# $a^4=0$ then $1-a$ is invertible in $R[x]/(d)R[x]$ [duplicate]

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Suppose $$a^4=0$$, for some $$a \in R[x]/(d)R[x]$$, then prove that $$1-a$$ is invertible.

I was thinking since $$a^4 = a \cdot a \cdot a \cdot a=0$$, this implies that $$a$$ has to be zero (?) . Now we have that $$1-a=1-0=1$$, and $$1$$ is invertible, since $$1 \cdot 1 = 1$$. Is it really that simple or am I making a logical error somewhere?

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• Use the geometric series for $\frac{1}{1-a}$ as inspiration. – Randall Oct 18 '18 at 19:46

If a ring has no zero divisors, you have that $$ab=0$$ implies either $$a=0$$ or $$b=0$$. In such a case, $$a^4=0$$ implies $$a=0$$. However, rings in general do have zero divisors. For example, $$2\not=0$$ but $$2^2=2\cdot 2=0$$ in $$\mathbb{Z}_4$$.

One way to show an element is invertible is to construct an inverse. For $$1-a$$ where $$a^4=0$$, our inverse is $$1+a+a^2+a^3$$. To see this multiply $$(1-a)(1+a+a^2+a^3)$$ and $$(1+a+a^2+a^3)(1-a)$$ out. Both give $$1$$.

This is an old trick based on the geometric series. Recall that $$\dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots$$ for any real number $$|x|<1$$. Formally, $$(1-x)(1+x+x^2+\cdots)=1$$. It then makes sense that $$(1-x)(1+x+x^2+x^3)=1$$ if $$x^4=x^5=\cdots=0$$.

• Great answer! I love the trick, another tool for the toolbox :) – Wesley Strik Oct 18 '18 at 20:01

In a ring, $$ab=0$$ does not imply that one of $$a,b$$ is equal to $$0$$. For example, consider $$2+2$$ in the ring $$\mathbb{Z}/4\mathbb{Z}$$.

To solve the actual problem, try to find some polynomials $$P,Q$$ so that

$$(1-a)P(a)=1+a^4Q(a)$$

• This is specifically polynomials under multiplication though. – Wesley Strik Oct 18 '18 at 19:48
• Doesn't matter much: $\mathbb{Z}_4[x]$ – Randall Oct 18 '18 at 19:49
• So for instance $X(X^2 +2) = 0$ if $d= X^2+2$ – Wesley Strik Oct 18 '18 at 19:49