2}p_3^{r_3} \cdots p_k^{r_k} and...
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