You have probably seen local rings defined this way: R is local if it has a unique maximal ideal.  That's fine if R is commutative.  Here is a more general definition.

R is local if it has a unique maximal left ideal.  Remember that the jacobson radical is the intersection of the maximal left ideals, and in this case J is our unique maximal left ideal.

J is automatically an ideal, so R/J is well defined.  Let y be an element of R representing a nontrivial coset of J.  In other words, y is nonzero in R/J.  Let y and J generate a larger left ideal, which must be all of R. Hence y has a left inverse in R/J.  This holds for every y, hence R/J is a division ring.

Conversely, let J be the jacobson radical and assume R/J is a division ring.  Suppose there is a maximal left ideal M beyond J, and let y lie in M-J.  Since xy = 1 in R/J, y and J span all of R.  With y and J in M, this is a contradiction, hence J is already a maximal left ideal, and it is the only maximal left ideal.

By symmetry, R/J a division ring implies J is the only maximal right ideal.  Thus a local ring R has a unique maximal left ideal, or a unique maximal right ideal, or a jacobson radical J such that R/J is a division ring.  The definitions are equivalent.

Asserting a unique maximal ideal is no longer sufficient.  Let R be the 22 matrices over the reals.  This is a simple ring whose only proper ideal is 0, yet it has many maximal left ideals, one for each line in the plane passing through the origin.  It is not a local ring.

Let R be a local ring.  If J is the maximal ideal, also serving as the jacobson radical, then the units of R map onto the units of R/J.  Everything in R/J is a unit, hence everything in R-J is a unit.  In other words, R minus its units is the maximal ideal.

Conversely, assume R minus its left invertible elements is an ideal J.  There can be no left ideals with elements outside of J, except for R, hence J is the only maximal left ideal and R is local.

If R minus its units is an ideal J, then J has no left invertible elements, else it would contain all of R, including the unit 1.  The left invertibles are all units, R minus its left invertibles is an ideal, and R is local.

If the nonunits form an ideal then they certainly form a closed subgroup under addition.  Conversely, let the nonunits form a subgroup J.  If x is in J and xy is a unit then x is also a unit, which is a contradiction.  Therefore xy remains a nonunit, and lies in J.  This makes J an ideal, and we are back to a local ring.

In summary, R is local iff R minus its units is closed under addition.

If all nonunits are nilpotent then R is local.  Let x be a nonzero nonunit with xn = 0.  The product (yx)*xn-1 = 0, and if yx is a unit then xn-1 has to equal 0.  This means yx is a nonunit, and is nilpotent.  The left ideal generated by x consists entirely of nilpotent elements.  This is a nil left ideal, and all such ideals live in J.  Every nonunit lies in J, and R is local.

Don't try the converse.  The localization of Z about 2 is local, but the nonunits are not nilpotent.  In fact the ring is an integral domain. A ring R is dedekind finite if x*y = 1 implies y*x = 1.  Thus a one sided inverse implies a unit.

Let R be local and suppose x is not right invertible.  If xy is right invertible then so is x, hence the right ideal is not right invertible.  The jacobson radical comprises those elements that are not right invertible.  Embed x*R in J, the jacobson radical.  Since J is a proper ideal, x cannot be left invertible either.  Therefore R is dedekind finite.

If e is an idempotent in a local ring R, let f = 1-e, the "other" idempotent.  Remember that e+f = 1, and e*f = 0.  Since e+f = 1 they cannot both lie in J, so let e be a unit.  Now e*f = 0, hence f = 0 and e = 1.  There are no nontrivial idempotents.

Again, don't try for the converse.  A ring like Z is dedekind finite, with no nontrivial idempotents, but it is not local. If R is commutative, and P is a prime ideal, the fractions of R with denominators outside of P form the localization of R at P, and are denoted RP.  The proper ideals in RP correspond to the ideals in R that lie in P.  P is of course the largest of these, containing all the others, hence PP is the maximal ideal in the local ring RP.  In fact it is called a local ring because it reflects R at a local level, i.e. a neighborhood of R about P.  That is where the term local ring came from.

Once you have a local ring by any means, you can extend it out to a complete power series.  Let R have a unique maximal ideal M, so that R/M is a division ring.  Let S be the formal power series R[[t]].  Let M and T generate the ideal J.  Nothing in J is invertible, because the constant term cannot come out 1.  Conversely, anything outside of J has a constant term that is a unit in R.  Use synthetic division to divide this into 1, giving the inverse power series.  S consists of units and J, and is a local ring.

