# A not constant unit in the ring $\mathbb{Z}/27\mathbb{Z}[X]$ and another one in $\mathbb{Z}/96\mathbb{Z}[X]$.

Could someone help me to find a unit (not constant) in the ring $$\mathbb{Z}/27\mathbb{Z}[X]$$ ? And another one in the ring $$\mathbb{Z}/96\mathbb{Z}[X]$$ ?

I have done the following:

Let we take a polynomial $$f\in\mathbb{Z}/27\mathbb{Z}[X]$$. $$f$$ will be a unit if there exists another polynomial $$g\in\mathbb{Z}/27\mathbb{Z}[X]$$ such that $$f\cdot{g} =1$$.

I also know that the polynomials from $$\mathbb{Z}/27\mathbb{Z}[X]$$ have their coefficients in $$\mathbb{Z}/27\mathbb{Z}=\{\bar0,\bar1,\bar2,...,\bar{26}\}$$ and $$\bar{a}=\bar{b}\leftrightarrow a\equiv b\pmod {27}$$

But how can I follow now?

Hint. Let $$R$$ be a ring and $$a \in R$$ be nonzero nilpotent, i.e., let $$a \neq 0$$ be such that $$a^n = 0$$ for $$n \geqslant 1$$.
Consider the polynomial $$f(X) = 1 - aX \in R[X]$$. What happens if you multiply with $$1 + aX$$? What if you multiply that with $$1 + a^2X^2$$? Then with $$1 + a^4X^4$$?...

A more general result:

A polynomial in $$R[x]$$ is a unit iff the constant coefficient is a unit in $$R$$ and the other coefficients are nilpotent.

• I don't know if I have understood well.. In $\mathbb{Z}/27\mathbb{Z}[X]$ if I take $\bar2$ as the constant (because it is a unit in $\mathbb{Z}/27\mathbb{Z}$ ) and $\bar3$ as the coefficient of $X$ (because it is nilpotent; $\bar3 ^3=0$), I mean, if I take $f(X)=\bar2 + \bar3 X$, is $f$ a unit in $\mathbb{Z}/27\mathbb{Z}[X]$ ??? Oct 14, 2021 at 18:39
• @User160: Yes. You can multiply $f(X)$ with a unit and it won't change its invertibility. So, multiplying with $\overline{14} = \overline{2}^{-1}$ gives you $g(X) = \bar{1} - \bar{3} X$. Now you can follow the procedure in given to see that $g(X)$ has an inverse, namely $(1 + \bar{3}X)(1 + \bar{9}X^2) = 1 + \overline{3}X + \bar{9}X^2$. Oct 14, 2021 at 18:43
• @User160: To be explicit, multiplying that last polynomial with $\overline{14}$ will give you the inverse of your $f(X)$. So we have $$(\bar{2} + \bar{3}X)^{-1} = \overline{14} + \overline{15}X + \overline{18}X^2.$$ Oct 14, 2021 at 18:52
• which last polynomial?? Oct 14, 2021 at 18:59
• @User160: The one in my comment before that: $1 + \overline{3}X + \bar{9}X^2$. Oct 14, 2021 at 18:59