 This lecture is part of an online algebraic geometry course on schemes and will cover some basic properties of schemes such as being reduced or integral or irreducible or connected. So suppose you've got a scheme S, then one property is it could be integral, which means that it is both reduced and irreducible. So what does it mean for a scheme to be reduced and irreducible? Well a scheme is reduced means that all are reduced. So you remember there's a stalk at each point which is a local ring. So these are local rings. And what does it mean for a ring to be reduced? Well this means it's nil radical is zero. And the nil radical is the elements with x to the n equals naught for some n. And what does it mean for a scheme to be irreducible? Well irreducible just means the underlying topological space is irreducible. So S is not the union of two proper closed subsets and S is not empty. And if a space is irreducible then it also has the property of being connected which is a bit stronger which means that the only sets that are open and closed are the empty set and the whole space S. And also we usually add the condition that S is itself not empty. So this gives a sort of forewarning of one of the problems with scheme theory in that there are huge numbers of definitions of various properties and it's a real headache trying to remember all the different properties and what the relations between them are. We don't get this problem nearly so much if we're looking at varieties. For example all varieties are automatically interval reduced irreducible and connected. So we can sort of ignore these properties completely because they're automatically satisfied. Anyway let's just have a few examples. So the first example let's look at the spectrum of kx modulo x squared. So this only has one prime ideal so spectrum is a point although you can think of it as point with a tiny tangent vector sticking out if you like and this is irreducible because its spectrum is just a point but it's not reduced. On the other hand if we look at the spectrum of kxy over xy this is essentially just the union of the x-axis and the y-axis. I guess we should also remember that it's got a sort of generic point here and another generic point there so it's got two generic points and this is reducible because it's the union of two proper closed subsets the x-axis and the y-axis and it's reduced because this ring has no null present elements which as we'll see in a moment is equivalent to the corresponding scheme being reduced. If you want an example of a disconnected scheme you can look at the spectrum of kx over x squared minus x and this is isomorphic to the product of k with itself so its spectrum just looks like two points and is not connected but is still reduced. So we're now going to examine these properties in a little bit more detail so first of all let's have a look at the property of being reduced so when is s reduced and we can ask when is the spectrum of r reduced so let's look at the case of an affine scheme and ask when it's reduced well the answer is when the nil radical of r is zero so this is quite easy to check so the spectrum of r is reduced just means that the nil radical of all localizations at prime ideals are zero and so we want to show that if all the local rings of r have vanishing nil radical then r has vanishing nil radical and this is quite easy so suppose that x is not equal to zero and x to the n equals zero for some x in r what we can do is we can pick some maximal ideal containing the annihilator of x that's all the things such that x y equals zero and since x is not zero this is a proper ideal so it's contained in some maximal ideal and then it's easy to check that x is none zero in r localized at p so r localized at p is not reduced so if the ring r has a non-zero nil radical then one of its localizations at primes has a non-zero nil radical conversely if rp is not reduced this means that x over a to the n equals naught for some x that's not zero in rp so and this implies that b x to the n equals naught for some b not in p by definition of something being zero in the localization so this is equal to zero in rp so b x to the n is equal to zero and on the other hand you can easily check that b x is not zero in is not zero so so r is not reduced so this shows that a ring is reduced having zero nil radical if and only if all its localizations at p are reduced in the sense of having zero nil radical so this indicates that being reduced is a sort of local property so the following are equivalent so first of all a scheme s is reduced secondly sp is reduced for all points p so this is a local ring and this being reduced means it's nil radical vanishes thirdly every open affine sub scheme is reduced and fourthly s has a cover by reduced open affine sub schemes and these equivalences are typical for a large number of properties that that there are many properties that hold if and only if every local ring has that property and if and only if every open affine sub scheme has that property and if and only if there's sufficient if s is covered by sub schemes of that property so you tend to say you might characterize this by saying being reduced is a local property in other words you can tell whether a scheme is reduced or not just by sort of looking locally everywhere near every point and seeing if that holds for examples of some properties that are not local note that being integral or connected or irreducible are not local properties for example if you just take a scheme consisting of the spectrum of a product of two fields it just consists of two points and and each of these points is integral connected and irreducible so so we can cover this scheme by open sets that have all these properties but the scheme itself is not integral or connected or irreducible so so these are sort of global properties you sort of have to look at the whole scheme all at once to see whether or not it has these properties next let's have a look at connectedness for an affine scheme say the spectrum of some ring and this is very easy to describe spectrum of r is connected it's just