The p-adic numbers are a local ring with maximal ideal generated by p.  The quotient is a field, the integers mod p.  This is a variation on the power series, having carry rules for addition and multiplication.

Finally, let K be a division ring and let S be the lower triangular nn matrices with constant diagonal.  Let J be the matrices of S with 0 diagonal.  J is nilpotent, with Jn = 0.  Thus J is in the jacobson radical.  The quotient S/J is K, a division ring, hence S is a local ring.

There are of course many other examples. A left module M is indecomposable if it is not the direct product of two smaller modules.  In many cases a module is uniquely a product of indecomposable submodules, just as an integer is uniquely a product of primes.  This is the krull schmidt theorem.

Assume M is decomposable, and write M as A*B.  Let S be the ring of endomorphisms of M.  Let e be the projection that maps M onto A, and let f be the projection that maps M onto B.  Note that e+f = 1, and ef = fe = 0.  These are orthogonal idempotents.  Therefore, if S has no nontrivial idempotents, M is indecomposable.

Conversely, let S have a nontrivial idempotent e, and let f = 1-e.  Let A = e(M) and let B = f(M).  For any x in M, apply e+f to x, and write x as the sum of something in A plus something in B.

Suppose z is in both A and B.  Since e is idempotent, applying e again does not change z.  Similarly, applying f does not change z.  Since ef = 0, z is mapped to 0, and z = 0.  The components are disjoint, and M = A*B.  A module is decomposable iff its endomorphisms contain nontrivial idempotents.

A left module M is strongly indecomposable if its endomorphisms, written on the right, form a local ring.  Such a ring has no idempotents, hence strongly indecomposable implies indecomposable, as one would expect from the terminology.

The Z module Z, with endomorphism ring Z, is indecomposable, but not strongly indecomposable, since the ring of endomorphisms is not local.

If R is a local ring, then R is a strongly indecomposable R module.  This because the endomorphisms of R are R, following the image of 1.  Apply this to any of the examples in the previous section.

If M is a simple left R module then the ring of endomorphisms is a division ring.  This is a local ring, hence M is strongly indecomposable.

Let M be the integers mod pn, acting as a Z module.  The endomorphisms are isomorphic to M itself, following the image of 1, thus a local ring.  M is strongly indecomposable. Let M be acc, dcc, and indecomposable.  Let E be the endomorphism ring of M.  We will prove E is local, hence M is strongly indecomposable.

Let f be an endomorphism in E.  By fitting's theorem, there is some n such that fn splits M into kernel and image.  Of course M is indecomposable, so one or the other is trivial.

If the image is trivial then f is nilpotent.  If the kernel is trivial then f is injective, and that makes f an automorphism.  An automorphism is a unit in E.  Therefore the nonunits of E are all nilpotent.  This makes E a local ring, whence M is strongly indecomposable.

Z is not strongly indecomposable, hence dcc is a necessary condition for this theorem.

With E as above, i.e. the ring of endomorphisms of M, let J be the jacobson radical of E.  Since M is indecomposable, acc, and dcc, E is local, with all its nonunits nilpotent, and M is strongly indecomposable. J is a nil ideal, but in this case J is also a nilpotent ideal.  Let l be the length of M, i.e. the length of its decomposition series (not including 0).  Let f be an endomorphism in J.  If, as a submodule of M, the image of f7 equals the image of f8, then applying f again and again produces the same set.  This contradicts the fact that f is nilpotent.  Images must decrease, and we can only take l steps down.  Therefore fl = 0.

Consider Jl.  Take l functions from J and apply them, one after the other.  Suppose f2, applied to f1, does not reduce the size of f1(M).  Apply f2 again and again, and we still have f1(M).  Since M contains f1(M), f2 is not nilpotent on M.  This is a contradiction, hence each new function takes a step down.  After l steps we have to be at 0.  Each product of l functions drawn from J becomes 0, and all finite sums thereof are 0.  Therefore Jl = 0, and J is nilpotent.

The earlier example of the p-adic numbers is a ring / module that is not dcc.  It happens to be strongly indecomposable, but its endomorphism ring is the p-adic numbers, and the jacobson radical is generated by p.  This is not a nilpotent ideal, or even a nil ideal.  The theorem requires acc and dcc.

If R is left artinian, R is local iff R has no nontrivial idempotents.

One direction is easy.  We already showed at the start of this chapter that a local ring has no idempotents.  So assume R has no idempotents (other than 0 and 1).  Recall that left artinian implies left noetherian.  Thus R is a left R module with acc and dcc.

