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The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.

The ring $k[[x,y]]$ is a local UFD of dimension 2; so its prime ideals are the zero ideal, the maximal ideal, and all the ideals generated by an irreducible element. They are all closed (all ideals in …

It's a standard result in commutative algebra that every noetherian integral domain is a UFD if and only if every prime ideal of height 1 is principal. When applied to the local rings of X this gives …

As a module, $A[[X]]$ is the product of a countable family of copies of $A$. It is known that the product of flat $A$-modules is flat if and only if the ring $A$ is coherent, that is, every finitely g …

I don't think that there are any really easy examples. In the famous paper of Beauville, Colliot-Thélène, Sansuc and Swinnerton-Dyer "Variétés stablement rationnelles non rationnelles" they construct …

Start with your favorite example of an affine irreducible variety $X$ that is not Cohen-Macaulay. Embed $X$ in $\mathbb A^N$, and call $c$ its codimension. Now take $c$ general polynomials that contai …

If $R$ is a commutative ring with $K_{0}(R)=\mathbb{Z}$, then $\mathop{\rm Spec} R$ is connected, because otherwise $R$ would split as a product, and $K_{0}(R)$ would contain a copy of $\mathbb{Z} \op …

I suppose that "additive" means that "additive over short exact sequences". If so, this is does not seem too hard, at least if $X$ is separated. By noetherian induction, you may assume that for all p …

The simplest example I can think of is $\mathbb C[x_1, x_2, x_3, x_4, x_5]/(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - 1)$. The proof I am going to give is almost certainly an overkill, a simpler approac …

There are many counterexamples to this. Suppose that $S$ is a smooth surface over $\mathbb C$. Let $T \to S$ be a finite morphism from another smooth surface $T$, and consider a factorization $T \to V …

No, it is not injective in general, unless $R$ is regular notherian. There are many counterexamples; for a simple one you can take the ring $R := \mathbb C[t^2, t^3] \subseteq \mathbb C[t]$, compute t …

This is not true in general. For example, assume that $P$ is a projective module on $R$ that is not free, but such that $P \oplus R$ is free (there are many such examples). Set $S= R \oplus P$, and gi …

No, this is false as soon as $n ≥ 3$. A second syzygy $M$ of the residue field $K$ gives a counterexample: each $M[1/X_i]$ is projective, hence free, and it is reflexive, so the second condition is sa …

The classes of the $X_i - a_i$ are easily seen to be a basis for $\mathfrak{m}/\mathfrak{m}^2$.

No. Take $B = \mathbb{C}[[x,y,z]]/(z^2 - xy)$ and $A = \mathbb{C}[[x,y]]$.