Elements with norm 1 in $\mathbb{F}_{25}^\times$

Question

We define $\mathbb{F}_{25}^\times := \mathbb{Z}[\sqrt{2}]/5$, and we want to determine the elements which have a norm of 1.

I am striking out here, and I think it's because I'm defining $\mathbb{Z}[\sqrt{2}]/5 = \{x + y\sqrt{2} | x, y \in \mathbb{Z}/5$

Now, the norm is going to be $x^2 - 2y^2$ and I want this to equal 1, but I don't think there are any solutions because I end up with the solutions $-1, -2, 2, 4$

I don't know where I'm going wrong here.

Answer 1

There seems to be a notation problem. The ring of integers of $\mathbf Q(\sqrt 2)$ is $\mathbf Z[\sqrt 2]$, and since $2$ is not a square mod $5$, the prime $5$ is inert, and $\mathbf Z[\sqrt 2]/(5) \cong \mathbf F_{25}$, so your notation at the beginning should be $(\mathbf Z[\sqrt 2]/(5))^* \cong (\mathbf F_{25})^*$. Then your question on the determination of the elements of norm $1$ in $\mathbf F_{25}/\mathbf F_{5}$ becomes purely Galois theoretic. As $2$ is not a square mod $5$, we have $\mathbf F_{25}=\mathbf F_{5}(\sqrt 2)$ (there is no confusion in writing $2$ for $2$ mod $5$), a cyclic extension of degree $2$ over $\mathbf F_{5}$. Hilbert's thm. $90$ tells us that the elements of norm $1$ are those of the form $\frac {a+b\sqrt 2} {a-b\sqrt 2}= \frac {a^2+2b^2}{a^2-2b^2} + \frac {2ab}{a^2-2b^2}\sqrt 2$, with $a, b \in \mathbf F_{5}$, not simultaneously null.

Answer 2

... also, the norm of $x$ is $x\cdot x^5=x^6$, because $x\to x^5$ is the Frobenius automorphisms of $\mathbb F_{25}$ over $\mathbb F_5$. The multiplicative group of $\mathbb F_{25}$ is cyclic, order $24$, so the group hom $N(x)=x^6$ surjects to the subgroup of order $4$, $\mathbb F_5$. The kernel has order $6$. The Hilbert Thm 90 argument gives a way to for-sure make many of them, as $x=y^5/y$, for any $y\in \mathbb F_{25}$, and, provably (also by elementary means here) these are all.