The polynomial x^3 − 2 has no roots in F7 and is therefore irreducible in F7[x]. Adjoin a root k to make the field F := F7(k), which will be of degree 3 over F7 and therefore of size 343. The multiplicative group F× is of order 2 × 9 × 19. Show:
(1) that −1 is a primitive square root of 1 in F;
(2) that k is a primitive 9th root of 1 in F;
(3) that −1 + k is a primitive 19th root of 1 in F;
(4) and then that k − k^2 is a primitive root in F, that is, a generator of the multiplicative group F×.
Totally stuck on this one I'm afraid. I imagine that the fact the multiplicative group is 2 x 9 x 19 has something to do with it, since we are showing a square root, 9th root and 19th root, but I really don't know how to do this. I tried squaring, etc, but it didn't work. Thanks
I assume this is homework, and if so, please add the
homeworktag.Hints:
1) $-1 = 6$ in $\mathbb F_7$ has the property that $(-1)^2 = 1$ so the order of $-1$ is either $1$ or $2$. Can you rule out the first possibility?
2) Since $k^3 = 2$, and $2^3 = 1$ in $\mathbb F_7$, we have that $k^9 = 1$. So, $k$ can have order $1, 3,$ or $9$. Can you rule out the first two possibilities?
3) Since $k^3 = 2, k^6 = 2^2 = 4,$ and so $k^7 = 4k$, consider $$(k-1)^{21} = (k^7 - 1)^3 = (4k-1)^3 = k^3 -(-k^2) +5k - 1 = 2 + k^2 -2k -1 = (k-1)^2$$ and so $(k-1)^{19} = 1$. What are the possible orders of $k-1$?
4) What is the order of the product of an element of order $9$ and an element of order $19$? Do you know any such elements? If $\alpha$ is the product, what is the order of $-\alpha$?