# Explicit description of the map with the nilradical as its kernel

The nilradical of a commutative ring $$R$$, denoted $$Nil(R)$$, is the set of all nilpotent elements in $$R$$. It can be shown that $$Nil(R)$$ is an ideal of $$R$$. Every ideal is equal to the kernel of some ring homomorphism: just consider the projection $$\pi : R \to R/I$$.

I'm wondering: is there a ring homomorphism whose kernel is $$Nil(R)$$ that is more explicit than the projection map $$\pi : R \to R/Nil(R)$$?

• What do you mean by “more explicit”? What you have given seems like the most obvious solution, so it doesn’t seem to mean “obvious” – rschwieb Jul 2 at 2:36
• I could have been more clear. I had something like the following in mind: suppose instead I was looking for a map whose kernel is $A_n$, the alternating group on $n$ elements. Of course, the projection $\pi : S_n \to S_n / A_n$ would be such a map, but I would say that the signature map $sgn : S_n \to \{1,-1\}$ (where $\{1,-1\}$ is viewed as a group under integer multiplication) is a more explicit map. – Charles Hudgins Jul 2 at 3:26

You're not going to get anywhere near something as crisp as the quotient $$S_n/A_n$$, because that is a highly special case: it's always order $$2$$ (for $$n >1$$) and cam be defined from the parity of the structure of the elements themselves.

In contrast, $$R/Nil(R)$$ is going to be quite wild by comparison. The quotient doesn't have a fixed order or any particularly special property, other than the fact it is a reduced ring.

But maybe the following is along the lines of what you're looking for. Let $$\mathcal C$$ be a family of prime ideals such that $$\bigcap \mathcal C=Nil(R)$$. Of course, it could be all prime ideals, but sometimes fewer will do.

You always have a homomorphism $$R\to \prod_{P\in\mathcal C}R/P$$, whose kernel is $$Nil(R)$$. By the first isomorphism theorem, $$R/Nil(R)$$ is a subring of the thing on the right, which is a product of domains. You could additionally even embed those domains in their fraction fields, and have a map of $$R/Nil(R)$$ into a product of fields.

Under special conditions, $$R$$ may permit selection of a finite $$\mathcal C$$ with the properties above, getting you an embedding into a finite product of fields: that's quite special.

If you know that there are finitely many maximal ideals that intersect to $$Nil(R)$$, then the Chinese Remainder Theorem actually says this injection is an isomorphism. In that case $$R$$ would be called a semilocal ring.

## Conclusion

So anyway, now I hope you can see the diversity of structure of $$R/Nil(R)$$. I don't know if it is "more explicit," but it does tell you more about the structure of the quotient.