Given a primitive 12-th root of unity, so its minimal polynomial is $$x^4-x^2+1$$ and hence the degree of its cyclotomic ring of integers is 4.
Recently I've learnt about quadratic field and ring of integers, $\mathbb{Z}[$$\sqrt{m}]$ where m congruent to mod 4 and how to find whether a number is a prime, irreducible by using norm. But as we know that there are degree 2.
How do I show that 5 is prime or not in the primitive 12-th cyclotomic ring of integers?
Thanks.
In questions like this an indispensable tool is the Dedekind-Hasse criterion: If the ring of integers of a number field $K$ is $\mathbb Z[\alpha]$, where $\alpha$ satisfies the monic polynomial $f(x)$, then the decomposition of a rational prime $p$ in $K$ is given by considering the decomposition of $f(x)$ modulo $p$. So, if $f$ is irreducible modulo $p$, then $p$ remains prime in $K$. In general if $f(x)\equiv\prod f_i(x)^{e_i} \pmod p$ where the $f_i(x)$ are distinct irreducible polynomials modulo $p$, then the prime $p\mathbb Z[\alpha]$ factors as the product of ideals $\prod (p,f_i(\alpha))^{e_i}$ in $\mathbb Z[\alpha]$.
In your case, the ring of integers of the given cyclotomic field is $\mathbb Z[\alpha]$ where $\alpha$ satisfies $x^4-x^2+1$.
From here, as Jyrki Lahtonen points out in the comments: $$x^4-x^2+1\equiv (x^4-2x^2+1)-4x^2=(x^2-1)^2-(2x)^2=(x^2-1-2x)(x^2-1+2x)\pmod 5,$$
so that $f$ is not irreducible $\pmod 5$ and so $5$ does not remain prime in $Z[\alpha]$.
My original incorrect argument: Looking at this modulo $5$, and making the substitution $y=x^2$, we are looking at $$y^2-y+1 \pmod 5 $$ which has no roots and hence is irreducible. Thus so is $x^4-x^2+1$ and so the prime $5$ remains prime in the given field extension.