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A derivation on a ring 𝑅 is a map 𝐷:𝑅→𝑅 satisfying 𝐷(𝑎+𝑏)=𝐷(𝑎)+𝐷(𝑏) and 𝐷(𝑎𝑏)=𝑎𝐷(𝑏)+𝐷(𝑎)𝑏.

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Does any derivation of commutative algebra preserve its nil-radical?

Then using the product rule for derivations many times, we see that $$ 0=D^n(x^n)=n! D(x)^n+Y, $$ where $Y$ is divisible by $x$. …
Vladimir Dotsenko's user avatar