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Of course, we all know great mathematicians who constantly make mistakes even now, and not because of foundations. In any case, it's not like "long dead Italian algebraic geometers" is a category of …

Saying that the étale topology is equivalent to the euclidean topology is vastly overstating the case. For example, if you compute the cohomology of a complex algebraic variety with coefficients in $\ …

[Edit:] The answer should be positive, that is, every profinite group appears as the fundamental group of a scheme. Here is a sketch of proof. First of all, I claim that for any finite group $G$ ther …

There is somewhere a theorem in Hartshorne's book saying that an ample divisor on a normal projective connected scheme of dimension at least 2 is connected. Now proceed by induction on $n$. If there i …

This is true if we assume that the vector bundles has constant rank (it is clearly false if we allow vector bundles to have different ranks at different points). Let $U_1$ be an open dense subset of $ …

The treatment in EGA is indeed intimidating, but in fact over a field the formula is not hard to prove. You only need $X$ and $Y$ to be separated schemes over $k$, and $\mathcal{F}$ and $\mathcal{G}$ …

In order to do geometry, you need to have some kind of global structure which has good local models (the "neighborhoods") and good gluing conditions. In algebraic geometry, the good local models are r …

The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory …

Suppose that $X$ is a smooth complete positive-dimensional algebraic space over $\mathbb C$. Then $\mathrm H^2(X, \mathbb Q)$ can not be 0. In fact, every algebraic space contains an open dense subsch …

If $L/K$ is a Galois extension, then one can define a site in which the objects are intermediate fields $K \subseteq E \subseteq L$ which are finite over $K$. Or, you can take finite étale algebras $A …

Any curve of large enough degree will do. Set $F:= E'\otimes E^{\vee}$; if $d$ is a very large integer, then $\mathrm H^1(F(-d)) = 0$. Take any curve $C$ of degree $d$, and suppose that $E\mid_C$ and …

When you are truly fluent in scheme theory, you don't know whether you are "thinking schemes" or "thinking varieties", the intuitions are merged together. As to learning, for most people starting wit …

This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties" J. Fac. Sci. Univ. Tokyo Sect. IA Math. 24 (1977), no. 1, 11-21, http:/ …

Artin's axioms do not apply in this case, because the stack is not limit-preserving. They only work with stacks that are locally finitely presented. In any case, it is easy to give examples of quasi- …

It it true for any $Y$: see Corollaire 21.12.7 of EGAIV.