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 Post subject: Re: Best book(s) for Complex Analysis (undergrad) Posted: Tue Jul 31, 2018 11:51 pm

Replies: 1
Views: 146

 I would suggest that you read the following books on Complex Analysis: ⋅  J. Marsden, M. Hoffman - Basic Complex Analysis : Although it is developed in a rather slow manner, the exposition is in my opinion very nice, and the book contains lots of examples as well as exercises that help one... Forum: Algebraic Geometry   Topic: Divisors and Picard Group

 Post subject: Re: Divisors and Picard Group Posted: Wed Mar 07, 2018 10:53 am

Replies: 5
Views: 1010

 I cannot see which is the Picard group of $X=Proj(\mathbb C[x,y,z]/(xy-z^2)]\subset \mathbb{P}^{3}$ I don't know either the answer. I have just proved that $Cl(X)=\mathbb{Z}$. Maybe you could share your computations. You could also take a look at [Hartshorne / II / Ex. 6.3], which is related to you... Forum: Algebraic Geometry   Topic: Divisors and Picard Group

 Post subject: Re: Divisors and Picard Group Posted: Sun Jan 28, 2018 10:49 pm

Replies: 5
Views: 1010

 Hi! Let me mention the following, which you may find helpful. ⋅  Recall the following general facts: On a variety $X$, say, over $\mathbb{C}$, it holds that $\text{CaCl}(X) \cong \text{Pic}(X)$. Moreover, if $X$ is normal, then Cartier divisors on $X$ correspond to (are... Forum: Algebraic Geometry   Topic: Divisors and Picard Group

 Post subject: Re: Divisors and Picard Group Posted: Thu Jan 18, 2018 1:03 am

Replies: 5
Views: 1010

 Hi! How is your question related to [Hartshorne / II / 6.5.2]? Could you please explain exactly at which point of this particular example you are stuck? Forum: Algebraic Geometry   Topic: Locally free sheaves

 Post subject: Re: Locally free sheaves Posted: Tue Dec 19, 2017 12:40 am

Replies: 1
Views: 398

 Hi! You can find an answer to your question in the following reference: [Q. Liu - Algebraic Geometry and Arithmetic Curves - Chapter 6 / Lemma 4.1 & Corollary 4.2] Forum: Algebraic Geometry   Topic: Geometric Genus

 Post subject: Re: Geometric Genus Posted: Sun Nov 12, 2017 1:01 am

Replies: 1
Views: 554

 Let $n = \dim X$. As $X$ is rational, (by definition) $X$ is birationally equivalent to $\mathbb{P}^{n}$, and since the geometric genus $p_{g}(X) (= P_{1}(X) = \dim H^{0} (X, \omega_{X} ) )$ is a birational invariant, we have that $p_{g} (X) = p_{g} (\mathbb{P}^{n})$. But the canonical s... Post subject: Exercise On Cohomology of Complex Spaces Posted: Sun Mar 05, 2017 12:49 am

Replies: 0
Views: 379

 Assuming the following result THEOREM : Let $X$ be a complex space of dimension $n$ and let $\mathcal{S}$ be any sheaf on $X$. Then $\mathrm{H}^{q}(X, \mathcal{S}) = 0 \, , \, q > 2n$ prove the following results ⋅  LEMMA : Let $X$ be a complex space of dimension $n$ such that $\mathr... Forum: Functional Analysis Topic: An exercise on Fréchet Spaces  Post subject: An exercise on Fréchet Spaces Posted: Sat Mar 04, 2017 10:16 pm Replies: 1 Views: 494  Let V,W be Fréchet spaces and let T be a Hausdorff space. Consider the diagram \[ V \overset{f}{\longrightarrow} W \overset{i}{\longrightarrow} T$where $i$ is a continuous, linear, injective map and $f$ is a linear map. Show that $f$ is continuous if and only if $i \circ f$ is continuous. Forum: Algebraic Structures   Topic: Isomorphism

 Post subject: Re: Isomorphism Posted: Thu Feb 23, 2017 2:02 am

Replies: 2
Views: 610

 Consider the composite map $R \overset{\varphi}{\to} R \overset{\pi}{\twoheadrightarrow} R/J$ Note that ⋅  $\text{Ker}( \pi \circ \varphi ) = I$, as $\text{Ker}( \pi ) = J$ and $\varphi(I) = J$ ⋅  $\text{Im}( \pi \circ \varphi ) = R/J$, as $\pi$ is surjective and $... Forum: Algebraic Geometry Topic: Numerically Proportional  Post subject: Numerically Proportional Posted: Mon Feb 20, 2017 10:42 pm Replies: 0 Views: 489  Let$X$be a smooth projective surface over$ \mathbb{C} $and let$H$be an ample divisor on$X$. Let$L$and$M$be two$\mathbb{Q}$-divisors on$X$which are not numerically trivial, and such that $L^2 = L \cdot M = M^2 = 0$Show that$L$and$M\$ are numerically proportional, i.e. \[ \exists \... Sort by:
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