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

I know (2x+1) is a unit. Is there any other units in $Z_4[X]$ if there are infinite units, could you generalize it in a certain from such as (2nx+1) s.t n is an integer?

user26857
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user88310
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2 Answers2

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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 $1$.

Jyrki Lahtonen
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  • This answer is worth remembering clearly! Just adding that if $u$ is a unit in a ring $R$ and $x$ is a nilpotent element in $R$ with $x^n = 0$, then $({u^{-1}}x)^n =0$, and $$1/(u+x) = u^{-1}/(1+({u^-{1}}x) = {u^{-1}}(1+({u^{-1}}x)+\cdots +({u^{-1}}x^{n-1}),$$ so $(u+x)$ must be a unit in $R$. I tried to answer the question in an elementary way below. Hope it is correct! – student Sep 01 '20 at 18:30
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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.

student
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    This is correct, and it is a nice addition to explain that the units are all of their own inverses. Just pointing out that it is unnecessary to include both signs in $\pm1$. After all $-1=3=1+2\cdot1$ is already included. Of course, if you only allow $h(x)$ with zero constant term, then you need to keep $\pm1$. – Jyrki Lahtonen Sep 01 '20 at 18:49
  • You are right!! It is unnecessary to include both signs $\pm 1$. – student Sep 02 '20 at 19:05