Find the roots of x^3 - 1 in F (sub 7)
Find the roots of x^3 - 1 in F (sub 5)
(There is a bar over the 1's in the above problems)
Do you know what a root of a polynomial is?
It is an element of the field for which the polynomial evaluated at that element yields 0.
I just checked every element of the field (all 5 of them) to see if the polynomial was 0 in the field or not. Only 1 was a root. That is how I knew.
Okay, here is some group-theory analysis (although this particular problem is so simple that it hardly needs this kind of thing).
Now, if is a prime, is a group under multiplication modulo Now, is a root in if and only if the subgroup generated by namely has order dividing 3. Since 3 is a prime, this means that either (i) is trivial or (ii) a cyclic group of order 3. If then and so by Lagrange there is no subgroup of order 3 and so the only solution to the equation is But for it is possible to have to have a subgroup of order 3 and so it turns out that there are two solutions modul0 7.
The conclusion is: has one root in (namely 1) and three roots in (nanely 1, 2 and
If there is anything not clear, please ask.
For as the abstractionist and I have both shown you already 1 is indeed the only root to that polynomial.
However, this is not the case for as theabstractionist was so kind to point out, both 2 and 4 are roots as well.
Please take the time to read people's responses carefully, they are not answering your questions for their own health.
(mod 7) so 1 is a root
(mod 7) so 2 is a root
(mod 7) so 4 is a root.
A polynomial of degree n can have at most n roots, so this is all of them as has degree 3 and we have listed 3 roots.
Clear?