# Tag Info

9

The empty scheme is initial in the category of schemes, and the zero ring is not a local ring, since it does not have a unique maximal ideal (it does not have any maximal ideal!). There is no special convention needed here--this all just follows from the general definitions. In particular, there is no issue with what the unique map out of the empty scheme ...

8

The empty scheme is the initial object. The zero ring is not a local ring. Conventionally local ring homomorphisms are between local rings, but we may extend the definition to general rings by defining “local” to mean that an element becomes invertible in the codomain if and only if it is invertible in the domain. Under this definition a ring homomorphism ...

6

The category of schemes has an initial object, which is affine, given by the spectrum of the zero ring, which empty. The zero ring is not a local ring; a local ring has to have a unique maximal ideal, and the zero ring doesn’t have any (it’s the only ring with this property), because it is not a field. If you take out the empty scheme then the resulting ...

5

Claim: For any flat and proper morphism of schemes $X\to S$ with $S$ connected and having geometrically connected, geometrically reduced fibers, we have that $\Gamma(X,\mathcal{O}_X)\cong\Gamma(S,\mathcal{O}_S)$. Proof: By properness and flatness, we have that $X\to S$ is both closed and open, so because $S$ is connected $X\to S$ is surjective. By Stein ...

5

I think definition group scheme is not “a scheme where underlying set has a group structure”, but “a group object in a category of schemes”. Here’s similar confusion that might happen: in a category of schemes, underlying set of the product $X_1, X_2$ of two schemes is not the same as product of underlying sets. (Even $X_1=X_2=\operatorname{Spec}\mathbb{Z}$ ...

4

First, we need to recognise the $A$-algebras that arise as the affine open subschemes of $\operatorname{Spec} A$. Here is a criterion: Proposition 1. An $A$-algebra $B$ corresponds to an open subscheme of $\operatorname{Spec} A$ if and only if there is a (finite) set $S \subseteq A$ with the following properties: For every $a \in S$, the induced ...

4

If you drop the completeness requirement, you have the inclusion: $$\mathbb{R}\hookrightarrow\frac{\mathbb{R}[t_\lambda|\lambda\in \mathbb{R}^{>0}]}{\langle t_\lambda t_\mu- t_{\lambda+\mu}|\lambda,\mu\in \mathbb{R}^{>0},t_\lambda |\lambda\geq1 \rangle}$$ Note the RHS is a quotient of a polynomial ring (not formal power series). The maximal ideal $m$ ...

4

The map $f$ is not open. We can describe $X$ as $$X = \left\{ ((x,y),[u,v]) \mid xu=yv \right\} \subset \mathbf A^2 \times \mathbf P^1.$$ The map $f$ is simply the projection $((x,y),[u,v]) \mapsto (x,y)$. Now consider the open set $U = \{ v \neq 0 \} \subset X$. I claim that $f(U)$ is not an open set of $\mathbf A^2$. If $(x,y)$ is a point of $\mathbf A^2$...

4

The equality $\operatorname{Spec}(\prod A_i) = \bigsqcup\operatorname{Spec}(A_i)$ is always false in the category of schemes if the set of indices $i$ is infinite and all the $A_i$ are $\neq 0$. Indeed $\operatorname{Spec}(\prod A_i)$ is quasi-compact (like any $\operatorname{Spec})$, whereas $\bigsqcup\operatorname{Spec}(A_i)$ is never quasi-compact. NB ...

3

The only general relation between the dimension of the real points and the complex points of a variety is that the former is not more than the latter. Depending on your definition of "degree" (for instance, number of intersections counted appropriately with a generic complementary dimensional linear subspace), the notion may be worthless. This is ...

3

Well, $f-g\in P$ for all $P$ means $f-g$ is in the intersection of all of the primes, i.e. the nilradical, which by assumption is zero.

3

There are several improvements you can make on Buchberger's Algorithm for computing Groebner bases, to the point where using other methods usually doesn't make sense. Most computer algebra systems use Groebner bases for their computations since they typically outperform any other method. I think one main reason Groebner bases are so revered is that they ...

3

(1) Much more generally (and therefore much easier!): Given an arbitrary topological space $Y$ there is a sheaf $\mathcal C_Y$ on $X$ characterized by the requirement that for any open subset $U\subset X$ we have $\mathcal C_Y(U)=\mathcal C(U,Y)$, the set of continuous functions $U\to Y$. Your $\mathfrak A^+$ is exactly the sheaf $\mathcal C_A$, where $A$ ...

2

There's two things going on here: If you take the scheme-theoretic intersection, $I(X\cap Y)= I(X)+I(Y)$ by definition and everything's fine. This is a common perspective to take as you get more experienced in algebraic geometry. Even if you don't take the scheme-theoretic intersection, the Hilbert polynomials of $R/\sqrt{I+J}$ and $R/(I+J)$ have the same ...

2

You have forgotten to consider the infinitely near point of order zero: $p$ itself. In which case, $m_p(C) = 2$ and $m_p(D) = 1$; hence, $$m_p(C)m_p(D) + m_x(\hat{C}) m_x(\hat{D}) = 2\cdot 1 + 1\cdot 1 = 3 = m_p(C\cap D).$$ (It's likely worthwhile to also mention that there are no further higher order infinitely near points since the tangent lines of the ...

