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Let $R$ be a commutative ring with unit. Consider $a ∈ R$ such that $a^2 = 0$. Show that $f (x) = ax + 1 \in R[x]$ is invertible in $R[x]$.

I have no idea how to do it, can you give me any tips?

Ilovemath
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  • Hint:the sum of a nilpotent and unit is invertible by the [method of simpler multiples](https://math.stackexchange.com/a/3224776/242). See also [here](https://math.stackexchange.com/a/19145/242) for a charaxcterization of units (invertibles) in polynomial rings. – Bill Dubuque Oct 31 '20 at 09:07

1 Answers1

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Hint: $(ax+1)(1-ax) = 1-a^2x^2 = 1$.

Wuestenfux
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