Let $A$ be a commutative ring and $N$ its nilradical. Assume $N$ is a prime ideal of $A$. Then it is known that Spec $A$ is irreducible for the Zariski topology. However the zero ideal may be not primary.

Is there a concrete example of this situation and what is the primary decomposition of the zero ideal? Does it decompose Spec $A$, even though it is irreducible?

  • $\begingroup$ @math54321 thanks, if you make this comment into an answer I'll accept it. By the way x^2 and xy are quotiented out, so no need to put them in the second primary ideal. $\endgroup$
    – V. Semeria
    May 14 at 17:48
  • $\begingroup$ That's true, but writing it as $(x^2, xy, y^n)$ makes it obviously primary to $(x,y)$ $\endgroup$
    – math54321
    May 14 at 17:55

1 Answer 1


This happens precisely when $A$ has embedded primes. The classic example is $A = k[x,y]/(x^2,xy)$: here the nilradical is $(x)$, and there are infinitely many primary decompositions of $0$ in $A$, arising from $(x^2,xy) = (x) \cap (x^2,xy,y^n)$ in $k[x,y]$ for any $n \ge 2$.


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