I am starting to read Milne's Etale Cohomology notes and in it they mention:
A morphism $\varphi :Y \rightarrow X$ is unramified if it is of finite type and if the maps $\mathcal{O}_{X,f(y)} \rightarrow \mathcal{O}_{Y,y}$ are unramified for all $y\in Y$. It suffices to check the condition for the closed points $y$ of $Y$.
I wasn't really sure how it suffices to check that the condition holds for closed points. This is a local definition so we can check it locally: consider a homomorphism of rings $\varphi :R\rightarrow S$. Then if $\mathfrak{p}$ is a prime ideal in $S$ which is contained in the maximal ideal $\mathfrak{m}$ then we can't really compare the fields $S_{\mathfrak{p}}/{\frak{p}}$ and $S_{\frak{m}}/{\frak{m}}$. I don't think there's a canonical morphism in either direction.
I've also searched around and I couldn't find a mention of this anywhere. Is this statement true? and if so why?
No one answered my question yet and I did some digging. Not quite sure if this is absolutely correct or not, but again if $\varphi: Y \rightarrow X$ is unramified at $y\in Y$ then this means we can find affine opens around $y \in U = \text{Spec}(A)$ and $\varphi(y) \in V = \text{Spec}(R)$ such that $\Omega_{A/R} = 0.$ Now $\Omega_{A/R}$ is just an $A-$module so we can check that it vanishes by checking it at the maximal ideals, and vanishing at the maximal ideals is then equivalent to checking that the map is unramified at the closed points.