E, the endomorphism ring, is isomorphic to R, according to the image of 1.  Since E has no nontrivial idempotents, R is indecomposable.  Combine this with acc and dcc, and R is strongly indecomposable.  That makes E a local ring.  Since R equals E, and E is local, R is local.

As above, the jacobson radical of R is nilpotent, with exponent bounded by the length of the composition series of R.  But this is true even if R is not a local ring. This is a generalization of the traditional krull schmidt theorem.  If you review that theorem, you'll see that the module M must satisfy acc and dcc.  If it does, then it splits into indecomposable summands in a unique way, up to order and isomorphism.  In this generalization, the acc and dcc requirements are waved, and instead, we insist that the summands be strongly indecomposable.  To see why this is a generalization, assume M is both acc and dcc.  M factors into a direct sum of indecomposable submodules.  Each of these summands is acc and dcc, and by the previous theorem, the summands are all strongly indecomposable.  So this theorem covers all the bases.  Well almost - except for the application of krull schmidt to nonabelian groups.  The endomorphisms don't form a ring, much less a local ring, thus groups cannot be strongly indecomposable.  But for modules, this is a nice extension of the original theorem.

Assume M is the direct sum of G1 G2 G3 etc, and H1 H2 H3 etc, where at least one of the direct sums is finite.  Each Gi is strongly indecomposable and each Hi is indecomposable.  I am following the notation and the structure of the krull schmidt proof, so you may want to review that now.

Assume, by induction, the assertion Wj-1 is true.  In other words, M is the direct sum of G1 through Gj-1, and Hj through Hl.  (Note that k or l could be infinite, i.e. an infinite direct sum.)

Let pi be the projection onto Gi, and let qi = pi for i < j, and the projection onto Hi for i ≥ j.  In other words, q reflects the decomposition of wj-1.

Remember that M is a (possibly infinite) direct sum.  The identity map on M is then the direct sum of its projections qi.

Compose this with pj and pj is the direct sum of qipj.  Since pi and pj are orthogonal, we can start the sum at i = j.

All these endomorphisms map M into Gj.  Restrict this equation to an endomorphism on Gj.  In this context, pj is the identity map.

Since Gj is strongly indecomposable, each qipj is an automorphism, i.e. a unit in the ring of endomorphisms, or it is nilpotent, i.e. a member of the jacobson radical.

Suppose every qipj is nilpotent.  Let x be nonzero in Gj and represent it as a finite sum of projections qi.  The finite sum over qipj maps x to x.  However, the sum of these endomorphisms still lives in the jacobson radical, and remains nilpotent.  This contradicts the fact that x always maps to x.  Therefore there is some i such that qipj is an automorphism when restricted to Gj.

Let's avoid the Christmas rush and reindex now.  Set i = j, and qjpj is an automorphism on Gj.

Remember that qj maps all of M onto Hj.  It therefore maps Gj into Hj.  As the head of an automorphism, qj is injective, and embeds Gj into Hj.  Write it as an exact sequence like this.

0 → Gj → Hj → Hj/Gj → 0

Start in the middle, at Hj, and work backwards.  Apply the projection pj, then qj.  This gives two functions, forward and reverse, such that qjpj is an automorphism on Gj.

0 → Gj ⇔ Hj → Hj/Gj → 0

Let Y be the submodule of Hj that is the kernel of Pj.  Each element of Gj becomes a unique coset of Y in Hj.  Hj is Y cross Gj, as a set and as a module.  This makes Hj a direct product, yet Hj is indecomposable.  Therefore Y is trivial, and Hj and Gj are isomorphic.

From here the proof proceeds as before.  Each Gj corresponds to Hj, and the two sequences must end together.  The representation of M as a finite direct sum of strongly indecomposable modules is unique. Artinian implies noetherian, but in some cases we can infer the converse.  For the rest of this chapter, rings are assumed commutative.

A module over a field is free with a fixed rank.  Such a module is acc iff it has finite rank iff it is dcc.

Let R be a noetherian ring where the product of finitely many maximal ideals equals 0.  Some of these maximal ideals could be duplicated in the product.  This is sufficient to prove R is artinian.

Let M1 be the first maximal ideal, with Q1 = R/M1.  Note that Q1, the first quotient, is a field.

M1 is a noetherian R module, and the product ideal M1M2 is a noetherian submodule of M1.  Similarly, M1/(M1M2) is a noetherian R module.  However, the action of R depends only on the coset of M2 in R.  In other words, M1/(M1M2) is essentially a Q2 = R/M2 module.  Since Q2 is a field, and M1/(M1M2) is acc, it is also dcc.

