# Reducible in $R[X]$ implies reducibility in $(R/I)[X]$

Let $$R$$ be an integral domain, $$I$$ a proper ideal of $$R$$, $$\pi:R \to R/I$$ the canonical projection. Let $$f=\sum_{i=0}^n a_iX^i$$ be a monic polynomial and $$\overline{f}=\sum_{i=0}^n \pi(a_i)X^i \in (R/I)[X]$$. Show that if $$f$$ is reducible in $$R[X]$$ then $$\overline{f}$$ is reducible in $$(R/I)[X]$$.

I am stuck on this exercise. I tried to show this by the contrapositive statement:

Suppose $$\overline{f}$$ is irreducible in $$(R/I)[X]$$ and suppose $$f=ab$$ in $$R[X]$$, I want to show that $$f$$ is irreducible. Since $$\overline{f}=\overline{a}\overline{b}$$ is irreducible in $$(R/I)[X]$$ , without loss of generality, we can assume $$\overline{a} \in \mathcal U((R/I)[X])=\mathcal U(R/I)$$. This means there is $$b \in R$$ with $$ab \in I$$. I don't know how to conclude from here that $$a$$ or $$b$$ is a unit. I would like any suggestions to complete my solution and also if another one has a straightforward proof, he or she is welcome to share it.

• If you get stuck on a contrapositive proof, try a direct proof. Have you looked at examples, like $x^2-x-2\in\mathbb Z[x]$? Look at it as a polynomial over $\mathbb F_p$. Nov 7 '14 at 5:28

You could proceed by proving the contrapositive. However, I think it is easier to do things directly here. Let $f(x) \in R[x]$ be a reducible monic polynomial of degree $n$. Then it admits a factorization $f(x)=g(x)h(x)$ where neither $g(x)$ nor $h(x)$ is a unit. Now let
$$g(x) = \sum_{i=0}^{\alpha}m_{i}x^{i}$$ and
$$h(x) = \sum_{j=0}^{\beta}n_{j}x^{j}.$$
where $\alpha+\beta = n$. The main idea here is that if $h(x)$ and $g(x)$ are not units in $R[x]$, then they are not units when reduced modulo $I$. To see this, first note that $I$ cannot contain any units, since it is a proper ideal of $R$. Next, note that $m_{\alpha}$ and $n_{\beta}$ are units, since $f$ is monic and $m_{\alpha}n_{\beta} = 1$ is the coefficient of the leading term. This further tells us that $\alpha, \beta \geq 1$, since $n_{\beta}$ is a unit, and $f(x)$ is reducible by assumption. Hence, $\pi(m_{\alpha}) \neq 0$ and $\pi(n_{\beta}) \neq 0$. Now note that $\pi(g(x)) = \pi(m_{\alpha})x^{\alpha} + \cdots$ and $\pi(h(x)) = \pi(n_{\beta})x^{\beta}+\cdots$. Since the leading coefficients don't vanish and $\deg(g(x)), \deg(h(x)) \geq 1$, it follows that $h(x)$ and $g(x)$ are not units in $(R/I)[x]$. Feel free to comment if you need further clarification on any part!