2

Let me compile the discussion from the comments to mark this as answered. If $X$ is a topological space, $x\in X$, $U\subset X$ an open subset containing $x$, and $\mathcal{F}$ a sheaf on $X$, then the stalk of $\mathcal{F}$ at $x$ is the same as the stalk of $\mathcal{F}|_U$ at $x$. Using your definition of the stalk as equivalence classes that agree on an ...

2

As $k$ is algebraically closed, the closed points of $X$ (resp. $X \times_Y X$) are the elements of $X(k)$ (resp. $X(k)\times_{Y(k)}X(k)$), and thus $\delta_{\pi}(k)$ is surjective iff $X(k) \rightarrow Y(k)$ is injective (that is set-theoretical), ie iff $\pi$ is injective on closed points. Note that if $k$ isn't algebraically closed, it is easy for $\pi$ ...

2

To compute the cohomology of $\mathcal{O}_C$ one can use the exact sequence $$0 \to \mathcal{O}_C \to \mathcal{O}_{L_1} \oplus \mathcal{O}_{L_2} \to \mathcal{O}_P \to 0$$ for $C = L_1 \cup L_2$, where $P$ is the intersection point of the lines and the maps are induced by the restriction of functions, and the exact sequence $$0 \to \mathcal{O}_L(-1) \to \... 2 Because fg=0, so in each fiber O_x, x \in U, f_xg_x=0 \in m_x. But O_x is a local ring with m_x its maximal ideal, so f_x \in m_x or g_x \in m_x 2 I'm not sure exactly what \cal F and \cal G are, but let's takes some topological space X, let \cal F be the sheaf of continuous complex functions on X and \cal G be the sheaf of continuous non-zero complex functions on X. Your definition then makes sense as a sheaf homomorphism. This map exp is a surjective map of sheaves, but its image as a ... 2 Exercise: X being reduced (in the sense that \mathcal{O}_X(U) is a reduced ring for all open U\subset X) is equivalent to \mathcal{O}_{X,x} being reduced for all points x\in X. Hint/Solution outline: With this in hand, since the stalk at \mathfrak{p}\in\operatorname{Spec} A is the same as the stalk at \mathfrak{p}_f \in\operatorname{Spec} A_f, ... 1 I agree that the proof has a gap, and this has been complained about on this website at least once before here (though there are some extra assumptions in that post which are not made here, so the proposed solution is not quite applicable for you). If you assume that the U_\alpha are affine, then this is proven at the Stacks Project in tag 01K4, though as ... 1 In Question 1 you ask what "growth of P_d" means. Well, P_d is a function of the natural number d, and (just like any such function) one can ask if is bounded above by a polynomial function of d. In this case one can prove that for any variety X of dimension n, the function P_d is bounded above by a constant multiple of d^n. But ... 1 About question I For n=1 all nonzero quadratic forms are nondegenerate, smooth and reducible. For all n\geq 1 the properties "nondegenerate" and "smooth" are equivalent. Quadratic forms satisfying these conditions are often called regular in the literature. For n\geq3 these properties "nondegenerate" and "smooth&... 1 There is a natural map WC(E/K) \to WC(E/K_v) just by viewing a homogeneous space C for E over K as being defined over K_v. The trick is to check that this map agrees with the restriction map on cohomology after we identify WC(E/k) with H^1(k, E). Following Silverman X3.6 let p_0 \in C(\bar{K}) and let c : \sigma \mapsto p_0^\sigma - p_0 be ... 1 You can get the other inequality from dimension theory. In particular, we have the following theorem Let \pi:X \to Y be a morphism of locally Noetherian schemes, and let p \in X, q\in Y be points with q=\pi(p). Then$$\operatorname{codim}_X p \leq \operatorname{codim}_Y q + \operatorname{codim}_{\pi^{-1}(q)} p$$As you note, if we assume \pi is ... 1 OK, here is an answer with an example. Let us consider the sample elliptic curve with affine equation over the rationals$$ y^2 = x(x-10)(x-17)\ , $$which has e_1,e_2,e_3 equal to 0,10,17 (in this order, say). Here are some rational points on the curve, and their maps in \Bbb Q^\times/(\Bbb Q^\times)^2. For psychological reasons, a class \bar a of ... 1 \Gamma’ is the subgroup generates by the prime divisors of \Delta. There are only finitely many such primes. So you have a finitely generated abelian group whose generators all have finite order (2) and so the group is itself finite. 1 Yes, if you do not require Y to be irreducible then this bound can be violated. The cheapest example is to take f: X \rightarrow Y surjective of whatever relative dimension you like, bigger than 0, let X' and Y' be the disjoint union of each of them with a point, and define f' to equal f on X and to map the new point on X to the new point ... 1 I think I got an easy answer that the schematic dominance for fibers is preserved when substituting B_\mathfrak{p} for B. For all f,f'\in B and the image \mathfrak{q}\in\operatorname{Spec} of \mathfrak{p},$$ \bigcap_{\lambda} \ker\left(M_{ff'}\otimes_A A_{\mathfrak{q}}/\mathfrak{q}A_{\mathfrak{q}}\to (N_{\lambda})_{ff'}\otimes_A A_{\mathfrak{q}}/\...

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