Start with M1M2, which is acc.  Multiply by M3 and find another acc module.  The quotient, (M1M2)/(M1M2M3) is also acc.  Yet the action depends only on the coset of M3, hence it is a Q3 module, and a vector space.  It therefore becomes dcc.

Repeat this process until the product of M1 through Mn becomes 0.  At this point M1 is dcc, and Q1 is dcc, hence R is artinian. Let R be a commutative noetherian ring with dimension 0.  Dimension 0 means a prime cannot properly contain another prime.  Since maximal ideals are prime, all primes are maximal.

Since R is laskerian, write a primary decomposition Q1 through Qn for 0.  Their intersection is 0, hence their product, which lies in the intersection, is 0.

The radical of each Qi is some prime Pi, which is maximal.  Let Q be one of these primary ideals with radical P.  Since P is finitely generated, some exponent e satisfies xe ∈ Q for all the generators x of P.  Multiply e by the number of generators and use this as your new exponent.  By the multinomial theorem, anything in P, raised to this exponent, lies in Q.  Thus each Pi, raised to a sufficiently high power, lies in Qi.  Since the product over Qi is 0, a finite product of maximal ideals is 0.  By the above, R is artinian.

Conversely, an artinian ring is noetherian (we already knew that), so let's show it has dimension 0.  Remember that rings are commutative for the rest of this chapter.  Mod out by a prime ideal P and find an artinian integral domain.  This is a field, hence P is maximal, and R has dimension 0.

A commutative ring is artinian iff it is noetherian and has dimension 0. Consider the collection of finite intersections of maximal ideals in an artinian ring R.  Let H be a minimal ideal in this collection.  Suppose a maximal ideal M is not part of the finite intersection that produces H.  Bring in M, and H does not get any smaller.  In other words, M contains H.  Since M contains the product of maximal ideals it contains one of them.  Thus M is part of the intersection after all.  Since H is a finite intersection, R has finitely many maximal ideals.  This is the definition of semilocal, hence every commutative artinian ring is semilocal.

By intersection, H is the jacobson radical.  Since R is artinian, H is also nilpotent. We are about to analyze the structure of a commutative artinian ring, but only after it has been reduced.  If R has a nil radical, mod out by this ideal and analyze the quotient ring.

As shown in the previous section, the jacobson radical of R is nilpotent, and doubles as the nil radical of R.  Therefore an artinian ring is reduced iff it is jacobson semisimple.

Here is an example that shows how unpredictable the nil radical can be.  Let R be an artinian ring with 0 nil radical.  Adjoin x and y to create polynomials in two indeterminants.  Let x3 = 0, and let y7 = 0.  The new ring is a finitely generated R module via the powers of x and y, thus an artinian R module, and an artinian ring.  Every prime ideal contains x and y, and all polynomials generated by x and y.  Call this ideal H.  If c is a nonzero constant from R, find a prime ideal P in R that does not contain c.  Extend this to the prime ideal comprising polynomials with coefficients in P.  This still misses c, hence c is not part of the nil radical.  With this in mind, a polynomial with c as constant cannot be in the nil radical, since we could subtract away everything except c.  Therefore H is the nil radical.  Reduce the ring and get R back again.

One can imagine adjoining several indeterminants with various relations to build a rather unpredictable nil radical.  Thus we will assume R is reduced.

As shown earlier, R is semilocal, with finitely many maximal ideals, and their product lives in the jacobson radical, which is 0.  Apply the chinese remainder theorem and write R as a direct product of rings, one component ring for each maximal ideal.  Each component is R mod a maximal ideal, which is a field.  Therefore R is a direct product of fields.  Furthermore, each component is a simple, indecomposable R module, and there are finitely many components, so by the krull schmidt theorem, the decomposition is unique.  R is a unique direct product of fields, up to order and isomorphism. What does a commutative artinian local ring R look like?  Its maximal ideal M is its jacobson radical, which is nilpotent.  The powers of M reach 0 in finitely many steps.

Remember that R has dimension 0, thus there are no prime ideals other than M.  Let H be an ideal of R.  Since rad(H) = M, H is M primary.  But don't assume H is automatically a power of M.  Start with a field F and adjoin x and y, such that x3 = y7 = 0.  We did this in the previous section to build a nil radical.  Let M be the ideal generated by x and y, which is maximal.  Every prime containing 0 pulls in x and y, hence M is the only prime ideal.  The ring is a finite dimensional F vector space, hence an artinian ring.  We have an artinian local ring, yet the ideal generated by x is not a power of M.