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10

One way to see it is to note that $\mathbb{P}^1\times\mathbb{P}^1$ maps onto $\mathbb{P}^1$ (by projection to its first factor) while $\mathbb{P}^2$ does not. In fact, any map $\mathbb{P}^2\to\mathbb{P}^1$ is constant. [Edit: Based on Georges's comment, I think I should explain why all maps $\mathbb{P}^2\to\mathbb{P}^1$ are constant. If ...


7

In French the meaning "bottle" is definitely never used, in particular because it is syntactically impossible! The meaning in French of flasque is exactly flabby and the terminology is very appropriate: any section on an open subset of a flabby sheaf can be extended to the whole space. A tougher sheaf would never tolerate that: just try with the sheaf ...


4

Yes, your interpretation is correct, but you better be careful about the size of your categories. Say you work with a universe $\mathbb U$, and you call small the elements of $\mathbb U$. Then, on one hand, the category of small topological spaces is a locally small category (meaning that the hom-sets are small). On the other hand, $\mathbf{Sheaves}$ is ...


4

The surface $\mathbb{P}^1\times\mathbb{P}^1$ has disjoint algebraic curves lying on it, like $\{a\}\times \mathbb{P}^1$ and $\{b\}\times \mathbb{P}^1$ (where $a\neq b\in \mathbb P^1$). The surface $\mathbb{P}^2$ does not have any disjoint algebraic curves lying on it: this is a very weak form of Bézout's theorem.


4

Over $\mathbb C$ the underlying topological spaces of the two varieties are not even homeomorphic. Indeed, we have $H_2(\mathbb P^2,\mathbb Z)\cong \mathbb Z$ whereas $H_2(\mathbb P^1\times \mathbb P^1,\mathbb Z)\cong \mathbb Z^2$. The first result is completely standard (Greenberg-Harper, Theorem 19.21) and the second results from Künneth (same book, ...


4

The modern definition is a distillation of hundreds of years of calculations of intersections of curves. A good intuitive description is that the intersection number $I_p(f,g)$ at $p$ of two curves $Z(f), Z(g)$ lying on $\mathbb A^2$ is the number of actual points of intersection near $p$ of the two curves obtained by slightly deforming the equations of the ...


3

Hartshorne often assumes schemes to be noetherian so that his definition of "coherence" is correct. The general and correct definition of coherence is a bit complicated; what Hartshorne defines as coherence is actually called "locally of finite type", which agrees with "locally of finite presentation" over noetherian schemes. It turns out that, over ...


3

You are looking for maps from $U_{\alpha\beta} = U_\alpha \cap U_\beta \to \mathcal{O}^*$ that naturally arise from the hypersurface. All you have given are the functions $s_\alpha : U_\alpha \to \mathbb{C}$ which vanish along $V$. What happens if you look at the quotients $s_\alpha/s_\beta$?


3

It might help to remember that in these cases, $\mathrm{Spec}(R)$ is the affine cone over $\mathrm{Proj}(R)$. So $\mathrm{Spec}(R\otimes S)$ is the product of the affine cones over $\mathrm{Proj}(R)$ and $\mathrm{Proj}(S)$, which is again a cone. Then we take $\mathrm{Proj}$ of it. In terms of the ambient affine and projective spaces, this is like going ...


3

You can get the answer by using the adjunction formula. First assume that $C$ is smooth. Then its normal bundle $N_C$ is the restriction of $\mathscr O (d_1) \oplus \mathscr O(d_2)$ to $C$. Then the adjunction formula (Hartshorne II.8.20) says that the canonical bundle of $C$ is $$K_C = (K_{\mathbf P^3})_{|C} \otimes \bigwedge^2 N_C = (\mathscr O(-4) ...


3

Smooth morphisms are locally of finite presentation by definition. Since $k[x_1,x_2,\dotsc]$ is not finitely generated (hence, not finitely presented) as a $k$-algebra, the morphism is not smooth. But it is formally smooth: If $B$ is a commutative $k$-algebra and $I \subseteq B$ is a nilpotent ideal (in fact, every ideal works here), then every $k$-algebra ...


2

The cohomology of the intersection of two surfaces can also be computed by writing resolutions. First of all, we have the ideal sheaf sequence $$ 0 \to \mathcal I \to \mathcal O_\mathbb P \to \mathcal O_C \to 0.$$ We also have the exact sequence $$ 0 \to \mathcal O(-d-e) \to \mathcal O(-d) \oplus \mathcal O(-e) \to \mathcal O \to \mathcal I \to 0. $$ Then ...


2

No, this is not true. The two classes $c_1(L)$ and $c_1(M)$ live in the abelian group $\operatorname{Pic}(M)$, but for many $M$ this group has rank larger than 1, and there are linearly independent ample classes in the group. For an example, let $M$ be the blowup of $\mathbf P^2$ in a point. Then $\operatorname{Pic}(M)$ is isomorphic to $\mathbf Z^2$, ...


2

Suppose $X$ is integral and noetherian. Let $x$ be a point in the support of $D$. Then $D$ is represented by a rational function $f$ at $x$. If the support has codimension $\ge 2$ at $x$, then, after shrinking $X$ around $x$ if necessary, $f$ is regular and invertible outside of a closed subscheme of codimension $\ge 2$. (2)-(3) When $X$ is normal, this ...


