4
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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 $\mathbb F_p$. $\endgroup$
    – Lubin
    Nov 7 '14 at 5:28
4
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Very clear answer, thanks! $\endgroup$
    – user16924
    Nov 7 '14 at 5:38
  • $\begingroup$ My pleasure, glad I could help! :) $\endgroup$ Nov 7 '14 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.