Here are some examples illustrating the genuine necessity of noetherian assumptions:
1) Every scheme with just one point is the spectrum of a local artinian ring?
This is true for every noetherian one point scheme and false for every non-noetherian one point scheme.
A non-noetherian (and thus non-Artinian) example : $\operatorname {Spec}(\mathbb Q[T_1,T_2,T_3,\cdots]/\langle T_1^2,T_2^2,T_3^2 \rangle )$
- A scheme has only finitely many irreducible components ?
A noetherian scheme has only finitely many irreducible components, but a non-noetherian scheme may have infinitely many irreducible components: this is the case for any disjoint union of infinitely many non-empty schemes.
3) Every scheme has a closed point?
This is true for every noetherian scheme (actually for any quasi-compact scheme), but there exist schemes without any closed point: Qing Liu, Chapter 3, Exercise 3.27, page 114.
4) Injective modules give injective sheaves?
If $I$ is an injective module over the ring $A$, then the associated quasi-coherent sheaf $\tilde I$ on $X=\operatorname {Spec}(A)$ is an injective sheaf of $\mathcal O_X$- Modules if $A$ is noetherian but is not necessarily injective for $A$ non-noetherian: SGA6, Exposé II, Appendice I Un contre-exemple de Verdier, page 195.
5) A finitely presented sheaf is coherent?
Given on a scheme $X$ a sheaf of $\mathcal F$ of $\mathcal O_X$-Modules, does the existence of an open covering $(U_i)$ of $X$ for which one has exact sequences $\mathcal O_{U_i}^{n_i}\to \mathcal O_{U_i}^{m_i}\to \mathcal F\vert _{U_i}\to 0$ imply that $\mathcal F$ is coherent?
The answer is yes if $X$ is noetherian (or even locally noetherian) but no in general: there exist non-noetherian rings $A$ such that the structural sheaf $\mathcal O_X$ on $X=\operatorname {Spec}(A)$ is not coherent!
6) A scheme is affine if all its quasi-coherent sheaves are acyclic ?
Serre's criterion is that a noetherian scheme $X$ is affine if and only if $H^p(X, \mathcal F)=0$ for all quasi-coherent sheaves $\mathcal F$ on $X$ and all $p\gt 0$.
This no longer holds if $X$ is not assumed noetherian: given a field $k$, any infinite disjoint sum $X=\coprod \operatorname {Spec}k$ satisfies the cohomology condition but is not affine since it is not quasi-compact.