# 5-adic order of $\zeta-f(\overline 2)\zeta^2+f(\overline 2)\zeta^3+\zeta^4$

Let $$\pi:$$(the 5-adic integers)$$\to \mathbb{Z}/5\mathbb{Z}$$ be the reduction map.

Let $$f:\mathbb{Z}/5 \mathbb{Z} \to$$ (the 5-adic integers) have the following properties $$\forall x,y\in \mathbb{Z}/5\mathbb{Z}$$

1. $$f(x)f(y)=f(xy)$$

2. $$f(x)^{p-1}=1$$ unless $$f(x)=0$$

3. $$\pi(f(x))=x$$

Also, let us adjoin a primitive $$5$$th root of unity, call it $$\zeta_5=\zeta$$, to the 5-adic integers.

I want to find the 5-adic order of $$\zeta-f(\overline 2)\zeta^2+f(\overline 2)\zeta^3+\zeta^4$$

I see that the order of $$1-\zeta^i$$ is $$\frac 1 4$$ for $$i\not\equiv0$$ (mod 5).

I also see that $$f(\overline 4)=-1$$ and that therefore $$f(\overline 2)^2=-1$$ and $$f(\overline 3)^2=-1$$. Perhaps this can be used to rewrite the sum $$\zeta-f(\overline 2)\zeta^2+f(\overline 2)\zeta^3+\zeta^4$$ in a "better" way.

• I would certainly write $f(\overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $\zeta+i^3\zeta^2+i\zeta^3 +\zeta^4$. – Lubin Dec 18 '18 at 2:19
• @Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help. – Pascal's Wager Dec 18 '18 at 2:25
• Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you? – Lubin Dec 18 '18 at 2:33
• @Lubin I think that's right. – Pascal's Wager Dec 18 '18 at 2:36
• Then I suggest defining $\sigma$ in the Galois group over $\Bbb Q_5$, $\sigma(\zeta)=\zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $\zeta$ by $\zeta^2$, $\zeta^3$, and $\zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.) – Lubin Dec 18 '18 at 2:41