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This is pretty simple too, but here it goes. I take R to be commutative and have 1.

If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0.

If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field.

So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) We Sometimes, but not always (see below), we can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0.Indeed, both (R/P)[X]/(X^k) and R are isomorphic to quotients of R[X]/(X^k) by the ideal P[X]/(X^k), because the latter is the kernel of the two epimorphisms

R[X] / (X^k) -->> (R/P)[X] / (X^k) (projection of coefficients)

R[X] / (X^k) -->> R (evaluation at p^{k-1})

Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X).

2 added 271 characters in body

This is pretty simple too, but here it goes. I take R to be commutative and have 1.

If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0.

If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field.

So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) Maybe we We can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0? . Indeed, both (R/P)[X]/(X^k) and R are isomorphic to quotients of R[X]/(X^k) by the ideal P[X]/(X^k), because the latter is the kernel of the two epimorphisms

R[X] / (X^k) -->> (R/P)[X] / (X^k) (projection of coefficients)

R[X] / (X^k) -->> R (evaluation at p^{k-1})

Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X).

1

This is pretty simple too, but here it goes. I take R to be commutative and have 1.

If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0.

If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field.

So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) Maybe we can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0? Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X).