Re: Algebra with splitting field and mod 103.

On Nov 12, 8:20 am, mina_wo...@xxxxxxxxxxx wrote:
Hello sir~

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

Find the [K : Z_103]

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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).
so, I need your advice.

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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).
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...;)

Regards
Tonio

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