Is $3$ a prime element of $\mathbb{Z[\eta]}$? 
How to check whether $3$ is a prime element or not in $\mathbb{Z[\eta]}$, where $\eta$ is a $17$th primitive root of unity.

Also in general how can we check an element is prime or not in $\mathbb{Z[\eta]}$, where $\eta$ is a primitive n-th root of unity
 A: This is a pleasing example of a nice technique. Let me explain:
The field $\Bbb Q(\eta)=K$ is of degree $16$ over $\Bbb Q$, since the $17$th cyclotomic polynomial $\Phi_{17}$ is $\Bbb Q$-irreducible. It’s well known — I’ll leave it to you to look this fact up — that the field gotten by adjoining a $p$-th root of unity is ramified only at $p$. In particular the decomposition of the ideal $(3)\subset\Bbb Z[\eta]=R$ is simply a product of the distinct prime divisors of $3$ in $R$: no exponents greater than $1$. Let’s write $(3)=\mathfrak p_1\cdots\mathfrak p_g$, where these are the divisors of $3$.
Now $R/(3)=R/(\mathfrak p_1\cdots\mathfrak p_g)=R/\mathfrak p_1\oplus\cdots\oplus R/\mathfrak p_g$, the last expression being the ring-direct sum of the finite rings $R/\mathfrak p_i$. (If there had been ramification, we would be summing up things of form $R/\mathfrak p_i^e$, where $e$ was the ramification index, and such rings are not fields.)
Because $\eta\not\equiv1\pmod3$ (if it were, then $\Phi_{17}(1)=0$ also, but this is not the case), and as a result there’s a primitive $17$-th root of unity in one (in fact all) of the fields $R/\mathfrak p_i$.
These finite fields are extensions of $\Bbb F(3)=\mathbb Z/3\mathbb Z$. What is the first possibility for a finite field of form $\Bbb F(3^n)$ that contains a $17$-th root of unity? The answer is, the first $n$ such that $17|(3^n-1)$, this last number being the order of the multiplicative group of $\mathbb F(3^n)$. In other words, the first $n$ such that $3^n\equiv1\pmod{17}$. You may have noticed already that $3$ is a primitive residue modulo $17$, i.e. that $3^{16}\equiv1\pmod{17}$, but this is not true for any smaller $n$; if not, all you need to check is that $3^8=81\times81\not\equiv1\pmod{17}$.
What have we seen? One of the fields $R/\mathfrak p_i$ must be equal (isomorphic, actually) to $\Bbb F(3^{16})$. But in view of the wonderful formula $\sum_ie_if_i=n=[K:\Bbb Q]$, where the $e$’s are the ramification indices and each $f_i$ is the degree of $R/\mathfrak p_i$ over $\Bbb F(3)$, you see that there is only one $\mathfrak p$, namely the first one. Thus the decomposition is $(3)=\mathfrak p$.
The upshot? Yes, $3$ is still irreducible in $R=\Bbb Z[\eta]$.
A: In general, I guess you want to look at the norm and hope it's a prime number. To check whether prime numbers stay prime, we can actually say precisely what happens. Namely, the splitting behaviour of $p$ in $\mathbb Z[\eta]$ is the same as that of the minimal polynomial of $\eta$ (i.e. the cyclotomic polynomial) modulo $p$.
Let's say $n$ is prime, e.g. $n = 17$. Then $p$ is unramified for all $p \neq n$, and the inertia degree is the order of $p \in (\mathbb Z/n)^{\times}$. On the other hand, $n\mathbb Z[\eta] = (1 - \eta)^{n-1}$.
For the case $p \neq n$, use the discriminant or the fact that $x^n - 1$ is separable modulo $p$. Moreover, if we decompose $x^{n-1}+\ldots+x+1 = g_1\cdots g_r$ into irreducible factors, then each $g_i$ is the minimal polynomial of a primitive $n$-th root of unity $\zeta\in\overline{\mathbb F_p}$, and $\mathbb F_p(\zeta) = \mathbb F_{p^m}$ for some $m\in\mathbb N$. Hence $\deg(g_i) = [\mathbb F_p(\zeta) : \mathbb F_p] = m$ is minimal such that $\mathbb F_{p^m}^{\times}$ has an element of order $n$, i.e. $n|(p^m-1)$, which is the same as $p^m \equiv 1 \bmod n$. We obtain the isomorphism $$\mathbb Z[\eta]/p \cong \mathbb Z[x]/(p,x^{n-1}+\ldots+x+1) \cong \mathbb F_p[x]/(x^{n-1}+\ldots+x+1) \cong \prod_{i=1}^r \mathbb F_p[x]/g_i,$$
which actually restricts to $\mathbb Z[\eta]/\mathfrak p \cong \mathbb F_p[x]/g_i$ for $\mathfrak p | p$, so $\deg(g_i)$ is indeed the inertia degree. In fact, the prime decomposition will be $p\mathbb Z[\eta] = \prod_{i=1}^r (p,f_i(\eta))$ for $f_i \equiv g_i \bmod p$.
It remains to answer the extremely difficult question what the order of $3 \in (\mathbb Z/17)^{\times}$ is. Apparently, it is indeed a primitive root modulo $17$, so $3$ will be prime in $\mathbb Z[\eta]$.
