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Math Help - Help Understanding a Proof: showing that p(x) is Irreducible in Z

  1. #1
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    Help Understanding a Proof: showing that p(x) is Irreducible in Z

    The task is to show that if polynomial p(x) is irreducible in ANY (does not need to be in all of them) Z_p [x], where p is a prime number, then p(x) is also irreducible in Z[x]. It is also assumed that the leading term's coefficient is 1.

    Now, I've found a proof close to this in a book, but I don't fully understand it. In this proof, it's shown that p(x) will be irreducible in Q, not Z, but I suppose it's easy to say that it follows that if p(x) is irreducible in Q, it will also be in Z.

    Here's the link to the proof, it's theorem 17.3.

    http://books.google.com/books?id=CnH3mlOKpsMC&pg=PA308&lpg=PA308&dq=&sourc e=bl&ots=npv4Nw4G-0&sig=fh8RF3wFUSCd2LhmtUzx6OBjX8A&hl=fi&ei=b5ZWTcq 8A87oObWPiIIF&sa=X&oi=book_result&ct=result&resnum =1&ved=0CBYQ6AEwAA#v=onepage&q&f=false

    My questions are:

    1) Why is deg g (x) <= deg g(x)? Why is this not simply deg g(x) = deg g(x)? The leading term's coefficent can't be 2,3,5,7... anyway, so it shouldn't disappear, or have I missed something?

    2) I don't really understand the contradiction there. What is the counter-assumption in the beginning? That f(x) is reducible in Z_p[x]?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Koaske View Post
    The task is to show that if polynomial p(x) is irreducible in ANY (does not need to be in all of them) Z_p [x], where p is a prime number, then p(x) is also irreducible in Z[x]. It is also assumed that the leading term's coefficient is 1.

    Now, I've found a proof close to this in a book, but I don't fully understand it. In this proof, it's shown that p(x) will be irreducible in Q, not Z, but I suppose it's easy to say that it follows that if p(x) is irreducible in Q, it will also be in Z.

    Here's the link to the proof, it's theorem 17.3.

    http://books.google.com/books?id=CnH3mlOKpsMC&pg=PA308&lpg=PA308&dq=&sourc e=bl&ots=npv4Nw4G-0&sig=fh8RF3wFUSCd2LhmtUzx6OBjX8A&hl=fi&ei=b5ZWTcq 8A87oObWPiIIF&sa=X&oi=book_result&ct=result&resnum =1&ved=0CBYQ6AEwAA#v=onepage&q&f=false

    My questions are:

    1) Why is deg g (x) <= deg g(x)? Why is this not simply deg g(x) = deg g(x)? The leading term's coefficent can't be 2,3,5,7... anyway, so it shouldn't disappear, or have I missed something?

    2) I don't really understand the contradiction there. What is the counter-assumption in the beginning? That f(x) is reducible in Z_p[x]?
    You're link doesn't show us the page.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    You're link doesn't show us the page.
    Weird... Well, in any case, I have included the theorem as a picture. The other picture is theorem 17.2 that is referred to in 17.3.
    Help Understanding a Proof: showing that p(x) is Irreducible in Z-theorem-17-2.jpg
    Help Understanding a Proof: showing that p(x) is Irreducible in Z-theorem-17-3.jpg
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  4. #4
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    there is an error in the text: one cannot conclude that deg(\overline{f}) = deg(f), because in fact, p may divide the leading coefficient of f. all the examples he gives in the text are with p for which this is not so.

    i cannot think of a situation in which deg(\overline{g}) < deg(g), if p does not divide the leading coefficient of f, it cannot divide the leading coefficient of g or h.

    you are correct about the counter-assumption. the argument is:

     f(x) = g(x)h(x) \rightarrow \overline{f}(x) = \overline{g}(x)\overline{h}(x), which is logically equivalent to the contrapositive, which is what he is actually trying to prove.

    this in turn, relies on the fact that the homomorphism Z-->Zp induces a homomorphism of Z[x] onto Zp[x].
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