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One of the possible explanations why considering schemes is important is the following. In some sense, schemes are like varieties, but "with nilpotents". What I mean is the following. Suppose you are intersecting two plain curves $y=0$ and $y=x$, and also intersect $y=0$ with $y=x^2$. Set-theoretically, in both cases you get the origin $(0,0)$. But really ...

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The article "What is ... a shtuka?" by David Goss might be helpful. My memory is that the Mumford article cited by Goss is also quite helpful, and (among other things) sheds light on why the person who invented Shtuka's (Drinfeld) also contributed fundamentally to the theory of integrable systems. For the relationship with Langlands, it helps first to ...

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A) If you take modern in a comprehensive sense, algebraic geometry has certainly thrown a new light on such classical topics as: 1) Pascal's theorem on the mystic hexagram (Fulton page 62). 2) Poncelet's porism on the closing of polygons inscribed in a conic: here is a sophisticated modern proof. 3) Steiner's problem ( the rigorous determination of the ...

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Since you tagged this question with number theory, it seems like you might be interested in the answer to this question over $\mathbf Q$ or other `arithmetic' fields. Let me just make a few comments about the problem over $\mathbf C$, to show that it is already extremely subtle there. In other words, the short answer to your question "is there an easy way?" ...

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Locally principal ideals turn into locally principal ideal sheaves, which are same thing as invertible sheaves equipped with an embedding $\mathcal L \hookrightarrow \mathcal O_X.$ Locally principal fractional ideals turn into locally principal fractional ideal sheaves, which are the same things as invertible sheaves equipped with an embedding $\mathcal L ... 3 Your$X$is the image of the morphism$\mathbb A^1\to\mathbb A^2$sending$t\mapsto (t,t^2)$. This corresponds to the$\mathbb C$-algebra map$\phi:\mathbb C[x,y]\to \mathbb C[t]$sending$x\mapsto t,\,y\mapsto t^2$. We have: $$X=V(\ker\phi)=V(y-x^2).$$ This algebraic set is irreducible, since$(y-x^2)\subset \mathbb C[x,y]$is a prime ideal. 2 I think that perhaps you're asking a slightly wrong question. If you don't have any motivation to learn schemes (e.g. from Vakil's notes), then probably that's not what you should be studying right now. Instead, try to learn some classical algebraic geometry, such as the basics of the theory of curves. Miles Reid's Undergraduate algebraic geoemtry is a ... 2 I believe it's the "Introduction to Algebraic Curves" of William Fulton. He let you download the book from his site. Here is the link: http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf 2 We need oracles that (a) test for any point${\bf p}\in{\mathbb R}^2$and any ellipse$E$whether${\bf p}\in {\rm int}(E)$, (b) compute for any ellipse$E$its center${\bf m}_E$, (c) compute for any two ellipses$E_1$and$E_2$the set$\partial E_1\cap\partial E_2$, consisting of$0$,$2$, or$4$points. (Degenerate cases will not be ... 2 I think it's only a matter of indexing. For example, according to the notation in Weibel's book we would have$\cdots\to \mathscr R_1\to \mathscr R_0\to \mathscr G\to 0$in the projective resolution, rather than defining$\mathscr R_0 = \mathscr G$. That is, in a projective (or free) resolution, the final map$\mathscr R_1\to\mathscr R_0$is typically ... 1 Let$p\in C_n$be a point with$y=0,$i.e.$p=(k,0)$for some$k=1,\ldots,2n$. In a small neighborhood$U$around$p$, all the terms$(x-1),\ldots,(x-2n)$, except for$(x-k)$don't change dramatically, so inside$UC_n$actually looks pretty much like$C_1$. 1 Here is a proof of the equivalence: Suppose that$Y$is normal, and let$f: X \to Y$be finite. Let Spec$A$be an open affine subset of$Y$. Since$f$is finite, we have that$f^{-1}($Spec$A)$is an open affine subset Spec$B$of$X$, with$A \to B$finite. Since$f$is birational,$A$and$B$have the same fraction field, say$K$. Thus$B$is a ... 1 Inspired by @ Semiclassical comment, I got one answer: suppose three ellipse are$a$,$b$and$c$; so there are three combinations if we draw two from the three:$(a,b),(a,c),(b,c)$test the number of interaction points between$(a,b), (a,c),\text{ or } (b,c)$, if the number is$0$, there is no common point between the three; if the number is$1$, then if ... 1 The following proof works if one assumes that the fiber$F$is proper. I am suspicious it might not be true if the fiber is not proper. Let$X = Z_1 \cup Z_2$where each$Z_i$is closed. Then since each fiber$\iota_y : F \hookrightarrow X$is irreducible, we have that either$\iota_y^{-1}(Z_i)$is of the same dimension as$F$and equal to all of$F$, or ... 1 Recall that any localization of a reduced ring is reduced. Since the Proj of a graded ring$R$is covered by open affines which are Spec's of degree zero parts of various localiations of$R$, if$R$is reduced then so is its Proj. In the case of blowing up Spec$A$along an ideal$I$, we have$R = A \oplus I \oplus I^2 \oplus \cdots,$and you can easily ... 1 Answer. Yes, there is an example. Let$k$be a field, let$R = k [a, b] / (a^2, a b, b^2)$,$A = R [x] / (a - b x, x^2)$,$B = R [y] / (b - a y, y^2); then $$A \otimes_R B \cong R [x, y] / (a - b x, x^2, b - a y, y^2) \cong k [a, b, x, y] / (a^2, a b, b^2, a - b x, x^2, b - a y, y^2)$$ but \begin{align} a & = (a - b x) + (b - a y) x + (a - b x) x y + ... 1 Yes. (And you don't needX$or$T$to be connected, just$U$.) To see this, pull-back$X$over$U$, and so reduce to the case$U = T$. So the question becomes: if$T$is connected and$X \to T$is finite etale of degree$d$, are there at most$d$sections$T \to X.$Suppose that$f$and$g$are two such sections: then these induce a morphism$f\times g: ...

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The last generator is superfluous, as $$(x^2yt^2 - xz^4) - y(x^2t^2 - yz^3) + z^3(xz - y^2) = 0$$ Thus $I = (xz-y^2, x^2t^2-yz^3)$ is generated by a regular sequence (indeed, $xz-y^2, x^2t^2 - yz^3$ are both prime elements).

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One can give a proof in terms of the "chord--tangent law" (i.e. the explicit description of the group operation via intersecting lines etc.) that avoids computation. Basically, one uses Bezout's theorem, together with a limiting argument in the Zariski topology to cover various degenerate configurations. I think this is explained in Miles Reid's ...

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The condition is that the polynomials generate an ideal of height $k$. This is much harder to check for $k>2$ than $k=2$, because height $1$ primes are principal in $\mathbb{C}[X,Y]$, so we just need to check that our polynomials have no common factors.

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As something of an aside: to get $\mathbb P^1\times \mathbb P^1$ from $\mathbb P^2$, you blow up the plane at two points, and then blow down the (proper transform of) the line joining the two blown-up points. Also, the section of Hartshorne Ch. V discussing ruled surfaces describes quite carefully how the invariants of the line bundle $\mathcal L$ (on ...

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