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Your conjecture 2 is false if you don't assume that $G$ is reduced (in positive characteristic there are affine group schemes that are not reduced). As to conjecture 3, it is hopelessly wrong (think …

Yes, this can happen. See for example https://projecteuclid.org/download/pdf_1/euclid.nmj/1118801622

It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial. Here is the argume …

No: if $B$ is a quotient of $A$, then $B \otimes_A B = B$, but $B$ is very rarely smooth over $A$. The correct homological characterization of smooth maps involves André-Quillen cohomology.

Sure. If you have an algebraic group $G$ defined over $\mathbb Q$ acting on a $\mathbb Q$-algebra $A$, the $\mathbb Q$-algebra of invariants $A^G$ is the equalizer of two usual homomorphisms of algebr …

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 …

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 …

Every such automorphism is contained in the automorphism group of the field or rational functions $F(t)$ over $F$, which equals $\mathrm{PGL}_2(F)$, and so is a finite group. [Edit:] upon further ref …

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 …

I suppose you intend to ask whether there is isomorphism as in the formula. The answer is negative, in general. For example, take $R = k[x]$, $I = (x)$, $M = R/I^2$. [Edit] What you may have in mind …

Edit: the first version of this was completely wrong, I hope this one works. Take a line $L$ in $\mathbb P^3$, and two general surfaces of degree $d > 3$ passing through $L$; it is not hard to see th …

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 …

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 …

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 $F_i$.