# Thread: modulo squares and cubes

1. ## modulo squares and cubes

If we look at the field F_7* modulo the squares, then we can say that this is isomorphic to Z/2 by sending the squares to 0, and the non-squares to 1 (at least we think this is the reason that it is isomorphic to Z/2).

Now we want to look at F_7* modulo cubes, we think this is isomorphic to Z/3, but we don't know why. We can send the cubes to 0, but how do we know which elements are mapped to 1 and 2?

2. Originally Posted by lizzy
If we look at the field F_7* modulo the squares, then we can say that this is isomorphic to Z/2 by sending the squares to 0, and the non-squares to 1 (at least we think this is the reason that it is isomorphic to Z/2).

Now we want to look at F_7* modulo cubes, we think this is isomorphic to Z/3, but we don't know why. We can send the cubes to 0, but how do we know which elements are mapped to 1 and 2?

Since $\displaystyle a^3\in\{-1,0,1\}\!\!\pmod 7$ , then $\displaystyle a^3\in\{-1,1\}\!\!\pmod 7$ for $\displaystyle a\in\left(\mathbb{F}_7\right)^*$ . Also, if $\displaystyle \phi: \left(\mathbb{F}_7\right)^*\rightarrow \mathbb{Z}_3$ is such a homomorphism, it must be that $\displaystyle \phi(ab)=\phi(a)+\phi(b)$ ,

where the product in the left side is modulo 7, whereas the sum in the right side is modulo 3.

But it must be $\displaystyle \phi(a)=\phi(b)\!\!\!\pmod 3\Longleftrightarrow \phi(a)-\phi(b)=0\!\!\!\pmod 3$ $\displaystyle \Longleftrightarrow \phi(ab^{-1})=0\!\!\!\pmod 3\Longleftrightarrow ab^{-1}=\pm 1\!\!\!\pmod 7$ $\displaystyle \Longleftrightarrow a=\pm b\!\!\!\pmod 7$ .

From here you can now build your function $\displaystyle \phi$ taking into account, of course, that it must be $\displaystyle \phi(\pm 1)=0$ since only $\displaystyle \pm 1$ are cubes in $\displaystyle \left(\mathbb{F}_7\right)^*$...

Tonio