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Show that the polynomial $f(x)=1-2x$ has an inverse in $\Bbb Z_{16}[x]$. (Hint: Solve for $g$ from the equation $fg=1.)$

If $g(x)=ax+b$, then $f(x)g(x)=1 \iff (1-2x)(ax+b)=1$ so $$ax+b-2ax^2-2bx=-2ax^2+x(a-2b)+b=1$$

I can get the $-2ax^2$ term to disappear by setting $a=8$, but this forces $a-2b=8-2b = 2(4-b).$

Now in order to get this $x(a-2b)$ term to disappear I need $a-2b$ to be a multiple of $16$. So $$2(4-b)=16k \iff b=4-8k$$

But varying $k$ I cannot get anything useful except for example when $k=1$ I have that $b=-4$ so $$(1-2x)(8x-4)=-16x^2+16x-4 \equiv -4 \equiv 12 \pmod{16}$$

How can I get rid of this $-4$ here? Also I don't think that $\deg(g)=1$ neccessarily.

geojaffa
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  • Same general method(s) in the linked dupe work here. – Bill Dubuque Aug 16 '22 at 17:52
  • Which one would you suggest of the $5$ answers? – geojaffa Aug 16 '22 at 17:57
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    The most general way there is viewing it as the invertibility of a unit + nilpotent (a special case of the method of simpler multiples), as described in my answer there. But you may also find other (more specific) answers helpful. Note here $2x$ is nilpotent since $(2x)^4 = 0\ \ $ – Bill Dubuque Aug 16 '22 at 18:06

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