Recognising to be a field I am sorry if this is too well-known. Is $\mathbb{Z}[\omega]$ a field ? Here $\omega$ is a non real $p^{\text{th}}$ root of $1$ and $p$ is a prime. And prove whatever your claim is.
On an unrelated topic I was curious to know that to show $\displaystyle\prod_{j=1}^{n}(x^2+j^2) +1$ is irreducible in $\mathbb{Z}[x]$ I saw somewhere it was proven that it cannot be factored into polynomials $f,g$ with coefficients in $\mathbb{Z}[\textrm{i}]$ where $\textrm{i}=\sqrt{-1}$ and from there they concluded it is irreducible in $\mathbb{Z}[x]$ too. Can someone tell me how is that true? And can we prove something irreducible for polynomials with coefficients in  $\mathbb{Z}[\omega]$ as mentioned above and thus conclude it is irreducible in $\mathbb{Z}[x]$ too? If not then what would be the required condition for a set $A$ such that we can replace $\mathbb{Z}[\textrm{i}]$ above by $A$ ? Please help me and sorry if this is too well known. Thanks in advance.
 A: $\newcommand{\ZZ}{\mathbb{Z}}$
Firstly, $\ZZ[\omega]$ is not a field for any $p$. To see this, note that $\ZZ[\omega] = \ZZ[x]/\Phi_p(x)$, where $\Phi_p(x)$ is the $p$th cyclotomic polynomial, which is a monic (irreducible) polynomial over $\ZZ$ with constant term $\pm 1$. In fact, it's just $x^{p-1} + x^{p-2} + \cdots + x + 1$. Now you can ask, is $1/2\in\ZZ[x]/\Phi_p(x)$? In other words, does there exist a polynomial $f\in\ZZ[x]$ such that $2f\equiv 1 \mod \Phi_p(x)$, or equivalently $2f-1\equiv 0\mod\Phi_p(x)$, or equivalently $\Phi_p(x)$ divides $2f-1$ in $\ZZ[x]$? In other words, does there exist a polynomial $g\in\ZZ[x]$ such that $g(x)\Phi_p(x) = 2f(x)-1$? If such a $g$ were to exist, than we can look at the equation mod 2 (ie, in $\mathbb{F}_2[x]$), to see that $g(x)\Phi_p(x) \equiv 1\mod 2$, which is to say that $\Phi_p(x)$ is invertible in $\mathbb{F}_2[x]$. This is of course false since it's a monic polynomial of degree $\phi(p) = p-1$ (and hence its reduction in $\mathbb{F}_2[x]$ is also degree $p-1$) and the only units of $\mathbb{F}_2[x]$ are the units of $\mathbb{F}_2$, namely 1. Note that this argument will work to show that $1/n$ for any $n\in\ZZ$ is not in $\ZZ[\omega]$.
You can say even more. Note that $\omega$ is integral over $\ZZ$, since it satisfies a monic polynomial with coefficients in $\ZZ$, namely $x^p-1$. It's a standard fact that the product and sum of elements over a ring (in this case $\ZZ$) are also integral over the same ring. Hence, $\ZZ[\omega]$ is integral over $\ZZ$. This shows that the ring extension $\ZZ\subset\ZZ[\omega]$ is an integral extension, and in particular satisfies the lying over property, which basically says that every prime ideal of $\ZZ$ can be extended to a prime ideal of $\ZZ[\omega]$. Hence, for any prime $p\in\ZZ$, there is a prime ideal $\mathfrak{p}$ of $\ZZ[\omega]$ which contains $p$, which of course implies that $p$ is not invertible in $\ZZ[\omega]$. These is all basic facts of algebraic number theory and everything I just said can be found here: http://en.wikipedia.org/wiki/Integral_extension
I don't know why your polynomial in your second question is irreducible. However,  if a polynomial is irreducible over a ring, then of course it must be irreducible over any smaller ring, since any factorization over a smaller ring gives a factorization over the larger ring.
