I find this picture quite beautiful, and some like it:
Reminds me of a much more interesting Ansel Adams shot with 8x10 of aspens.
Here is what I've been doing recently:
Updating my econ blog, which I usually limit to friends.
Reforming the Banking system - Not.
Trying to solve a problem related to this issue in Algebra (a subject I have not touched for longer than I care to remember).
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.
There are some beautiful women here. I had overlooked them before. It's been a long time since anything much caught my eye on SG and was letting my subscription expire. Interesting in different ways:
Reminds me of a much more interesting Ansel Adams shot with 8x10 of aspens.
Here is what I've been doing recently:
Updating my econ blog, which I usually limit to friends.
Reforming the Banking system - Not.
Trying to solve a problem related to this issue in Algebra (a subject I have not touched for longer than I care to remember).
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.
There are some beautiful women here. I had overlooked them before. It's been a long time since anything much caught my eye on SG and was letting my subscription expire. Interesting in different ways: