0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Apr 11 '19 at 4:22
3
$\begingroup$

HINT

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

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

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. $\ gh\equiv 1\cdot x\pmod{\!3}$

$\endgroup$

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.