equivalent to saying r has no what are they called idempotence x so and so idempotence just means x squared equals x it is no idempotence x other than naught and one and of course we should also add the condition that r is not the zero ring which is a zero ring is an exception to almost everything and this is easy to see suppose x squared equals x with x not equal naught or one then um then the spectrum of r is the union of two disjoint open sets where a is the set of ideals such that a is in p and this is the set of ideals such that b is in p and you can see that um uh sorry that x is in p and one minus x is in p and x times one minus x is equal to naught so every prime ideal is contained in one of these and no prime ideal can contain x and one minus x because that would mean it was was zero so these are both open sets so the spectrum of x so the spectrum of r is not connected um on the other hand if the spectrum of if if the spectrum of x if the spectrum of r is a disjoint union of two sets a and b then by the sheaf property we can just choose an element a to be one on the set a and zero on the set b uh sorry there shouldn't be a there should be x and then one minus x is zero here and one here and you can see that x times one minus x is zero everywhere so it's equal to equal to zero um when I say it's one here I mean it's image in all the local rings is the identity um so uh next let's have a look at uh the property of being integral so when is the spectrum of r integral and the answer is when r is an integral domain um so this sort of splits the property of being an integral domain up into two apparently different properties because you remember saying s was integral means it has to be reduced and irreducible and irreducible so saying something is an integral domain is you're really combining two quite different properties into one property um so let's just show let's just prove this um first of all if if r is an integral domain it's very easy to check it's both integral and and sorry it's both reduced and irreducible so we just have to show that if a b equals naught in r where r is reduced and irreducible then a equals naught or b equals naught um um so since a b equals naught this means that the spectrum of r is the union of primes containing a and primes containing b and these are closed subsets so um since r is irreducible so the spectrum of r is irreducible implies um it is either this set or this set we may as well suppose that it's this set so we can suppose um that all primes contain a um well as a result from commutative algebra that this implies a is nilpotent so a is equal to zero as r is reduced um so um all we've got to do is to um explain why if all primes contain a then a is nilpotent which we'll draw in the next sheet um just have a quick warning if you're looking at coordinate rings of algebraic sets then you may remember that an algebraic set is irreducible if and only if its coordinate ring is an integral domain so it's easy to absolutely mindedly think that being irreducible is equivalent to being an integral domain that's only true if you don't have nilpotent elements once you start allowing schemes that have nilpotent elements you've got to remember that being that that that a space being irreducible does not imply its coordinate ring is an integral domain you also have to check that it's reduced anyway um we'll just finish off that proof by recalling the following standard result from commutative algebra that the nil radical of ring r is equal to the intersection of all prime ideals and this is fairly short so I'll just recall the proof on the one hand um if x the n equals naught then x is in all primes because if a product of elements is in a prime then one of the elements is so if x the n is in a prime ideal then x must be in it on the other hand suppose x the n is not equal to naught for all n then we look at the following multiplicative subset one x x squared and so on and we notice that since x the n is not equal to naught m intersection i so intersect the zero ideal is equal to zero now pick a maximal element of the set of ideals i with naught contained in i and i disjoint from m now this exists by Zorn's lemma and um we're actually using the fact that x the n is none zero because we have to show this set of ideals has at least one element to apply Zorn's lemma and the one element is is the ideal zero so so if if actually n was zero for some n we couldn't find the maximal element of this set of ideals because it would be empty so anyway we've we found a maximal ideal and now it's easy to check i is prime this is part of the general thing that if you take any collection of ideals then the maximal elements of that set very often turn out to be prime ideals so so um x is not in some prime so this shows the nil radical is the intersection of all prime ideals this is actually again fairly typical of scheme theory and we have some geometric result about schemes like saying um the spectrum of a ring is integral if and only if the ring is an integral domain and this turns out to depend on some funny result of commutative algebra so scheme theory involves enormous amounts of commutative algebra like this um next we should look at irreducible subsets of a scheme and these turn out to be exactly the closures of closed points notice this fails for varieties because um the closure of a point of a variety is always just a point and you don't get all the irreducible subsets and we would just do the case of affine schemes the spectrum of r it's often easy to reduce the case of affine schemes just by using the fact that any scheme is covered by open affine schemes so i won't bother with the details of this so all you do is you let c be an irreducible closed subset of the spectrum of r and we want to find a prime ideal in other words a point of spec of r whose closure is c so we just put i equals the intersection of all ideals in c and i is prime because if ab is an i and a is not an i and b is not an i then c is the union of two closed subsets we can take the set of ideals p such that a is not in p or the set of prime ideals