2

On any projective variety there are no nonconstant functions, so $\Gamma(X)$ just consists of constants and is therefore isomorphic to $k$.


2

The Picard groups of both varieties are not isomorphic (this is more advanced) : We have $\operatorname {Pic} (\mathbb P^2)\cong \mathbb Z$ whereas $\operatorname {Pic} (\mathbb P^1\times \mathbb P^1)\cong \mathbb Z^2$ (see here for a very general context)


2

For the canonical bundle $K$ on these varieties we have the self-intersection numbers $$K^2_{\mathbb P^1\times \mathbb P^1}=8$$ and $$K^2_{\mathbb P^2}=9$$


2

Recall that in set theory one constructs the cartesian product $\prod_p X_p$ of sets as a set of functions in $p$ with values in $\bigcup_p F_p$ such that each $p$ has some value in $F_p$. In set theory, every object is a set and we can take unions of arbitrary sets (even if this does not make sense at all from a non-formal mathematical perspective... what ...


2

Proposition: Let $R$ be a local ring, and $M, N$ $R$-modules with $M \otimes_R N \cong R$. Then $M, N \cong R$. Proof: By the answers to this question, $M, N$ are finitely generated and projective. Since $R$ is local, $M, N$ are in fact free, say $M \cong R^m, N \cong R^n$. Then $R \cong M \otimes N \cong R^m \otimes R^n \cong R^{mn}$, so $mn = 1 \implies m ...


2

Take $M=\{ax+by=0\} \subset\mathbb{A}_k^2$. The pre-image under the projection of $V(y-xz)$ to the affine plane is contained in the union of two lines $L:\{x=0,y=0\}\cup \{a+bz=0,y-xz=0\}$ (convince yourself that these are lines). On the complement, you have $x\neq 0$ and thus $z=y/x$ makes sense. So the complement looks like the affine plane minus the line ...


2

You're right that the range of the map you want to write down changes with $(p)$. In fact this happens even in the case of $\mathbb Z$: Ellenberg doesn't have space to write down many details, but the map he describes is $\mathbb Z\to \mathbb Z/(p), n\mapsto n\pmod p$. Why this map, rather than the one to $\mathbb C$ you suggested? Simply because the latter ...


2

Hint: Can you construct a ''fat'' projective line of some sort? Namely, remember that $\text{Spec } k[x]$ is the affine line and we get projective space by gluing together two affine lines. Can you do something similar by fattening up the lines you glue? My example was the following. Take $\text{Proj } k[x,y,z]/z^2$. I believe that working on affines, ...


2

In Lazarsfeld, Positivity in Algebraic Geometry I, Section 2.1, one can find a proof the following: Theorem 2.1.33: For any line bundle $L$ on a normal variety $X$ such that $\kappa(X,L)>0$ there is an associated rational map $\phi_L : X \dashrightarrow Y$ such that $\operatorname{dim} Y=\kappa(X,L)$ and (after passing to a resolution of $\phi$) the ...


2

I don't know what your definition is for "quasiaffine subset", but consider an example like $\{(x,0)\mid x\neq 0\} \subset \mathbb{A}^2$. The former can be identified with $\mathbb{A}^1 \setminus \{0\}$, which is quasi-affine (in fact, it's even affine). But it's not open or closed in $\mathbb{A}^2$. Definitions really matter here, but hopefully this ...


2

If you restrict to schemes over $k$ for some fixed field $k$, and consider skyscraper sheaves of $k$-modules (rather than just abelian groups), then I think that the following slight variation on Georges' answer gives a way of viewing skyscraper sheaves as quasi-coherent sheaves. Let $(X,\mathcal O _X)$ be a locally ringed space over $k$ and $x \in X$. ...


1

Yes, this is correct. Notice that your hom sheaf is the internal hom in the category of étale sheaves on $S$ resp. the subcategory of finite étale schemes over $S$. Recall that the category of finite $\pi_1(S,s)$-sets is equivalent to the category of finite étale schemes over $S$. Using the equivalence, the problem becomes simply: If $X,Y$ are finite ...


1

Since $F[X]$ is a PID, and thus the ideal generated by $S$ is thus generated by a single polynomial $f$, you have that $V(S)=V(f)$, and so $U$ is principal open.


1

I'm assuming you believe that $\hom_{\mathrm D(k)}(\mathrm R f_\ast \mathcal F[i],k)=\hom_{\mathrm D(k)}(k,\mathrm R f_\ast \mathcal F[i])^\vee$. Now let $f:\mathcal A\to \mathcal B$ be a left-exact functor on abelian categories. It is well-known (see Corollary 10.5.7 of Weibel, for example) that there are natural isomorphisms $\mathrm H^i(\mathrm R ...


1

I think the plane with doubled origin is enough. From the construction you see two copies of $\mathbb{A}^2$ which intersect in a copy of $\mathbb{A}^2 - 0$.


1

Note that $z=x+i y$ is symmetric to itself if and only if $F(x, y)=0$. That is, $z$ is symmetric to itself if and only if $z$ lies on the curve defined by $F$. Also since $F$ has real coefficients, the pair $(z,z_0)$ is symmetric if and only if $(z_0,z)$ is symmetric. Finally the map $z\mapsto z_0$ defined locally around a smooth point on $F$ which fixes ...



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