 So this lecture is part of an online commutative algebra course and will be about artinian rings. So the main theorem we're going to prove is that all artinian rings are notarian. In fact, we prove a rather slightly more precise version. So the following four conditions are equivalent for a ring. So first of all, we have the condition that the ring r is notarian and nought is a product of maximal ideals. So just write that as pi max product or maximal. Second condition is that r is notarian and all primes are maximal. So this condition here actually has a name. When we do dimension theory of rings, we will see that this condition that all primes are maximal just says r is zero dimensional. So so artinian rings are just the zero dimensional notarian rings. The third condition says that r has finite length as an r margin. And you remember from last time that this is the same as saying that r is artinian and notarian. And the fourth condition says that r is artinian. And we're first going to prove the three easy implications, which is that one implies two, two implies three and three implies four. And then we'll have to prove that four implies one, which is not so easy. So first of all, let's just do one implies two. So we've got to show that if r is notarian and nought is a product of maximal ideals then then then all primes are maximal. Let's just write nought is equal to m1, m2, mn as a product of maximal ideals. Now if p is prime, so let's choose any prime ideal p. Well this is contained in the prime p because nought is obviously contained in p. Now if a p contains a product of two ideals, then it contains one of them. So p contains some mi. However, this mi is maximal. So p, if it contains a maximal ideal, it must obviously be equal to mi. So every prime ideal is maximal. So we've shown that one implies two. Next we'll show that two implies three. So two was r is notarian and primes are maximal. And three was finite length. So suppose r is not a finite length. What we do is we choose i maximal. So r over i is not a finite length. Now we show i is prime. So this is yet another special case that if we pick an ideal that's maximal in some set of ideals then it always turns out to be prime. Well to prove it's prime we may as well put s to be the ring r over i and we want to prove we want to show that s is an integral domain. In other words, a b equals nought implies a equals nought or b equals nought in s of course. And we know that s over a and s over b have finite length if a and b are both not equal to zero. And s does not have finite length by assumption. And now we notice that we have a map from s over b onto a s which is just multiplication by a. And this thing here has finite length. We just sort of said so down here. And as this map is onto this also has finite length. Well now we can look at the following short exact sequence. We get nought goes to a s goes to s goes to s over a s goes to zero. And now we've just shown that a s is finite length and s over a s is finite length because we said so down here. So we find that s is finite length which is rather unfortunate because we chose s to be not a finite length. So this is a contradiction. And this contradicts the assumption that we could find an ideal such that r over i is not a finite length. So r can't have finite length. So that doesn't hold. So we've shown that condition two implies condition three. And now condition three implies condition four we did last time because you remember we showed that any finite length module is artinian and notarian. So we've shown that one implies two implies three implies four. And we now come onto the really hairy part of the proof. So we want to show that an artinian ring is one in which nought is a product of maximal ideals and also one in which it's notarian. Well, we're first going to prove that an artinian ring implies that nought is a product of maximal ideals. So this proof that we're about to give us extremely hairy and probably the hairiest proof we have in the course. You'll not expect to understand it. I'm just giving it as a sort of example of a fairly typical complicated proof in commutative algebra. So it goes as follows. Choose j minimal among ideals that are the product of maximal ideals. So this means jm equals j for all maximal ideals m. Because if jm was less than j, then this would be a smaller ideal that wasn't a product of maximal ideals. And now this implies that j squared equals j. And to see this, we just notice that j is equal to j for any maximal ideal. And we also notice that j is a product of maximal ideals. So these two facts imply that j squared equals j. And now we're going to suppose that j is not equal to zero. So what happens if j is not equal to zero? Well, then we choose i minimal. Again, using the fact that the ring is artinian so that ij is not zero. I guess we're sort of using the artinian condition here. And we're also using it here. And this is one reason why the proof is particularly tricky because we need to use the artinian condition twice in two different ways. And this implies that i times j is not equal to naught for sum i and i. So the ideal i must equal i because we chose the ideal i to be minimal. And if i wasn't equal to the ideal generated by little i, then we could replace it by the smaller ideal i. Now we notice that ij times j, which is equal to ij, is equal to zero. And we can get that by observing that j squared is equal to j to get that inequality. And it should be not equal to zero, of course. And ij is not equal to zero there. So ij is equal to i