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6

(Let me answer the case of coefficients in $\mathbf Z$ (or $\mathbf Q$ or $\mathbf R$ or $\mathbf C$.) I think it should only take a bit of universal coefficient wizardry to deduce the answer in general, but it is too late here for that.) First, by the Lefschetz hyperplane theorem and Poincaré duality, we have an isomorphism $$ H_i(\mathbf P^n, \mathbf Z) ...


5

By the Lefschetz Hyperplane theorem, the inclusion $i : X \hookrightarrow \mathbb{CP}^n$ induces a map $i^* : H^q(\mathbb{CP}^n, \mathbb{Z}) \to H^q(X, \mathbb{Z})$ which is an isomorphism for $q \leq n - 2$ and injective for $q = n - 1$. Recall that $$b_q(\mathbb{CP}^n) = \dim H^q(\mathbb{CP}^n, \mathbb{Z}) = \begin{cases} 1 & q\ \text{even}, 0 \leq q ...


4

There is no uniform answer to your question: the numbers $l(3P)$ and $l(4P)$ you are asking about depend on the curve $C$ and on the point $P$. a) We always have $l(2P)=1$ Indeed, $1\leq l(2P)\leq 2$ for a curve of positive genus while for genus $g\geq 2$ the existence of a point $P$ with $l(2P)=2$ characterizes hyperelliptic curves. However a ...


3

(i) If $a\neq0$ the prime ideals you want are the maximal ideals $(x-a,y-b)\; (b\in \mathbb C)$ and $(x,y-b)\; (b\in \mathbb C)$ about which you already know PLUS the two prime but not maximal ideals $(x)$ and $(x-a)$. (ii) If $a=0$ the prime ideals you want are the maximal ideals $(x,y-b)\; (b\in \mathbb C)$ PLUS the prime ideal $(x)$ Thinking ...


3

I think the following works. In the noetherian case, you can just take $V_i = U_i$ below. Let $S$ be qcqs, let $X \to S$ be separated and of finite type, and let $X = X_1 \cup \cdots \cup X_r$ be a decomposition of $X$ into finitely many irreducible components. Let $U_i = X \setminus \bigcup_{i \ne j} X_j$; note this is a non-empty open subset of $X$ ...


3

This does follow from the Hodge decomposition, which lets you view the first de Rham cohomology as the set of harmonic 1-forms. The map you are discussing can be viewed as $\Omega^{1,0} \oplus \overline{\Omega^{1,0}} \rightarrow \mathscr{H}^{(1)}$. This is then a canonical map, and you can see it is an isomorphism pretty easily.


3

In response to user264059's comment, the decomposition you presented in your original post is generally called the Hodge decomposition, so it is not really clear what you are looking for here. (A more direct proof of the Hodge decomposition than what?) That said, there are several approaches. The theory of elliptic PDE's gives a proof of the Hodge ...


2

We’re looking in the local, complete situation above $\ell$ at the $p^m$-torsion points of $E$ for all $m$. What does it mean to say that $T_p(E)^{G_v}\ne0$, where $G_v=G_{K_v}$, the Galois group of an algebraic closure of $K_v$ over $K_v$? It would mean that there was a consistent sequence of $p^m$-torsion points of $E$, in particularly infinitely many of ...


2

Yes, the canonical morphism $p: C^n\to C^{(n)}$ is flat by miracle flatness. More precisely that aptly named theorem states that a morphism $p:X\to Y$ of varieties over a field is flat whenever $X$ and $Y$ are regular and all the fibers of $p$ have dimension $\dim X-\dim Y$ ($=0$ in our case). For a proof of miracle flatness see Matsumura, 23.1, page 179, ...


2

We show that$$\dim_\mathbb{C} \mathfrak{m}^k/\mathfrak{m}^{k+1} = k+1$$if $k < m$. Let $p = [a, b, c]$ (we can assume $c \neq 0$). Then we can identify $\mathcal{O}_p$ with $\mathbb{C}[x, y]/(f)$, we can identify $p$ with $(a/c, b/c) \in \mathbb{C}^2$, $f(a/c, b/c) = 0$. In this case, we have$$\mathfrak{m} = \left\langle x - {a\over{c}}, y - ...


2

I'm not sure this answers your first question, but here's one way to prove $k[x,y,z]/(y - x^2, z - x^3) \cong k[x]$. The intuition is that $$ k[x,y,z]/(y - x^2, z - x^3) \cong k[x,x^2,x^3] = k[x] $$ but I'm guessing you want something more rigorous. First, let's consider the following lemma. Lemma. Let $R$ be a unital commutative ring and $R[x]$ be the ...


2

There is a unique $k$-linear ring homomorphism $\phi:k[x,y,z]\to k[t]$ such that $\phi(x)=t$, $\phi(y)=t^2$ and $\phi(z)=t^3$, and the ideal $I=(y-x^2,z-y^3)$ is conttained in its kernel, as its generators are. If $f=\sum_{i=0}^na_ix^i$ is an elemntt of of $k[x]$ which is in $I$, then $\phi(f)$ is zero. But $\phi(f)$ is just $\sum_{i=0}^na_it^i$, whose ...


1

We may restrict to appropriate affine coordinate patches to check zeros and poles. Indeed, away from $z = 0$, taking the affinization $z=1$, the curve is defined by $$y^2 = x(x-1)(x-\lambda).$$Then it is clear $y/z$ indeed has zeros of degree $1$ at$$p = (0, 0),\text{ }q = (1, 0), \text{ }r = (\lambda, 0),$$and that these are the only zeros of the function ...


1

To prove the indicated statement from scratch, we first sketch a proof of another result that's called Dickson's lemma: For any infinite sequence of distinct $\alpha_0, \alpha_1\in \mathbb{N}^n$, there exist $i < j$ such that $\alpha_i \leq \alpha_j$ (that is, each component $\alpha_i^k \leq \alpha_j^k$.) Assume the result holds for $(n-1)$-tuples. ...


1

The ideals $J_0 \subset J_1 \subset J_2 \subset \cdots$, and $k[x_1, \dots, x_n]$ is Noetherian.


1

Let's observe that if $a \neq 0$ the ideals $(x)$ and $(x-a)$ are coprime, by CRT you obtain: $$\frac{\mathbb{C}[x,y]}{x(x-a)} \simeq \frac{\mathbb{C}[x,y]}{(x)} \oplus \frac{\mathbb{C}[x,y]}{(x-a)} \simeq \mathbb{C}[y]\oplus \mathbb{C}[y]$$ Now is easy to look at the prime ideals of $\mathbb{C}[x]\oplus \mathbb{C}[x]$ and I think you can easely ...



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