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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?

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1 Answer 1

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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.

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  • $\begingroup$ 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 $\mathbb{Z}/27\mathbb{Z}[X]$ ??? $\endgroup$
    – User160
    Oct 14, 2021 at 18:39
  • $\begingroup$ @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 $(1 + \bar{3}X)(1 + \bar{9}X^2) = 1 + \overline{3}X + \bar{9}X^2$. $\endgroup$
    – Aryaman Maithani
    Oct 14, 2021 at 18:43
  • $\begingroup$ @User160: To be explicit, multiplying that last polynomial with $\overline{14}$ will give you the inverse of your $f(X)$. So we have $$(\bar{2} + \bar{3}X)^{-1} = \overline{14} + \overline{15}X + \overline{18}X^2.$$ $\endgroup$
    – Aryaman Maithani
    Oct 14, 2021 at 18:52
  • $\begingroup$ which last polynomial?? $\endgroup$
    – User160
    Oct 14, 2021 at 18:59
  • $\begingroup$ @User160: The one in my comment before that: $1 + \overline{3}X + \bar{9}X^2$. $\endgroup$
    – Aryaman Maithani
    Oct 14, 2021 at 18:59

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