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lola1990
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Homework Statement
Homework Equations
The Attempt at a Solution
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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!
Opposite Eisenstein's criteria is a mathematical test used to determine whether a polynomial with integer coefficients has a specific form that guarantees it is irreducible.
The criteria states that if a polynomial has the form of anxn + an-1xn-1 + ... + a1x + a0, where an, an-1, ..., a1, a0 are integers and an is prime, then the polynomial is irreducible.
A polynomial is irreducible if it cannot be factored into two or more polynomials with integer coefficients. In other words, it cannot be broken down into simpler terms.
Opposite Eisenstein's criteria is useful because it provides a quick and easy way to determine whether a polynomial is irreducible. This can save time and effort when working with polynomials in various mathematical contexts.
Yes, Opposite Eisenstein's criteria only applies to polynomials with integer coefficients and a specific form. It cannot be used for polynomials with non-integer coefficients or those that do not follow the specified form.