Jacobson Radical, Artinian Implies Noetherian

Artinian Implies Noetherian

The Levytsky (biography) Hawkins theorem states that a left artinian ring is also left noetherian. First we're going to prove a lemma about modules over a semiprimary ring.

Let R be semiprimary and let M be a dcc (or acc) R module. Let J be the jacobson radical of R.

Let M₀ = M. Let M₁ = J*M, the span of elements of J times elements of M inside M. Let M₂ be J²*M, and so on. This descending chain of submodules stops, because J is nilpotent.

Let F be a quotient module at level k in this filtration. Note that F is an R module.

Let c be an element of F, a coset representative. Now c is spanned by elements of J^k*M, and if we multiply by J on the left, c winds up in M_k+1. Thus F is killed by J. This makes F a well defined R/J module.

Now R/J is semisimple, and any module over a semisimple ring is semisimple, hence F is semisimple. This makes F a direct sum of simple modules.

The submodule M_k is dcc (or acc), and so is the quotient module F. The sum is finite, and F is the finite direct product of simple modules.

Let F₁ be M₀/M₁. This is a finite product of simple modules, and by induction, M₁ has a finite decomposition series. Add on the modules in F₁, and M has a finite decomposition series. This makes M both noetherian and artinian.

If R is semiprimary, and M is a left R module, M is left artinian iff M is left noetherian.

Now for the golden result. View the ring R as a left R module. If R is artinian it is semiprimary, and by the above it is noetherian. In other words, left artinian implies left noetherian.

In general, R is left dcc iff it is left acc and semiprimary.

Without One

If a ring does not contain 1, then the proof breaks down, and you can build an artinian ring that is not noetherian. Start with any abelian group having this property, and let the product of any two elements be 0, whence every subgroup becomes an ideal.

Consider the group of rational numbers between 0&1 with powers of p in the denominator. Add rational numbers mod 1, and let the product of any two numbers be 0. In an ideal H, find a fraction with the greatest power of p in the denominator, and this generates the ideal, bringing in all such fractions, and all fractions with lesser powers of p. Ideals can get larger forever, with higher powers of p in the denominator, but they can't get smaller forever.