Show $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$ (inverting unit + nilpotent)

Question

Show that $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$.

Once you have found the inverse like here, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = $1$ imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?

Answer 1

If $R$ is a commutative ring: the units in $R[x]$ are the polynomials whose constant term is a unit, and whose higher order coefficients are nilpotent. You can apply this directly to your example.

Answer 2

To find an inverse polynomial for that holds $p(x)(4x^2+6x+3)=1$ so it has to be $3y=1\mod 8$ [edit: For more context on $y$, see the comments below]. So $y=3$ and the polynomial might look like this:

$p(x)=(ax+3)$

Then $(4x^2+6x+3)(ax+3)=4ax^3+(6a+12)x^2+(3a+18)x+9$. Now it has to be $4a\equiv 0\mod 8$ and $6a+12\equiv 0\mod 8$ and $3a+18\equiv 0\mod 8$.

Is there such an $a$?. Yes indeed. For $a=2$ we have $8\equiv 0\mod 8$

$24\equiv 0\mod 8$ and $24\equiv 0\mod 8$.

If we would fail to find this $a$ in this step, we would have to try with $p(x)=(ax^2+bx+3)$ and proceed as above, which gets more and more complicated.

So it is $(4x^2+6x+3)(2x+3)\equiv 1\mod 8$

Answer 3

Hint: As in the hinted paper, a possible ansatz would be

$(4x^2+6x+3) (ax+b) = 4ax^3+(4b+6a)x^2+ (6b+3a)x+3b=1$.

This requires $4a\equiv 0\mod 8$ (so $a$ must be even), $4b+6a\equiv 0\mod 8$, and $6b+3a\equiv 0\mod 8$ and $3b\equiv 1\mod 8$ (so $b=3$).

The cases left are $a$ even with $b=3$.

Answer 4

Write $$ 4x^2+6x+3 = 3(4x^2+2x+1) = 3((2x)^2+(2x)+1) = 3 \frac{(2x)^3-1}{2x-1} = \frac{-3}{2x-1} $$ Therefore, $$ \frac{1}{4x^2+6x+3} = \frac{2x-1}{-3} = 3(1-2x) = 3-6x = 2x+3 $$

Answer 5

Using $ $ simpler multiples, $ $ we can invert $\,\ a \:\!-\:\! f,\,$ for invertible $\,a,\,$ say $\,\color{#0a0}{ab = 1},\,$ & nilpotent $\,f,\,$ say $\color{#c00}{f^n = 0},\,$ by inverting a simpler multiple $\, a^n-\color{#c00}{f^n} = \color{#0a0}{a^n},\,$ which has obvious inverse $\,\color{#0a0}{b^n},\,$ i.e.

$\ \ \ \ \ \ \ \ \ \color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \overbrace{\dfrac{1}{a-f} = \dfrac{a^{\large n-1}\!+\!\cdots\! +\! f^{\large n-1}}{\!\!\!\!\!\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}}}^{\large \text{check via cross multiply}} =\, \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})$ $\!\begin{align}{\rm so}\ \ &\color{#0a0}{3(3)=1},\, \color{#c00}{f^{\large 3} = 0}\ \Rightarrow\ \dfrac{1}{3-f} = \dfrac{3^{\large 2}+\,3f\,+\, f^{\large 2^{\phantom{|^{|^|}}\!\!\!\!\!}}}{\color{#0a0}{3^{\large 3}-\color{#c00}{f^{\large 3}}}}\ \ \, =\, \ \ \color{#0a0}{3^{\large 3}}(3^{\large 2}\! +3f + f^{\large 2}) = \bbox[5px,border:1px solid #c00]{2x+3}\\[.1em] &{\rm because}\ \ \ \color{#c00}{2^{\large 3}\mid f^{\large 3^{\phantom{ |^|}}}} {\rm by}\ \ 2\mid f = -6x-4x^2\,\ \ [\,{\rm to\ invert}\ \ 3\!-\!f = 3\!+\!6x\!+\!4x^2 \in \smash[b]{\Bbb Z_{\large \color{#c00}{2^{\Large 3}} }]}\end{align}$

Remark $ $ We may view this as a constructive proof that divisors of units are units, i.e. if $\,d\mid u\,$ unit, then $\,dd' = u\,$ so $\,\dfrac{1}d = \dfrac{d'}u.\,$ Above we chose $\,u = a^n-f^n,\,$ a unit multiple $\,a-f,\,$ which is "simpler" since it has $ $ obvious $ $ inverse, because $\,u= a^n\,$ for unit $\,a,\,$ by $\,f^n = 0$.

The same idea shows how to invert all cyclotomics $\Phi_{k}(f)$ for $\,k\ge n,\,$ since they have simpler multiple $\,1-f^k = 1.\,$ There are simple algorithms for recognizing polynomials of cyclotomic form (i.e. divisors of $\,1-x^k)$ and products of such, so the above method generalizes widely.

This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses. Analogous methods may be employed for for computing remainders via mod arithmetic.

Generally we can use standard (nonconstructive) methods of commutative algebra to give quicker proofs, e.g see here, which mentions how it is related to Euclid's proof that there are infinitely many primes (and generalizations). Compare also the methods used here to prove that a polynomial is a unit iff its constant term is a unit and all other coefficients are nilpotent (the method of proof there can be made constructive - analogous to above)

Answer 6

Lets show that $\frac{1}{3+6x+4x^2}=3+2x$ in $\Bbb Z_8[x]$ by our division method that we learned in primary school (I couldn't make this shape tough: |____ ): $$ \begin{array}{c|l} 1&\hline 3+6x+4x^2\\ -(9+18x+12x^2)&\hline\color{red}{3+2x}\\ 6x+4x^2&\hline\\ -(6x+4x^2)&\hline\\ 0&\hline \end{array} $$