such that b is not in p and this fails you can't do this because c is irreducible so so this is not possible because c is irreducible um so we found our prime ideal i and now all we do is we have to check that c is the closure of i i guess strictly speaking it should be the closure of the set whose only element is i and we can check that c is contained in the closure of i because i bar is just the prime ideals containing i on the other hand i is contained in the closure of c because for any open set of the form df with i in df we have f is not in i so you remember there's a basis for the open sets consisting of the of the sets df which are just the prime ideals not containing f well if f is not in i so f is not in j for some j in the closed set c because i is the intersection of all elements in c so um this element j is in df so we've shown that for any neighborhood of i we can find an element of c in that neighborhood so we've just shown that i is contained in the closure of c and now c is closed so i is contained in c and again because c is closed we use the fact that c is closed again we find that the closure of i is now contained in c so c is contained in the closure of i my closure is contained in c so c is the closure of the prime ideal or the rather the closure of the set whose one element is the prime ideal i um we'll finish off by doing an example to illustrate all this stuff about connected components and irreducible sets and so on what we're going to do is we're going to look at the group ring of a group g which is just cyclic of order four so we can take the integer group ring or we can take the rational group ring or we can take the group ring over the complex numbers say so the group ring is just going to be you can think of this as just being z of x over x the four minus one um so this is representative of an abelian group we could do the same thing for any abelian group but this is the simplest case that gives something interesting going on so what does the spectrum of this look like well it is four components so its spectrum looks like this and it is four prime ideals we can have x minus one x plus one x minus i x plus i because actually the four minus one factorizes into the product of these four irreducible factors um and um all of these correspond to idempotent so we can have the idempotent one plus x plus x squared plus x cubed over four which is one here and not here and not here and not here and there are four other irreducible idempotents in fact there are 16 idempotents all together but there are four irreducible ones so here we get one minus x plus x squared minus x cubed over four here we get one um minus i x plus minus x squared plus i x cubed over four here we get one plus i x minus x squared minus i x cubed over four there's a 50 50 chance i've got some of the signs wrong here by the way um so what we see is that these four components actually correspond to the four complex representations of the group g so that's the four ways for g to act on a one dimensional vector space it can act as multiplication by one minus one i or minus i um well if we try doing this over the rational numbers well it doesn't factorize like that we only get three prime ideals x plus one x minus one and x squared plus one so now we have the spectrum as three components rather than four and we've got a map of rings here so we get a map of spectra in the opposite direction which obviously does this so here two components have been combined into one component when we sort of reduce our field from the complex numbers to the rationals now what happens over um z of g well here things get rather more complicated what we end up with is we get sort of you try and draw a picture of the spectrum it looks like this so this is a copy of the spectrum of z and this is a copy of the spectrum of z and this is a copy of the spectrum of z i and they all have generic points and the map takes these things to the generic points and then there are also lots of other points because on this copy of the spectrum of z we have a point three x minus one here three x plus one and here we get three x squared plus one and here we get a similar point for five but here the we sort of get two points for five because we've got an ideal five x minus two and five x plus two and so on so you remember the spectrum of z of i some of the prime split and at this point here something funny goes on because we only get one ideal which is the ideal generated by two and x minus one so here we now have three components three connected components and these correspond to the three rational representations of g and here we only have one component the scheme is actually now connected but we have three irreducible components rather bad terminology here so we have one irreducible component there one irreducible one there and one irreducible one there and the spectrum is the union of these three irreducible components and so this describes the rational representation of g and this describes the complex representations of g so what does this say about the integer representations of g well as you see what it says is that integer representations of g are really rather complicated and this is why people do representation theory over fields not over arbitrary rings in general because the representation theory of a group over a ring starts looking very very complicated as you can see from the spectrum just finish off by pointing out that we also if you look at this point here there's a sort of non reduced ring associated with this so if you take the ring z g and quotient it up by the prime two of z this actually becomes the field f2 where you take polynomials in x and then quotient out by x to the four minus one which is equal to f2 of x modulo x minus one to the four so we get a not a non reduced ring so this is one of the reasons why modular representation theory where you try working over a finite field becomes rather complicated because you start picking up non reduced rings and okay that's all about these properties next lecture we will discuss some further properties related to being notarian