The sum of two nilpotent elements of a commutative ring is nilpotent. This can be checked by a direct calculation using the binomial theorem. In fact, this calculation shows the stronger statement $x^n=y^m=0 \Rightarrow (x+y)^{n+m-1}=0$.


But we can also give a more sophisticated proof: If $x,y$ are nilpotent, they are contained in every prime ideal. Hence, the same is true for $x+y$. Hence, $x+y$ is nilpotent: otherwise, the localization at $x+y$ would be non-zero and therefore have a prime ideal, but this corresponds to a prime ideal in the given ring not containing $x+y$. (In short: The set of nilpotent elements is the intersection of all prime ideals, hence closed under addition.)


The general existence of prime ideals [is equivalent](https://mathoverflow.net/questions/98549/) to the Boolean Prime Ideal Theorem and therefore the proof above is not constructive. The proof shows nothing about the nilpotence exponent of the sum. On the other hand, it is quite elegant and it is really a no-brainer if you are used to commutative algebra. Moreover, it can be made "more constructive" (not really constructive, as Matt F. points out), or at least provable in $\mathsf{ZF}$, as follows:


We restrict our attention to the subring generated by $x,y$. This ring is countable. The same is true for the localization at $x+y$. [There is](http://www.sciencedirect.com/science/article/pii/016800728390012X) a constructive proof that every non-trivial countable commutative ring has a maximal ideal, hence has a prime ideal. And now we may proceed as before.


So we have two constructive proofs: (a) the direct calculation using the binomial theorem, (b) the proof using prime ideals. The **question** is: Assume that we know the proof (b), is there a general method how to produce the proof (a) from it? Perhaps even including the stronger statement about the nilpotence exponent? This is just a toy example for the general question how to get rid of prime ideals in proofs in commutative algebra where we would expect to have, or already know, more direct proofs. I have only picked this toy example because I hope that the general method can be easily explained with it.


Another toy example: How to produce the [direct proof](https://math.stackexchange.com/questions/10400) of $I+J=A \Rightarrow I^n+J^n=A$ for ideals $I,J \subseteq A$ from the [proof using prime ideals](https://mathoverflow.net/a/59087/98306)? A more sophisticated example can be found [here](https://math.stackexchange.com/questions/919933/the-kernel-of-r-to-a-otimes-r-b-is-nil), where I have no idea how a direct calculation looks like (perhaps I will ask this in a separate question).


I know that Thierry Coquand and Henri Lombardi have worked on related questions, but after some skimming through their work I couldn't find an answer to my question.



How to construct a constructive proof from a non-constructive proof using prime ideals?
Since this is somewhat hidden in the comments, let me give the following answer:


* The statement that the sum of two nilpotents is nilpotent is so basic that it seems to be used in the construction and the verification of the Zariski locale/topos/lattice. I don't think that constructive algebra can prove this without circular arguments.
* However, the statement $I+J=A \Rightarrow I^n+J^m=A$ can be proven by working in the lattice of radical ideals: $$\sqrt{I^n+J^m}=\sqrt{I^n} \vee \sqrt{J^m}=\sqrt{I} \vee \sqrt{J}=\sqrt{I+J}=A.$$
Whenever one uses the open subset $D(I)$ in a proof, one may simply replace it by the radical ideal $\sqrt{I}$


