What is an example of a formal scheme that is not algebraizable?

Recall that, if $X$ is a locally noetherian scheme and $Z$ is a closed subset (of the underlying topological space), then one can form the formal completion of $X$ along $Z$ which is sometimes denoted $X_{/Z}$. This is a formal scheme whose underlying topological space is $Z$.

What is a formal scheme that is not of this form?

**Update:** Emerton and Francesco Polizzi suggested several examples that arise in the study of deformations of varieties with trivial canonical bundle. It'd be nice to see some more elementary, explicit examples as well.

**Update 2:** In comments, Francesco Polizzi mentioned that further examples can be found in [Hironaka-Matsumura, "Formal functions and formal embeddings" J. math. soc. Japan 20, Theorem 5.3.3 ] and [Hartshorne, Ample subvarieties of algebraic varieties, p. 205].

**This is too long to fit into comments:**

~~@FP: Thanks! I'm not sure I quite follow the argument for non-algebraizability in the book. Sernesi states that, if $X \to \text{Spec}(\bar{A})$ is an algebrization, then $X$ would admit a non-trivial line bundle "since $X$ is of finite type over an integral scheme." Furthermore, he states that this line bundle can be chosen to "correspond to a Cartier divisor whose support does not contain $X_{s}$ [the special fiber] and has nonempty intersection with $X_{s}$." (note: The notation $X$, $X_s$ is different in the text.)~~

It is not clear to me why such a line bundle exists: $\mathbb{A}^n$ is a finite type scheme over an integral scheme that has no non-trivial line bundles.

~~I understand how this shows that there is no algebraization by a ~~*$\bar{A}$-projective* scheme, but why is there no algebraization by an arbitrary scheme?

I was a little nervous about the argument (Raynaud has an example of a family of Abelian varieties over a nodal curve with non-projective total space), but my concern was needless.

Here is one argument. Let $X_0/\mathbb{C}$ be an algebraic $K3$-surface. We assume algebraizability and derive a contradiction. The statement about existence of non-algebraic deformations is very strong: In fact, there exists a 1st order deformation $f_1 \colon X_1 \to \text{Spec}(\mathbb{C}[t]/(t^2))$ with the property that the restriction of any line bundles $L_1$ on $X_1$ to $X_0$ is numerically trivial. We use this deformation to derive a contradiction.

By definition, there exists a morphism $f_1 \colon \text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ with property that the versal deformation restricts to $X_1$. Now factor this morphism as $\text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[t]]) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ (by lifting the images of $x_1, \dots, x_{20}$ under $f_1^{*}$).

If $X_{t} \to \text{Spec}(\mathbb{C}[[t]])$ is the restriction of the versal deformation, then the generic fiber is an algebraic $K3$-surface, hence admits an ample line bundle. The total space $X_{t}$ of the family is regular, so it is possible to extend this line bundle to a line bundle $L_{t}$ on $X_{t}$. But then the restriction of $L_{t}$ to the special fiber is not numerically trivial (by flatness); however, no such line bundle can even lift to 1st order. Contradiction.