As I said before, a number field K with *r* real conjugates and *s* pairs of proper complex ones has *r* + *s* – 1 independent units, in the sense that if (*u*1^*a*1)(*u*2^*a*2)…(*u*(*r*+*s*-1)^*a*(*r*+*s*-1)) = 1 where the *a*(*i*)’s are integers, then all *a*(*i*)’s are zero. While this doesn’t apply to every set of *r* + *s* – 1 units, it does apply to some set of that size but not to any set of *r* + *s* units.

What I’m more concerned with in this post is the number of roots of unity in K, as a function of *n* = [K:**Q**]. First, note that every number field has at least two roots of unity, 1 and -1. Also note that **R** only has these two roots of unity, so if K is a subfield of **R**, it has no additional roots of unity. Since every K for which *r* > 0 can be regarded as a subfield of **R**, it follows that if K has more than two roots of unity, *n* = 2*s*.

Now, let L be the subfield of K generated by roots of unity. L has degree *m* over **Q**, where *m* < *n*. In fact *m* divides *n*, since K is a vector space over L, so that K is additively the same as L^*l* = **Q**^*lm* for some* l* and *lm* = *n*. If L has *k* roots of unity, then the multiplicative group of roots of unity can be denoted as C(*k*), which is just a way of writing **Z**/*k***Z** in multiplicative notation.

Since C(*k*) has an element corresponding to 1 in **Z**/*k***Z**, call it *z*(*k*), L is in fact the subfield of K generated by *z*(*k*); this is because every other root of unity in K is a power of *z*(*k*). The conjugates of *z*(*k*) are other *k*th roots of unity. Furthermore, *z*(*k*) is not a root of unity of any order less than *k*, or else there are fewer than *k* roots of unity in K. The name for such a root is a primitive *k*th root of unity. It’s not difficult to see that *z*(*k*)^*c* is a primitive *k*th root of unity iff *c* and *k* are coprime. So the number of conjugates of *z*(*k*) is at most phi(*k*), defined to be the number of positive integers less than *k* that are coprime to *k*, or, equivalently, the number of equivalence classes mod *k* that are coprime to *k*. In fact the two are equal, but that requires a technical field theoretic proof I don’t want to get into.

So if K has exactly *k* roots of unity, then phi(*k*) must be a divisor of [K:**Q**]. Also, *k* can never be odd, because we can always pair off roots of unity with their negatives. After all, if *z*^*m* = 1, then (-*z*)^2*m* = 1, and *z* = –*z* iff *z* = 0.

The converse, mind you, isn’t always true. phi(6) = 2, but **Q**(SQRT(*n*)) doesn’t have 6 roots of unity unless *n* = -3. In fact there are only two roots of unity in **Q**(SQRT(*n*)) unless *n* = -1 or -3, where it’s understood that if *n* = *mb*^2 then we only ever write **Q**(SQRT(*m*)). Moreover, in **Q**(SQRT(*n*1), SQRT(*n*2)) the number *k* must satisfy phi(*k*) = 1, 2, or 4.

If *k*1 and *k*2 are coprime then an equivalence class mod *k*1*k*2 is determined by its equivalence class mod *k*1 and its class mod *k*2, and is coprime iff it’s coprime to *k*1 and *k*2, so that phi(*k*1*k*2) = phi(*k*1)phi(*k*2). That implies that phi(*k*) is the product of phi(*p*^*m*) over all *p*^*m* dividing *k* such that *p*^(*m*+1) does not divide *k*; and phi(*p*^*m*) is just the number of positive integers less than *p*^*m* coprime to *p*, which is (*p*-1)*p*^(*m*-1).

In other words, if phi(*k*) divides 4, then every prime *p* dividing *k* must be one more than a divisor of 4 – i.e. be 2, 3, or 5 – and also the exponents of 2, 3, and 5 are at most 3, 1, and 1 respectively. It’s not especially hard to then compute that *k* has to be 1, 2, 3, 4, 5, 6, 8, 10, or 12, where the odd values are not possible from a previous result.

If *k* = 2, then the only roots of unity are -1 and 1, which is the general case. If *k* = 4, then K contains *i* = SQRT(-1) and some other square root. If *k* = 6, then K contains SQRT(-3) and some other square root. If *k* = 8, then K = **Q**(SQRT(-1), SQRT(2)) which is generated by primitive 8th roots of unity. If *k* = 12, then K = **Q**(SQRT(-1), SQRT(3)) which is generated by primitive 12th roots of unity. If *k* = 10, then K can’t be written as **Q**(SQRT(*n*1), SQRT(*n*2)); the only proof I can think of requires Galois theory, in which case the proof is that the field of 10th roots of unity’s Galois group is C(4) while **Q**(SQRT(*n*1), SQRT(*n*2))’s Galois group is always C(2) * C(2). In particular, **Q**(SQRT(*n*1), SQRT(*n*2)) has no roots of unity apart from 1 and -1 whenever *n*1 and *n*2 are both different from -1 and -3.

[…] This is for those of you who feel insufficiently challenged. […]

best rated dehumidifier…[…]Roots of Unity « Abstract Nonsense[…]…

In fact, Jiu-jitsu practitioners and soccer players in Brazil often have a large meal of acai berry pulp before a competition. viagra http://www.fcroggwil.ch