lola1990
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Homework Statement
Homework Equations
The Attempt at a Solution
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The discussion centers on proving the irreducibility of the polynomial \( f(x) = a_n x^n + ... + a_1 x + a_0 \) under specific conditions related to prime divisibility. It is established that if a prime \( p \) divides the coefficients \( a_i \) for \( i = n, n-1, \ldots, 1 \) but does not divide \( a_0 \), and if \( p^2 \) does not divide \( a_n \), then \( f(x) \) is irreducible over the integers. The approach involves reducing the polynomial modulo \( p \) and analyzing the leading coefficients of the factors \( h(x) \) and \( g(x) \).
PREREQUISITESThis discussion is beneficial for mathematicians, algebra students, and educators focusing on polynomial theory and irreducibility criteria in advanced algebra courses.
lola1990 said:Homework Statement
Let [itex]f(x)=a_{n}x^{n}+...+a_{1}x+a_{0}[/itex]. Let p be a prime and suppose [itex]p~|~ a_{i}[/itex] for i in n,n-1...1 but p does not divide [itex]a_{0}[/itex]. Show that if [itex]p^{2}[/itex] does not divide [itex]a_{n}[/itex], then f(x) is irreducible.Homework Equations
The Attempt at a Solution
Let f(x)=h(x)g(x) with h(x),g(x) in Z[x], and reduce mod p so that [itex]a_{0}=h(x)g(x)[/itex]. We have that if the leading coefficient of g(x) is [itex]g_{r}[/itex] and the leading term of h(x) is [itex]h_{s}[/itex] with r+s=n, p divides either coefficient but not both (because then the product would be divisible by [itex]p^{2}[/itex]). Also, p does not divide the constant term of either polynomial. WLOG, suppose p divides [itex]g_{r}[/itex] but not [itex]h_{s}[/itex]. Now, I want to find a coefficient of f(x) so that I can force [itex]h_{s}[/itex] to be divisible by p, but I'm not sure how... help!