I'm in touch with the Chair of the Math department where I went to college, who is a world class mathematician and a very nice man. We, in some sense, though at different stages of lfie, had the same mentor. Today's exercise (this week's) is to see if my sorry lawyer-ass self can still prove a problem based on this disucssion:
The statement is true. A finite field with q elements (where q must
be a power of a prime p) has the property that its elements are all the
roots of the equation x^q - x = 0. (There are q roots and they are
distinct). The sum of the roots, which is the same as the sum of the
elements in the field, is the negative of the coefficient of the second
highest power of x occurring (true for all monic equations over
any field, if the equation has a complete set of roots in the field).
Hence, if q > 2, the sum of the elements
in the field is 0, since x^{q-1} occurs with coefficient 0.
If q = 2, so that the field is just Z/2Z and has two
elements, {0, 1}, the sum of the elements is 1. But in this field, 1 = -1,
so the sum can also be correctly described as -1.
If q is odd, there is a simpler argument: each nonzero element of the
field can be paired with its additive inverse. When one adds all the
elements up, the paired elements cancel, and so the sum is 0.
This argument does not work when q = 2^n. In these fields,
each element is its own additive inverse, so one can't arrange the
elements in pairs with only 0 left over. The other argument shows
that the sum is 0 anyway, except when n = 1.
For fields of characteristic 2, one can argue differently as follows. The
additive group is an abelian finite group in which every element has
order 2. The classification theorem for finitely generated abelian groups
then implies that it is a finite copy product of copies of Z/2Z, say n of them.
So the group can be thought of as all possible n-tuples with
entires 0 and 1. When one adds them all up, one can think about
what happens coordinatewise. In each coordinate, 0 will occur
2^{n-1} times and 1 will occur 2^{n-1} times, which is an even
number of times if n > !. Therefore the sum is 0 in each coordinate,
and must be (0, 0, 0, ..., 0), the zero element of the abelian
group and of the field. That leaves only the easy case where
the field is Z/2Z itself.
The point is to prove there is a cylic x, in essence. His comment to me: Good luck!
Which means he thinks it's beyond my reach, at least after years away from the Department.
* * *
Yesterday was the first day this year when the weather approached what would be normal. It was beautiful. Perhaps low 70s. 55 down by the Lake.
Sometimes we can't see down the road, but the view tells us to head in that direction:
Sotomayor and Beethoven.
Glass Steagall and the Bank Holding Company Act
Who I'd like to fuck, I'm guessing, as of today. It's been awhile and I'm not sure I've had this type of thought in ages (what type I won't bore you with, though suffice it to say that saneness is a much more important character trait than I realized in youth, when it was a challenge....):
She would do as well, but is not quite as nice as near as I can tell:
The statement is true. A finite field with q elements (where q must
be a power of a prime p) has the property that its elements are all the
roots of the equation x^q - x = 0. (There are q roots and they are
distinct). The sum of the roots, which is the same as the sum of the
elements in the field, is the negative of the coefficient of the second
highest power of x occurring (true for all monic equations over
any field, if the equation has a complete set of roots in the field).
Hence, if q > 2, the sum of the elements
in the field is 0, since x^{q-1} occurs with coefficient 0.
If q = 2, so that the field is just Z/2Z and has two
elements, {0, 1}, the sum of the elements is 1. But in this field, 1 = -1,
so the sum can also be correctly described as -1.
If q is odd, there is a simpler argument: each nonzero element of the
field can be paired with its additive inverse. When one adds all the
elements up, the paired elements cancel, and so the sum is 0.
This argument does not work when q = 2^n. In these fields,
each element is its own additive inverse, so one can't arrange the
elements in pairs with only 0 left over. The other argument shows
that the sum is 0 anyway, except when n = 1.
For fields of characteristic 2, one can argue differently as follows. The
additive group is an abelian finite group in which every element has
order 2. The classification theorem for finitely generated abelian groups
then implies that it is a finite copy product of copies of Z/2Z, say n of them.
So the group can be thought of as all possible n-tuples with
entires 0 and 1. When one adds them all up, one can think about
what happens coordinatewise. In each coordinate, 0 will occur
2^{n-1} times and 1 will occur 2^{n-1} times, which is an even
number of times if n > !. Therefore the sum is 0 in each coordinate,
and must be (0, 0, 0, ..., 0), the zero element of the abelian
group and of the field. That leaves only the easy case where
the field is Z/2Z itself.
The point is to prove there is a cylic x, in essence. His comment to me: Good luck!
Which means he thinks it's beyond my reach, at least after years away from the Department.
* * *
Yesterday was the first day this year when the weather approached what would be normal. It was beautiful. Perhaps low 70s. 55 down by the Lake.
Sometimes we can't see down the road, but the view tells us to head in that direction:
Sotomayor and Beethoven.
Glass Steagall and the Bank Holding Company Act
Who I'd like to fuck, I'm guessing, as of today. It's been awhile and I'm not sure I've had this type of thought in ages (what type I won't bore you with, though suffice it to say that saneness is a much more important character trait than I realized in youth, when it was a challenge....):
She would do as well, but is not quite as nice as near as I can tell: