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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?


marked as duplicate by rschwieb abstract-algebra Oct 18 '18 at 19:51

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  • 3
    $\begingroup$ Use the geometric series for $\frac{1}{1-a}$ as inspiration. $\endgroup$ – 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 $x^4=x^5=\cdots=0$.

  • $\begingroup$ Great answer! I love the trick, another tool for the toolbox :) $\endgroup$ – 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


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

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