[MUSIC] How to see these, consider a number, a n, which is 1 plus 2 plus 4 plus, and so on, plus 2 to the power, n. Let F denote my forbiness map. So, this takes x to x to the power p. I claim that for X in F P to the power of two to the power of N. F to the power of N of X is the same thing as for F to the power am of x for any M greater than N. And it recalls as a which sends X to X to the power of is identity M. One F q. So our F to the power 2 to the power n plus l is identity one F Phi to the power of two to the power of n for any L greater or equal than zero. This implies that there exists an automorphism phi of the field to L, such that phi restricted to F, phi to the power 2 to the power n, is equal to F to the power a n for any n, are greater or equal than 0. And clearly, cannot be above the map. And so for from that, F to the power of K is different from. So phi as F to n, to power, to the power 1 plus 2 plus, and so on, plus 2 to the power n, plus, and so on. But this is, of course, very informal. The rigorous conclusions we can draw from this is that our Galois group is not a cyclical group generated by the forbiddens map. So the Galla group of let's see f b bar over f beam is not generated by the forbiddens group And also, that we don't have a bijective Galois correspondents like we have for finite field extensions. Also, the correspondents, no bijective correspondents Between subgroups of the Galois group and subextensions. Indeed the fixed field of the [INAUDIBLE] map is of p. The same as the fixed field of the whole [INAUDIBLE] group. But these are not the same. [INAUDIBLE] and the whole Galois group coincide. [SOUND] Our next subject is the root of unity. This example is also known as extensions. So consider a number n, which is prime to the characteristic of K. And consider the polynomial Pn, which is just X to the power n- 1. Why do I want n to be prime to the characteristic? Just because under this condition, this polynomial has no multiple roots. So it has exactly n root which form subgroup of k bar. [INAUDIBLE] of K bar, a cyclic multiplicative subgroup of K bar is. This is denoted by mu n. So mu n is just the group of nth roots of unity in k bar tap. One distinguishes the primitive roots of unity. These are those which are not the fruit of unity for sum G which is smaller than N. For some we can't trick this rose of the set of those roots is denoted by u n star. And we know, since our group is a cyclic group, that all And of unity are powers of a single one. [COUGH] And of course the primitive roots of unity, Are of the forum, z to the power a where A is prime to n. And what we also know that the number of those primitive roots is the Euler number phi of n. The cardinality of mu n term is the oiler number file n. Now this is just the number of numbers between one and n, which are 100 n. So set [SOUND] Phi-n equal to the product of x minus alpha, where alpha is primitive. So a priori, this is a polynomial with coefficients in k bar of x. Go to the nth cyglotomic polynomial. For example, phi 1 is, of course, our x minus 1 5 2 is x squared minus 1 divided by x minus 1. So it's x plus 1. X squared minus 1 is a product of x minus. [INAUDIBLE] range is overall. Second roots of unit here and we have to divide by X minus, a non primitive second root of unit too which is one. And so we obtain X+1 by the same reason, P3 is x cube minus 1 over X -1. This is x squared + x + 1. P4 is x4- 1 divided by the product of x-. So x plus 1 times x minus 1 so this is x squared plus one. Phi 5 by the same reason is x 4 plus x 3 plus x squared plus x plus 1 and so on. And the proposition which I shall leave as an exercise are firms that in general pn is the product of five g for od divide in n. Possible equal to n of course. Secondly, Phi n has coefficients in the field so Q if the characteristic of K is zero, and if p if characteristic of K is p If the characteristic of k is 0, Phi n is a polynomial with integral coefficients. And e if it is p then the n polynomial is the reduction modulo p of the z and polynomial over Zed. So if this is p then phi n is the reduction mod p of the nth cyclotomic polynomial over Zed. [SOUND]