Re: Q[t]/<t^2+1> has square roots of -1
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 07 Oct 2008 05:00:29 GMT
In article <p6qle4dkcbv678cqthpno9dqahed1rfm3o@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
On Tue, 07 Oct 2008 00:38:52 -0400, quasi <quasi@xxxxxxxx> wrote:
On Mon, 6 Oct 2008 21:29:18 -0700 (PDT), lite.on.beta@xxxxxxxxx wrote:
Please tell me if this is correct reasoning:
Q[t] polynomails with rational coefficients
<t^2 + 1> ideal generated by t^2 + 1
let R:= Q[t]/<t^2 + 1>
Since (t^2+1) - 0 is in <t^2 + 1> (*)
We have t^2 + 1 = 0 in R (**)
t^2 = -1
Thus -1 has square roote -t and t in R.
Is my solution good?
Yes.
Is there a quicker easier way to go from (*) to (**)?
No.
But let me point out that there is a slight abuse of notation when you
use the letter "t" to denote both
the indeterminate of Q[t]
and
the image of that indeterminate in R.
Still, it's a convenient, standard abuse, just so long as you realize
that to be technically correct, you should either use another letter,
for example t-bar, or alternatively, represent the image of t in R by
p(t), say, where p is the natural map from Q[t] to R.
or by t + < t^2 + 1 >,
so it's clear you're talking about the coset containing t.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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