because i was a minimal ideal with this property. So here we know i is minimal. And ij is not equal to zero. So ij must be equal to i because ij is contained in i. And this implies that ij equals i for sum j and j. And why does this follow? Well, this follows from this fact here together with the fact that i, the ideal i is generated by i. And if ij is equal to i, we just notice that's just equivalent to saying that i times j minus one is equal to zero. And now we notice that j is contained in all maximal ideals. And why is this? Well, j is contained in all maximal ideals because it's contained in the ideal j. And the ideal j must be contained in all maximal ideals because otherwise j times sum maximal would be a smaller product of maximal ideals. And as you can see from the fact that we're nearly at the bottom of the page, we've nearly finished. So j minus one is a unit. And that just follows from the line above because if j is in all maximal ideals and j minus one is in no maximal ideals, so it must generate the unit ideal. Now, if j minus one is a unit and i times j minus one is equal to zero, these two facts imply i equals zero. And this is a contradiction. And it contradicts the fact that i times j is, can't be zero because i times j is none zero. And so we get a contradiction. Well, what do we get a contradiction to? Well, we get a contradiction to our assumption up here that j is not equal to zero. And if j is not equal to zero leads to a contradiction, this finally leads us to the conclusion that j is equal to zero. And since j was a product of maximal ideals, this implies naught is a product of maximal ideals, which is what we wanted. So that's the difficult part of the proof. Well, that's shown that an artinian ring, naught is a product of maximal ideals. Now we finish off with a relatively easy part of the proof. So we're going to show that an artinian ring is now notarian using the fact that naught is a product of maximal ideals. Well, we put naught is equal to m1, m2, up to nn by the result above. Well, it's not above, it's actually over to the left, but whatever. And now we look at r, which contains m1, which contains m1, m2, contains m1, m2, m3, and so on up to contains m1, up to mn, which is equal to zero. Now we look at each of these factors. So each factor m1, up to mi over m1, up to mi plus one is a vector space over the field r over mi plus one. So we've broken up r into a sort of extension of things and each quotient is a field. And now we use the fact that r is artinian implies all these fields have finite dimension. Because if one of these fields had infinite dimension, then it would have an infinite decreasing sequence of vector subspaces. And this would give an infinite decreasing sequence of ideals in r. So we're actually using the fact that r is artinian for a third time because you remember we used it twice in proving this inequality here. So each factor has finite length. So r has finite length. So r is notarian because we saw earlier that finite length implies notarian. Incidentally, it's sort of tempting to think at first sight that this factor is actually isomorphic to r over mi plus one, but it need not be it. It is in general a vector space of dimension bigger than one. And you can see that by just looking at the following explicit example. So as we take r to be k x, y over x squared and x, y, y squared. So r has a basis of one x, y and the ideal M, the maximum ideal M is just generation by x and y and is a two dimensional vector space. And if we now look at r contained in M, contained in M squared. Well M squared is equal to zero. And we see this factor is r over M, which has dimension one. But this factor is M over M squared now has dimension two over r over M because it is a basis of x and y. So these factor spaces here may indeed be vector spaces of dimension bigger than one in general. So having proved this result, we can now sort of classify artinian rings. So here we have a corollary. Artinian ring is a product of artinian local rings. And this just follows from a version of the Chinese remainder theorem. So an artinian ring has the property that nought is equal to M1 to the k1 up to times Mn to the kn for the maximum for maximal ideals M1 up to Mn. By the Chinese remainder theorem, this just means we can write r is equal to r over M1 to the k1 times r over M2 to the k2 and so on. And these are all artinian local rings. And furthermore, we see these artinian local rings of only one prime ideal. The possible prime is the maximal ideal because all primes are maximal and you can see this is the only maximal ideal. So this allows us to draw a picture of the spectrum of r. The spectrum of r is just a finite set, finite discrete set, and with each of these points correspond to the maximal ideal. This again illustrates the fact that artinian rings are a sort of analog in ring theory of finite sets or finite dimensional vector spaces. Incidentally, the converse isn't true. If the spectrum is a finite discrete set, this doesn't imply the ring is artinian, although it's pretty close. It does if the ring happens to be notarian, but if you look at something like the ring k of x1, x2, and so on, with an infinite number of variables that we had earlier, and we quotient it out by x1 squared, x2 squared, and so on, then this is the unique maximal ideal M. And we saw earlier the spectrum is just a point, and yet it's not notarian, so it's not artinian either. So artinian rings sort of look like finite sets of points if you look at their spectrum, but the converse isn't quite true.