Just to make sure I get it right.

is a field.

We want to express as a polynomial of degree less than 2.

Thank you!

Also, I have to find the roots of

but 2 is not a perfect square in

has no root.

Is that right????

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- Nov 7th 2008, 04:31 PMvincisonfireQuotient Ring
Just to make sure I get it right.

is a field.

We want to express as a polynomial of degree less than 2.

Thank you!

Also, I have to find the roots of

but 2 is not a perfect square in

has no root.

Is that right???? - Nov 7th 2008, 08:40 PMwhipflip15
Correct.

Also note that in your field reduces to just . - Nov 7th 2008, 09:42 PMvincisonfire
But reduce to 1 only if t=x. Therefore there might be roots even if x is not a root. For example, in this field, has two roots that are x and -x.

Yes it's a cheap example but it's late and it still gives the idea. x is not a variable anymore in this field. 'Polynomials' becomes roots.

Anyway, good night. - Nov 8th 2008, 04:30 PMThePerfectHacker
- Nov 9th 2008, 12:22 PMluzerne
Another (similar) way is to see that the discriminant is a square in the field since -->

Then I used the usual formula for quadratic functions and got (since 1/2 = 3 in Z/5Z), which is the same answer as TPH since in this field