If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$?

They key idea is that each successive polynomial ring extension $\,D\subset D[x]\,$ is factorization inert, i.e. the ring extension introduces no new factorizations, i.e. if $\, 0\ne d\ \in D\,$ factors in $\,D[x]\,$ as $\,d = ab\,$ for $\, a,b\in D[x]\,$ then $\,a,b\in D.\,$ From this one easily deduces that the requisite factorization properties extend to $\,R[x_1,x_2,\cdots\,].\,$ The same ideas work for arbitrary inert extensions.

Remark $\ $ Paul Cohn introduced the idea of inert extensions when studying Bezout rings. Cohn proved that every gcd domain can be inertly embedded in a Bezout domain, and every UFD can be inertly embedded in a PID. There are a few variations on the notion of inertness that prove useful when studying the relationship between factorizations in base and extension rings, e.g. a weaker form where $\, d = ab\,\Rightarrow\, au, b/u\in D,\,$ for some unit $\,u\,$ in the extension ring.

Let $A = R[X_1,\ldots, X_n]$ and $B = R[X_1,\ldots, X_n, \ldots ,X_N]$. Write $B = A[Y_1,\ldots,Y_k]$ to ease notation.

Now $B / pB \cong (A/pA)[Y_1,\ldots, Y_k]$. Since $p$ is irreducible in the UFD $A$, $pA$ is a prime ideal. So $A/ pA$ is an integral domain. So $B / pB$ is also an integral domain. So $p$ is irreducible in $B$.