Theorem: Let \(I\) be a proper ideal in the integral domain \(R\) and let \(p(x)\) be a non-constant polynomial in \(R[x]\). If \(p(x)\) is reducible in \(R[x]\) then the image of \(p(x)\) in \((R/I)[x]\) is also reducible. But we will show the contrapositive of it Contrapositive Statement of the Theorem: Let \(I\) be a proper ideal in the integral domain \(R\) and let \(p(x)\) be a non-constant polynomial in \(R[x]\). If the image of \(p(x)\) in \((R/I)[x]\) can not be factored into two polynomials of a smaller degree then \(p(x)\) is irreducible in \(R[x]\) For that first, we need to prove this theorem Theorem: Let $I$ be an ideal of the ring $R$ and let $(I)=I[x]$ denote the ideal of $R[x]$ generated by $I$ (the set of polynomials with coefficients in $I$ ). Then $$R[x] /(I) \cong (R / I)[x]$$ Proof: There is a natural map $\varphi: R[x] \to (R / I)[x]$ given by reducing each of the coefficients of a polynomial modulo $I$. The definition of addition and multip...