# 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 $\mathbb F_p$. Commented Nov 7, 2014 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 $$\pi(h(x)) = \pi(n_{\beta})x^{\beta}+\cdots$$. The leading coefficients are themselves units, since
$$\pi(m_{\alpha})\pi(n_{\beta}) = \pi(m_{\alpha}n_{\beta}) = \pi(1) = 1.$$

As $$\deg(g(x)), \deg(h(x)) \geq 1$$, 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!

• Very clear answer, thanks! Commented Nov 7, 2014 at 5:38
• My pleasure, glad I could help! :) Commented Nov 7, 2014 at 5:41
• The following sentence is wrong because $R / I$ possibly contains nilpotent elements: "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 $(R/I)[x]$". Commented Jan 7, 2023 at 18:32
• @MetinErsinArıcan yes, the sentence is less precise than it should be and needs amending. The correct reasoning would be to note that the leading coefficients of the reductions of $g(x)$ and $h(x)$ in $R/I$ are units, since $f$ is monic, and so this rescues things. I'll edit the answer in a little bit. Commented Jan 7, 2023 at 19:58