# $f(x) = x$ in $(\mathbb{Z}/6\mathbb{Z})[x]$ factors as $(3x+4)(4x+3)$

This question has a lot of parts so I'll post each part separately.

First, show $$f(x) = x$$ in $$(\mathbb{Z}/6\mathbb{Z})[x]$$ factors as $$(3x+4)(4x+3)$$

I am trying long division. I cannot divide $$x$$ by either $$(3x+4)$$ or by $$(4x+3)$$ mod 6. Anything multiplied by 3 is either 3 or 0 mod 6. And 4,2,0 mod 6 when a number is multiplied by 4.

• We need to exercise a bit of care in discussing things like "irreducibility" in a ring with zero divisors. Bill Dubuque dug up resources in an old answer. Not sure whether those articles give everything you want to know? – Jyrki Lahtonen Apr 11 '19 at 4:22

HINT

Note that $$(3x+4)(4x+3) = 12x^2 + 25x + 12 \equiv x \pmod{6}$$

• So, I plug x=0 in $12x^2+25x+12=0$ (mod 6) so it is reducible? – Dom Apr 11 '19 at 3:00
• @Dom it certainly factors... – gt6989b Apr 11 '19 at 3:03
• @Dom it has nothing to do with substituting $x=0.$ – Thomas Andrews Apr 11 '19 at 3:04
• @Thomas Andrews Yes, so 12 is congruent to 0 and 25 to 1 so we are left with x. – Dom Apr 11 '19 at 3:08

More conceptually, note that \,\begin{align}f \equiv 3\\ e\equiv 4\end{align}\, obey \,\begin{align}e^2\equiv&\ e,\ \ e\!+\!f \equiv 1\\ f^2\equiv&\ f,\ \ \ \ \color{#c00}{e\,f\,\equiv 0}\end{align}\ \ [orthogonal idempotents], so

$$(ex\!+\!f)(fx\!+\!e)\, \equiv\, \color{#c00}{ef}\, x^2 + (e^2\!+\!f^2) x + \color{#c00}{ef}\,\equiv\, x\qquad\qquad$$

The same will occur with the idempotents arising from any CRT direct product decomposition.

Essentially what occurs is that the trivial factorization $$x \equiv x\cdot 1\,$$ becomes nontrivial by permuting the factors $$\,x,1\,$$ in each product component, as below

$$\underset{}{\overset{\Large \bmod 2:}{\phantom{I^{I^{I^{I^{I^I}}}}}}}\underbrace{\overbrace{(3x\!+\!4)}^{\Large \equiv\ x }}_{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Large\bmod 3:\ \ \ \ \ \equiv\ 1}\,\underbrace{\overbrace{(4x\!+\!3)}^{\Large \equiv\ 1}}_{\Large \equiv\ x}\qquad$$

where we have that $$\, x\equiv gh \equiv x\cdot 1\pmod{\!2}\,$$ vs. $$\ gh\equiv 1\cdot x\pmod{\!3}$$