Irreducibility of $1+\sqrt{-3}$ in $\mathbb{Z}[\sqrt{-3}]$ We were having a discussion on the irreducibility of $1+\sqrt{-3}$ in $\mathbb{Z}[\sqrt{-3}]$
A classmate pointed out that $1+x$ is irreducible in $\mathbb{Z}[x]$, hence, putting $x=\sqrt{-3} \Rightarrow 1+\sqrt{-3}$ is irreducible in $\mathbb{Z}[\sqrt{-3}]$
It was followed by a discussion whether it is a valid claim, and these are the major arguments that came up:
For: By the definition of polynomial rings (in Gallian), because $x$ is a placeholder, we can!
Against: But we can't since $x$ is a placeholder for an element in the ring R itself.
Can someone help us understand the situation better?
 A: This is a nice question, which reflects understandable confusion. Let me first make some remarks which will help clarify some crucial terms in the question. The first thing we need to clarify is what it means to "substitute" an element of $\mathbb{Z}[\sqrt{-3}]$ for $X$.
Note that $\mathbb{Z}[\sqrt{-3}]$ is the image of a surjective ring homomorphism from $\mathbb{Z}[X]$, namely the so-called evaluation morphism
$$\varphi \colon \mathbb{Z}[X] \to \mathbb{Z}[\sqrt{-3}]$$
which sends $f(X) \in \mathbb{Z}[X]$ to $f(\sqrt{-3})$. Your question then may be rephrased as: suppose $f(X) \in \mathbb{Z}[X]$ is irreducible. Is $\varphi(f) = f(\sqrt{-3})$ necessarily irreducible in $\mathbb{Z}[\sqrt{-3}]$?
The answer is no. Indeed, consider $f(X) = X^{2}-3X+3$. Then $f$ is irreducible (by, say, Eisenstein's criterion at $3$), but
$$f(\sqrt{-3}) = -3 - 3\sqrt{-3} + 3 = -3\sqrt{-3} = (\sqrt{-3})^{3},$$
which is clearly not irreducible, since $\sqrt{-3}$ is not a unit in $\mathbb{Z}[\sqrt{-3}]$.
The ''issue'' is that (the absence of a) factorization in $\mathbb{Z}[X]$ does not exert control over factorization $\mathbb{Z}[\sqrt{-3}]$. In particular, since $\varphi$ is surjective with kernel $\langle X^{2}+3 \rangle$, $\mathbb{Z}[\sqrt{-3}] \cong \mathbb{Z}[X]/\langle X^{2}+3 \rangle$, so elements of $\mathbb{Z}[\sqrt{-3}]$ behave like elements of $\mathbb{Z}[X]$ subject to the extra relation that $X^{2} = -3$. This relation can change the factorization shape of an element of $\mathbb{Z}[X]$ considerably in $\mathbb{Z}[\sqrt{-3}]$, as you can see.
On the other hand, since $\mathbb{Z}[\sqrt{-3}]$ is a quotient of $\mathbb{Z}[X]$ in the manner described above, factorizations in $\mathbb{Z}[X]$ do descend to $\mathbb{Z}[\sqrt{-3}]$. That is, if $f(X)$ factors as $g(X)h(X)$ in $\mathbb{Z}[X]$, then because $\varphi$ is a ring homomorphism, we must have
$$f(\sqrt{-3}) = \varphi(f) = \varphi(g)\varphi(h) = g(\sqrt{-3})h(\sqrt{-3}).$$
Even here though, we cannot say necessarily that $f$ reducible implies $\varphi(f)$ is reducible, since it is possible that $\varphi(g)$ or $\varphi(h)$ may be a unit in $\mathbb{Z}[\sqrt{-3}]$.
A: Thinking of the $x$ as a place-holder will take you down a path of confusion. Think about $x$ as a formal symbol. The polynomial $1+x$ is irreducible indeed, but that does not mean any element of the form $1+t$ is (is $4=1+3$ irreducible?).
A: This logic does not really hold. We could equally render $3+2x$ irreducible in $\mathbb Z[x]$, but if we then put $x=\sqrt2$ we find $3+2\sqrt2=(1+\sqrt2)^2$.
With an irreducible polynomial in $x$ you can assert that there is no universal factorization for all $x$, but this may still allow for factorizations at some particular values of $x$.
A: The problem is this is that $\mathbb Z[\sqrt{3}]$ has stronger properties than just $\mathbb Z[x]$. For example $\sqrt{3}^2 = 3$, but $x^2\neq 3$. Now suppose $\alpha\in \mathbb Z[x]$ is reducible $\alpha=\beta\gamma$. Then obviously $\alpha(\sqrt{3})=\beta(\sqrt{3})\gamma(\sqrt{3})$. This still does not imply that $\alpha(\sqrt{3})$ is reducible in $\mathbb Z[\sqrt{3}]$, as it might be that $\beta(\sqrt{3})=1$.
Eg.: $\beta(x) = x^2-2$, $\gamma(x) = x$, $\alpha(x)=x^3-x$. But then $\alpha(\sqrt{3})=\sqrt{3}$ is not reducible.
But on the other hand if $\alpha(x)$ is reducible it might very well be that $\alpha(\sqrt{3})$ is reducible (similar to how if $n\in\mathbb Z$ is irreducible we can still have that $n\,\mathrm{mod}\,m$ is reducible).
For example take $\alpha(x) = x^2+5$, which is clearly irreducible. But $\alpha(\sqrt{3})=3+5=8$ is clearly reducible.
