Re: Algebra with splitting field and mod 103.

Tonico <Tonicopm@xxxxxxxxx> wrote:
On Nov 12, 8:20 am, mina_world@xxxxxxxxxxx wrote:

K is the splitting field of {7x^3 - 4x^2 + 2x - 5} over Z_103.
Find the [K : Z_103]

I think... 7x^3 - 4x^2 + 2x - 5 = (x-1)(7x^2 + 3x + 5)

If I can show that 7x^2 + 3x + 5 =/= 0 (mod 103),
K = Z_103(1) = Z_103 so, [K : Z_103] = 1. right ?

But I can't show that 7x^2 + 3x + 5 =/= 0 (mod 103).

An idea: the general equation for roots of quadratics is true in any
commutative ring (and thus in any field) with characteristic not equal
to 2, so apply here: the roots of 7x^2 + 3x + 5 mod 103 are:

x_1,2 = (100(+/-)Sqrt[9 - 140])/14 ==> the expression within the
square root is -131 = 28 (mod 103).

No, mod 103: -131 = -28 = 75 = 3(5)^2 = bb <-> 3 = (b/5)^2

Thus the problem reduces to checking if 3 is a square (mod 103).

If now you know something about the symbol of Lagrange, quadratic
reciprocity and stuff, find out whether 28 is a square mod 103,
otherwise I'm afraid you'll have to check by hand all the squares
mod 103...well, you actually need only to check 51 of them...;)

SIMPLER Since -3 == 10^2

if 3 == c^2

then -1 == (10/c)^2 contra p = 4n+3 (see [1])

[or -> -1 == 1 via (prior)^51 & Fermat's little Theorem]

--Bill Dubuque

[1] http://google.com/groups?selm=y8zmz0c2srh.fsf%40nestle.csail.mit.edu
.

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