Find a polynomial of degree > 0 in $\mathbb Z_4[X]$ that is a unit.

Solution 1:

All the polynomials of the form $$1+2p(x)$$ are units. This is because $2p(x)$ is nilpotent, and elements of the form $1+n$, $n$ nilpotent, are units in any ring.

These are all the units of $\mathbf{Z}_4[x]$. This follows from the fact that if $u(x)$ is a unit, then it must remain a unit after being reduced modulo two. But the only unit of $\mathbf{Z}_2[x]$ is the constant $1$.

Solution 2:

We claim that the units in ${{\bf Z}_4}[x]$ are polynomials of the form $\pm{1}+2h(x)$ where $h(x)\in{{\bf Z}_4}[x]$, and each such unit is its own multiplicative inverse. Consider $\pm{1}+2h(x)$. Then, we simply check that $$(\pm{1}+2h(x))^2 =1+4h(x)+4h^2(x)=1.$$ Conversely, if $f(x)$, and $g(x)$ are units in ${{\bf Z}_4}[x]$, we can separate even and odd coefficients of each unit, and write $$f(x)=2f_1(x)+r_1(x),$$ and $$g(x)=2g_1(x)+s_1(x).$$ Then, since $f(x)g(x)=1$, $$2(f_1(x)s_1(x)+g_1(x)r_1(x))+r_1(x)s_1(x)=1.$$ Now if $r_1(x)s_1(x)$ is non-constant, its leading term would have an odd coefficient of positive degree, contradicting $fg=1$. This forces $r_1=s_1=\pm1$. Plugging this back into the last identity displayed above we get $$\pm2(f_1(x)+g_1(x))+1=1.$$ This implies that $$2f_1(x)=-2g_1(x)=2g_1(x).$$ This implies in turn that $f(x) =g(x)$ proving that every unit is its own multiplicative inverse.