You're link doesn't show us the page.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]?