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A polynomial is reducible in R[x] implies it is reducible in (R/I)[x]

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 multiplication in these two rings shows that $\varphi$ is a ring homomorphism. The kernel is precisely the set of polynomials each of whose coefficients is an element of $I$, which is to say that $\operatorname{ker} \varphi=I[x]=(I)$, proving the theorem. $\blacksquare$

Now we come back to the proof of the theorem

Proof of Contrapositive Statement of the Theorem: Suppose $p(x)$ cannot be factored in $(R / I)[x]$ but that $p(x)$ is reducible in $R[x]$. This means there are monic, nonconstant polynomials $a(x)$ and $b(x)$ in $R[x]$ such that $p(x)=a(x) b(x)$. By the theorem above, reducing the coefficients modulo $I$ gives a factorization in $(R / I)[x]$ with nonconstant factors, a contradiction. Hence $p(x)$ is irreducible in $R[x]$ $\blacksquare$

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