A closed point immersion is unramified

The image of $i$ is the set of primes of $k[t]$ containing the kernel $(t)$ of $k[t]\to k$, namely $\{(t)\}=V(t)$; in particular $(t)$ is the image point. The assertion to be verified has four parts: ${\mathrm{\Omega }}_{X\mathrm{/}Y}=0$, formal unramifiedness of $i$, local finite type of $i$, and the failure of openness.

Vanishing of ${\mathrm{\Omega }}_{Y\mathrm{/}Y}$: apply [F2] to the identity morphism $f={\mathrm{i}\mathrm{d}}_Y$ with $X=S=Y$; for every ${\mathcal{O}}_Y$-module $\mathcal{F}$ the universal property gives ${\mathrm{Hom}}_{{\mathcal{O}}_Y}({\mathrm{\Omega }}_{Y\mathrm{/}Y},\mathcal{F})\cong {\mathrm{Der}}_Y({\mathcal{O}}_Y,\mathcal{F})$. A $Y$-derivation $D$ of ${\mathcal{O}}_Y$ annihilates the image of the structure map of the identity, namely all local sections of ${\mathcal{O}}_Y$, so ${\mathrm{Der}}_Y({\mathcal{O}}_Y,\mathcal{F})=0$ and hence ${\mathrm{Hom}}_{{\mathcal{O}}_Y}({\mathrm{\Omega }}_{Y\mathrm{/}Y},\mathcal{F})=0$ for every $\mathcal{F}$. Taking $\mathcal{F}={\mathrm{\Omega }}_{Y\mathrm{/}Y}$ and the identity endomorphism as the element of the Hom set shows that the identity of ${\mathrm{\Omega }}_{Y\mathrm{/}Y}$ is zero, so ${\mathrm{\Omega }}_{Y\mathrm{/}Y}=0$.

The conormal sequence of the closed immersion $i$, taken over the base $S=Y$, reads $\mathcal{I}\mathrm{/}{\mathcal{I}}^2\to i^{\ast }{\mathrm{\Omega }}_{Y\mathrm{/}Y}\to {\mathrm{\Omega }}_{X\mathrm{/}Y}\to 0$ and is exact by [F1]; since ${\mathrm{\Omega }}_{Y\mathrm{/}Y}=0$ by step 1.2, the middle term $i^{\ast }{\mathrm{\Omega }}_{Y\mathrm{/}Y}$ is the zero module, and exactness at ${\mathrm{\Omega }}_{X\mathrm{/}Y}$ then forces ${\mathrm{\Omega }}_{X\mathrm{/}Y}=0$: the image of the zero module is $0$, so ${\mathrm{\Omega }}_{X\mathrm{/}Y}=\mathrm{im}(0)=0$. In the affine model $A=k[t]$, $I=(t)$, $B=k[t]\mathrm{/}(t)=k$ the same conclusion is the algebraic conormal sequence with middle term $B{\otimes }_A{\mathrm{\Omega }}_{A\mathrm{/}A}=0$.

Not open: suppose the image $V(t)$ of $i$ were an open subset of $\mathrm{Spec}k[t]$. By [F9] the closed subsets are exactly the vanishing sets $V(J)$ for ideals $J\subseteq k[t]$, so there would be an ideal $J$ with $V(J)=\mathrm{Spec}k[t]\setminus V(t)$, that is, $V(J)$ misses exactly the point $(t)$. The zero ideal $(0)$ is a prime of $k[t]$ by [F10] and $(0)\mathrm{\ne }(t)$, so $(0)$ is not the point missed by $V(J)$; hence $(0)\in V(J)$, which by definition means $J\subseteq (0)$, so $J=(0)$. But then $V(J)=V((0))=\mathrm{Spec}k[t]$, since every prime of $k[t]$ contains $0$; this contradicts $V(J)=\mathrm{Spec}k[t]\setminus \{(t)\}$, because $(t)$ is a prime of $k[t]$ while $\mathrm{Spec}k[t]\setminus \{(t)\}\mathrm{\ne }\mathrm{Spec}k[t]$. Therefore the image is not open, and $i$ is not an open immersion.

Formal unramifiedness: by [F3], ${\mathrm{\Omega }}_{X\mathrm{/}Y}=0$ is equivalent to $i$ being formally unramified; combined with step 2.1 this gives the formal unramifiedness of $i$ without any finiteness hypothesis.

The general closed immersion: let $i\text{\hspace{0.17em}⁣}:X\to Y$ be any closed immersion. By the global argument of steps 1.2 and 2.1 with $Y$ in place of the affine line — ${\mathrm{\Omega }}_{Y\mathrm{/}Y}=0$ by [F2], and the conormal sequence over the base $Y$ by [F1] — one gets ${\mathrm{\Omega }}_{X\mathrm{/}Y}=0$, hence formal unramifiedness of $i$ by [F3].

For local finite type, pass to an affine chart $\mathrm{Spec}A\subseteq Y$: the restriction of a closed immersion to an open subscheme of the target is again a closed immersion by [F8], because the image becomes the intersection with the open set and the surjection of structure sheaves restricts; by [F7] that chart is $\mathrm{Spec}(A\mathrm{/}I)\to \mathrm{Spec}A$ for an ideal $I\subseteq A$, and $A\mathrm{/}I$ is a finitely generated $A$-algebra by [F6], so the affine-local condition of [F5] is satisfied on that chart.

Local finite type: the morphism $i$ is affine, and its coordinate map $k[t]\to k$ is surjective with $k=k[t]\mathrm{/}(t)$, so $k$ is a quotient of the polynomial algebra $k[t]$, hence a finitely generated $k[t]$-algebra by [F6], and the affine-local condition of [F5] is satisfied (the single chart $Y$ itself suffices). By [F4] the map $i$ is therefore unramified, being locally of finite type and formally unramified by step 3.1.

Hence by [F4] every closed immersion is unramified under the locally finite type convention, being locally of finite type by step 4.1 and formally unramified by step 3.2.

For the displayed example this gives ${\mathrm{\Omega }}_{\mathrm{Spec}k\mathrm{/}{\mathbb{A}}_k^1}=0$ by step 2.1, unramifiedness by step 4.2, and non-openness by step 2.2; the example is thus an immersion that is closed but not open and still unramified, and the general statement of steps 3.2, 4.1 and 5.1 covers all closed immersions.