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If p and q are prime numbers such that p is not a quadratic residue mod q. Show that if pq=-1 mod 4 then the polynomial f(x)=x^2-q is irreducible in F_p[x].
A polynomial is irreducible in F_p[x] if it cannot be factored into polynomials of lower degree with coefficients in the field F_p. In other words, it cannot be broken down into simpler factors in the field F_p.
Proving that a polynomial is irreducible in F_p[x] allows us to determine the structure and properties of the field F_p[x]. Additionally, it can help us determine if a polynomial has roots in a certain field or if it is a prime polynomial.
The polynomial f(x)=x^2-q is significant because it is a quadratic polynomial with a leading coefficient of 1, making it a monic polynomial. Additionally, it is a primitive polynomial, meaning it generates the field F_p[x]. Proving its irreducibility in F_p[x] is important in understanding the properties of this field.
There are a few different methods to prove the irreducibility of a polynomial in F_p[x]. One approach is to use the Eisenstein's criterion, which states that if a polynomial is of the form x^n + a_n-1x^n-1 + ... + a_1x + a_0, and there exists a prime number p that divides all coefficients except a_0 and p^2 does not divide a_0, then the polynomial is irreducible in F_p[x]. Another method is to use the rational root theorem to show that f(x)=x^2-q has no rational roots, and therefore cannot be factored.
Yes, f(x)=x^2-q can be irreducible in F_p[x] for all prime numbers p. This can be proven using the quadratic reciprocity theorem, which states that for any prime numbers p and q, if the Legendre symbol (p/q) is equal to -1, then x^2-q is irreducible in F_p[x]. Since there are infinitely many prime numbers that satisfy this condition, f(x)=x^2-q can be irreducible for infinitely many prime numbers.