I'd prefer examples that are unusual or nonstandard [...]

Ok then I finally have a chance to present some of my interests ;-).

Many people claim that schemes are only hausdorff in trivial cases. This is wrong. Namely, $Spec(A)$ is hausdorff if and only if $A$ is $0$-dimensional. More generally, a scheme $X$ is hausdorff if and only if it is $0$-dimensional and separated (in the scheme sense), and here you can replace $0$-dimensional by $T_1$ and separated by quasi-separated, if you wish. This shows that here the analogies of these notions in scheme theory and general topology become actually equivalences. Every compact hausdorff scheme is already affine (nice exercise, using just general topology and $Spec(R \times S) = Spec(R) \coprod Spec(S)$). Every locally compact totally disconnected hausdorff space can be made into a scheme; take the constant sheaf $\underline{\mathbb{Z}/2}$. Hausdorff schemes or more general sheaves on Stone spaces also play a role in the classification of certain algebraic rings. For example, every countable ring in which the elements satisfy $a^4=a$ is isomorpic to $\{f \in C(X,\mathbb{F}_4) : f(E) \subseteq \mathbb{F}_2\}$ for some closed subset $E$ of a Stone space $X$.

Now here is another example: In topology, the linearity of vector bundle maps is declared fiberwise (instead of working with vector space objects and declaring linearity in a functorial way). This is wrong in algebraic geometry: A $S$-endomorphism of the affine $n$-space $\mathbb{A}^n_S$, which is linear on every fiber $\mathbb{A}^n_{\kappa(s)}$, does not have to be linear.

It surprised me when I heard that infinite product is not an exact functor on the category of sheaves: If $A_i \to B_i$ are surjective, then $\prod_i A_i \to \prod_i B_i$ does not have to be surjective. The reason should be that the canonical map $(\prod_i A_i)_x \to \prod_i ({A_i}_x)$ is not an isomorphism. But to know, I don't know of any explicit example for the $A_i \to B_i$. Hints?

I'd prefer examples that are unusual or nonstandard [...]

Ok then I finally have a chance to present some of my interests ;-).

Many people claim that schemes are only hausdorff in trivial cases. This is wrong. Namely, $Spec(A)$ is hausdorff if and only if $A$ is $0$-dimensional. More generally, a scheme $X$ is hausdorff if and only if it is $0$-dimensional and separated (in the scheme sense), and here you can replace $0$-dimensional by $T_1$ and separated by quasi-separated, if you wish. This shows that here the analogies of these notions in scheme theory and general topology become actually equivalences. Every compact hausdorff scheme is already affine (nice exercise, using just general topology and $Spec(R \times S) = Spec(R) \coprod Spec(S)$). Every locally compact totally disconnected hausdorff space can be made into a scheme; take the constant sheaf $\underline{\mathbb{Z}/2}$. Hausdorff schemes or more general sheaves on Stone spaces also play a role in the classification of certain algebraic rings. For example, every countable ring in which the elements satisfy $a^4=a$ is isomorpic to $\{f \in C(X,\mathbb{F}_4) : C(Ef(E) \subseteq \mathbb{F}_2\}$ for some closed subset $E$ of a Stone space $X$:X$.

Now here is another example: In topology, the linearity of vector bundle maps is declared fiberwise (instead of working with vector space objects and declaring linearity in a functorial way). This is wrong in algebraic geometry: A $S$-endomorphism of the affine $n$-space $\mathbb{A}^n_S$, which is linear on every fiber $\mathbb{A}^n_{\kappa(s)}$, does not have to be linear.

It surprised me when I heard that infinite product is not an exact functor on the category of sheaves: If $A_i \to B_i$ are surjective, then $\prod_i A_i \to \prod_i B_i$ does not have to be surjective. The reason should be that the canonical map $(\prod_i A_i)_x \to \prod_i ({A_i}_x)$ is not an isomorphism. But to know, I don't know of any explicit example for the $A_i \to B_i$.

I'd prefer examples that are unusual or nonstandard [...]

Ok then I finally have a chance to present some of my interests ;-).

Many people claim that schemes are only hausdorff in trivial cases. This is wrong. Namely, $Spec(A)$ is hausdorff if and only if $A$ is $0$-dimensional. More generally, a scheme $X$ is hausdorff if and only if it is $0$-dimensional and separated (in the scheme sense), and here you can replace $0$-dimensional by $T_1$ and separated by quasi-separated, if you wish. This shows that here the analogies of these notions in scheme theory and general topology become actually equivalences. Every compact hausdorff scheme is already affine (nice exercise, using just general topology and $Spec(R \times S) = Spec(R) \coprod Spec(S)$). Every locally compact totally disconnected hausdorff space can be made into a scheme; take the constant sheaf $\underline{\mathbb{Z}/2}$. Hausdorff schemes or more general sheaves on Stone spaces also play a role in the classification of certain algebraic rings. For example, every countable ring in which the elements satisfy $a^4=a$ is isomorpic to $\{f \in C(X,\mathbb{F}_4) : C(E) \subseteq \mathbb{F}_2\}$ for some closed subset $E$ of a Stone space $X$:

Now here is another example: In topology, the linearity of vector bundle maps is declared fiberwise (instead of working with vector space objects and declaring linearity in a functorial way). This is wrong in algebraic geometry: A $S$-endomorphism of the affine $n$-space $\mathbb{A}^n_S$, which is linear on every fiber $\mathbb{A}^n_{\kappa(s)}$, does not have to be linear.

It surprised me when I heard that infinite product is not an exact functor on the category of sheaves: If $A_i \to B_i$ are surjective, then $\prod_i A_i \to \prod_i B_i$ does not have to be surjective. The reason should be that the canonical map $(\prod_i A_i)_x \to \prod_i ({A_i}_x)$ is not an isomorphism. But to know, I don't know of any explicit example for the $A_i \to B_i$.