It is widely known that if a univariate polynomial $f(x)= \sum_{i} a_ix^i$ is irreducible over a finite field $\mathbb{F}_p$ for some prime $p$ not dividing the leading coefficient of $f(x)$ then the polynomial is irreducible over $\mathbb{Z}$.
My question is does this generalise to extensions of $\mathbb{F}_p$ and $\mathbb{Z}$? Does this hold for bivariate polynomials, and in general, polynomials in $n$ unknowns?
We have the following general theorem.
Thm. Let $R$ be an integral domain, and let $\mathfrak{p}$ be a prime ideal of $R$.
Let $f\in R[X]$ be a primitive non constant polynomial (primitive= a common divisor of the coefficients of $f$ is necessarily a unit) satisying the following conditions:
(1) the leading coefficient of $f$ does not belong to $\mathfrak{p}$
(2) the polynomial $\bar{f}\in A/\mathfrak{p} [X]$ is irreducible.
Then $f$ is irreducible.
(Side remark : Note that $R$ is not supposed to be a UFD, so it may happen that $f$ is reducible in $K[X]$, where $K=Quot(R)$.)
You can have several generalizations according the choice you are making for $R$ and $\mathfrak{p}$.
For example, if $R=\mathcal{O}_L$ is the ring of integers of a number field, then $R/\mathfrak{p}$ will be a finite field.
If you take $R=\mathbb{Z}[T_1,\ldots,T_n]$ and $\mathfrak{p}=(p)$, you will obtain a generalization for polynomial in several variables.
Note that we even have a less known generalization (but less easy to apply since $A/\mathfrak{a}$ may be not a integral domain)
Thm. Let $R$ be an integral domain, and let $\mathfrak{a}$ be an ideal of $R$.
Let $f\in R[X]$ be a primitive non constant polynomial (primitive= a common divisor of the coefficients of $f$ is necessarily a unit) satisying the following conditions:
(1) the class of leading coefficient of $f$ in $A/\mathfrak{a}$ is nonzero and is not a zero divisor.
(2) the polynomial $\bar{f}\in A/\mathfrak{a} [X]$ is irreducible.
Then $f$ is irreducible.