Thread: What is zero
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Feb5-04, 12:32 PM
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Originally posted by Hurkyl
The proof that [itex]x^2 - 1[/itex] has only two roots (1 and -1) depends on the fact that you're working over a field; that is division always works (for nonzero things). When I introduced this new thing, [itex]h[/itex], we are no longer working over a field, so this proof fails.
What does it mean to be "working over a field?" I thought that the coefficients of the polynomial had to be members of a field.

Originally posted by Hurkyl
Basically, I'm suggesting you do the same thing with these numbers. Just like you use the fact [itex]i^2 = -1[/itex] to simplify the expressions for ordinary complex numbers, you can use the fact [itex]\alpha^2 = 2\alpha - 2[/itex] to simplify expressions for complex numbers written in this new way.
I get it. That's kind of interesting.

Originally posted by matt grime
... the Riemann Sphere ... extends the complex plane to include a point at infinity, but this is basic algebra here so we don't touch it.
Does one need the Riemann sphere to define 0 as a member of C?

Originally posted by matt grime
... take the integers mod 8, ... here ... we say 4 and 2 are divisors of zero. This is not a good place to do arithmetic - we can't divide.
How can we have divisors without being able to divide? What is this new (to me) definition of divisor? Why do you only list 4 and 2? I thought that 0/anything = 0. Is it that 0/2 = 4 and 0/4 = 2? Is that true? If so, that's wierd.

Originally posted by matt grime
... {p} is the set of all multiples of p, that is all polynomials of the form p(x)q(x) for some omthe poly q(x).
What is a "omthe poly?"

Originally posted by matt grime
This irreducibility means that when we form the mod p analogue for polynomials - ie two polys r and s are defined to be equivalent if p divides r-s, or equivalently there is another poly h with r = s+h*p -- that we get a field.
When we say "divide," do we always assume that there is no remainder? So, basically, by saying "p divides r-s," that is another way of saying that "p is a factor of r-s?"

Originally posted by matt grime
notice that we can define a map to C by sending x to i. This is acutally an isomorphism - that is R[x]/{p} and C are actually the same field.
This looks like Chinese to me (and I am not oriental).

Originally posted by matt grime
Ian Stewart's book on Galois Thoery is a good place to learn about this - unlike most university level texts this emphasizes the examples and works inside C most of the time.
Thanks for the reference. I will try to motivate myself through it (first I have to motivate myself to the library).

Originally posted by lethe
try to declare j a root of [itex]e^x=0[/itex] over the reals. you won't get a field.
I realize that this addresses my question with a counterexample, so I do not dispute it. But, for the sake of future discussion, will infinite order polynomials be relevant? If so, then this seems like an example to show that not even the complex field is alg closed.