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I'm not so good on the scheme-theoretic language, so let me embed $F_n$ as the affine variety $\text{res}\_{n,n}(X^n + ..., b_{n-1} X^{n-1} + ...) y = 1$ one dimension up. Then a morphism $k[a_0, ... …

Let me elaborate a little on Pete's answer from a perspective which I briefly outlined in this blog post. The point of looking at affine schemes as opposed to commutative rings is that the category o …

I guess I should stop being so cryptic. Here is the full solution. Using Newton's sums it is possible to compute the coefficients of the polynomial $$P(x) = (x - x_1)...(x - x_n)$$ provided that $ …

Let $x = (x_1, ... x_d)$ and change coordinates so that $L(x) = x_1$. Since $C_0$ is irreducible it is, in particular, not divisible by $x_1$. Then, as mdeland says, staying on the hyperplane $x_1 = 0 …

When $K$ is not algebraically closed, one useful choice of replacement for the naive notion of an affine algebraic subset of $K^n$ is an affine algebraic subset of $\bar{K}^n$ which is a union of orbi …

The object you've defined is not the group of automorphisms of $\mathbb{P}^n$; among other things, it is a group-valued functor, not a group. Here is a simpler example of this sort of thing: In any c …

If you remember the tensor product then in nice cases $X \mapsto \text{QCoh}(X)$ is fully faithful, even for some nice stacks; this is a generalization of Tannaka-Krein duality due to Lurie in Tannaka …

If the map from $M_{1, 1}$ to the $j$-line can be said to have a degree, that degree should be $\frac{1}{2}$, which makes everything work out. The reason is that its fibers are generically not a finit …

This is long and may not at all address your question, but here's a claim I'd like to defend: taking pullbacks of universal things is an abstract-nonsense operation, but taking pushforwards is no …

I don't know if this is what you're looking for, but here's a heuristic argument for this sort of thing being true in great generality. This should be a comment but it got long. It's not hard to con …

Write $BGL_n$ for the classifying stack of rank $n$ vector bundles; in principle it is not necessary to know what a $GL_n$ torsor is in order to say this. $E$ is represented by a map $f : S \to BGL_n$ …

No, e.g. $A$ could be $\mathbb{C}^{\mathbb{N}}$ where $t$ is the sequence $(1, 2, 3, ...)$. This is not even Noetherian or countable-dimensional over $\mathbb{C}$. A slight modification of this constr …

If you know that genus is a birational invariant, you can explicitly write down some maps: $x_0 x_3 - 2x_1 x_2 = 0$ is birational to $\mathbb{P}^2$ via the substitutions $x_0 = RS, x_3 = 2T^2, x_1 = …

The projective closure is given by $\frac{X^k - Y^k}{X - Y} - Z(X - Y)^{k-2} = 0$. There is an obvious isomorphism from $\mathbb{P}^1$ given by $$(R : S) \mapsto \left( R(R - S)^{k-2} : S(R - S)^{k-2 …

This is true if $X$ is geometrically irreducible by the Lang-Weil bound, which gives us that the size of $|X \cap Y|$ is $(c(X \cap Y) + O_c(p^{-1/2})) p^{\dim (X \cap Y)|}$ where $c(X \cap Y)$ is the …