I am trying to answer the following question about the ring of polynomials over $\mathbb Z\_{10}[x]$:

* Prove that no linear polynomial in $\mathbb Z\_{10}[x]$ is a unit.

So far I have determined that:

* $f(x)$ is a unit if $f(x)g(x)=1$.
* The units of $\mathbb Z\_{10}[x]$ cannot be the zero divisors $(0,2,4,6,8)$ so we are left with $(1,3,7,9)$.
* $1$ is its own multiplicative inverse. Similarly, $9\cdot9\cdot9=81\equiv1(\mathrm{mod}10)$. Finally, $3\cdot7=21\equiv1(\mathrm{mod}10)$ so $(1,3,7,9)$ are in fact units.
* $\deg(f(x)g(x))=\deg(f(x))+\deg(g(x))$ so if $\deg(f(x))=1$ then $\deg(g(x))=-1$ (as $f(x)g(x)=1$ implies that $\deg(g(x))=-1$). The only unit in $\mathbb Z\_{10}[x]$ whose multiplicative inverse is also an inverse is $1$. This implies $ax+b=1$.

This is where I get stuck. I have written “As $a$ and $b$ must be $(1,3,7,9)$ it is clear from inspection that $ax+b\equiv1(\mathrm{mod}10)$ is impossible” but I don't think this is a convincing argument as it doesn't even convince myself.