 So this is Algebraic Geometry Lecture 10 where we will be giving the proof of the Laskan-Nurta theorem and giving an example of it. So the Laskan-Nurta theorem states that for modules, finitely generated modules over notarian rings, nought is an intersection of primary sub-modules. Recall that n is primary means that m over n is co-primary, m over n, this is so short we can get it onto one slide if we cheat slightly. So it has two steps, first of all any sub-module and the proof of this is very easy, the proof you can just write down in two words, notarian induction. So what this means is the sub-modules of m have this sort of notarian property that any non-empty subset has a maximal element because that's true because m is finitely generated over a notarian ring and notarian induction sort of means you prove things by considering a maximal counter example. So this means suppose n is a maximal sub-module among those that are not a finite intersection of irreducibles. Now such a module, if the theorem isn't true, such a module must exist by the notarian induction. On the other hand, either this module is irreducible in which case we're done because it's an intersection of itself or it's not irreducible in which case it's the intersection of two largest sub-modules and by induction each of these two largest sub-modules must be a finite intersection of irreducibles so we're done. So step one is just completely routine, step two takes a little bit more effort, it says any irreducible sub-module n is primary and to prove this we can reduce to the case zero just by quotienting out by n and so we have to show that we want to show that if nought is irreducible so we want to show if the sub-module nought is irreducible this implies n is co-primary. Well that's quite easy because if m is not co-primary this implies there are two associated primes p and q so m has sub-modules of the form r over p and r over q, this is a bit sloppy it doesn't mean r over p is a sub-module, it means that r over p is isomorphic to a sub-module but it's fairly clear what it's meant and the intersection of these two sub-modules is zero because the annihilator of any non-zero element in this sub-module is p and the annihilator of any non-zero element in this sub-module is q so they must have intersection zero so this implies nought is not irreducible. Okay so there's the proof, ask a note missing out one or two details but it's an enormous improvement on the hundred pages that Laska originally took. Now let's look at an example of this, let's look at the ideal generated by x, y and y squared in the ring of polynomials in two variables over x set y squared equals zero well y squared equals zero is just the same as y equals zero but it's not quite the same and somehow it's a sort of slightly thicker version of y squared equals zero so it's sort of the y axis doubled up like this in some informal sense, I mean it is just the y axis but you should think of it as being a sort of rather thick version of the y axis. So the intersection of these two sets is going to look like this, well the intersection will just be the y axis except near the origin you should think of it as being slightly thicker it's got a sort of double point here that's strictly speaking this doesn't really make sense because a double point of a fine two space is really the same as a single point but informally we should think of this point as sticking up an infinitely small distance along the y axis so geometrically it's really just an irreducible set but informally there's a little extra bit here and you can detect that extra bit by looking at the primary decomposition of this idea so a primary decomposition of the y just corresponds to the algebraic subset consisting of the x axis and the ideal x y squared just corresponds to the algebraic set consisting of a point except you can sort of think of this point as being a slightly slightly bigger than a point it's really a sort of double point sticking up a bit so the metric example is something called an embedded component so so this ideal here is called an embedded component and it's pretty obvious why that is true geometrically so so this point here is really embedded in the in the line here um so we notice this primary decomposition is not unique but it's that there are some rather trivial ways in which it can be none unique you can just add extra um primary ideals but it's also none you none unique in a rather more serious way for instance that the ideal x the ideal x y y squared is not only equal to the intersection of the ideal y and x y squared so it's equal to the intersection of these it's also into it's also equal to the intersection of y and x plus y y squared um so x plus y y squared um is a different ideal from this and it's also co-primary