Are Irreducible Polynomials Like Prime Numbers?!?

In Abstract Algebra, what does it mean for a polynomial to be irreducible over an integral domain or field? Let D be an integral domain. A polynomial f(x) ∈ D[x] that is neither zero not a unit (invertible element) of D[x] is irreducible over D if, whenever f(x)=g(x)*h(x), where g(x),h(x) ∈ D[x], then g(x) or h(x) is a unit in D[x]. For example, f(x)=2x^2+4=2(x^2+2) is irreducible over the field of rationals ℚ but reducible over the integral domain of integers ℤ. The reason is that 2 is a unit in ℚ[x] (with inverse 1/2) but not a unit in ℤ[x] (since 1/2 is not an integer). Also, x^2 + 2 is not a unit in either of these polynomial rings. If D = F is a field, then a nonconstant f(x) ∈ F[x] is irreducible over F if f(x) cannot be expressed as the product of two polynomials of lower degree. The Reducibility Test for degrees 2 and 3 over a field is also important. Let F be a field. If f(x) ∈ F[x] and the degree of f(x) is 2 or 3, then f(x) is reducible over F if and only if f(x) has a zero in F (so f(x) is irreducible over F if and only if f(x) does NOT have a zero in F).

#AbstractAlgebra #FactoringPolynomials #factoring #PolynomialRing

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