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

I think this is true. There is a homomorphism $R \otimes_A Hom_R(M,N) \rightarrow Hom_A(M,N)$, given from the obvious $R$-module structure of $Hom_A(M,N)$. To check that this is an isomorphism we can …

I don't think this can happen. The division algorithm works over arbitrary rings, with unique quotient and remainder, as long as you are dividing by a monic polynomial. This implies that if $g$ is mon …

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 …

I am looking for a reference for the following generalization of the Weierstrass preparation theorem for formal power series. Suppose that $A$ is a noetherian complete local ring with residue field $k …

Take $A = \mathbb Z$, $M = \mathbb Q$, $B = \mathbb Z/p\mathbb Z$, were $p$ is a prime.

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 $F_i$.

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 …

Those invariant polynomials are called multisymmetric functions. There are several papers on them; you could start with J. Dalbec, Multisymmetric functions, Beiträge Algebra Geom. 40(1) (1999), 27-51 …

It is true for locally free sheaves of algebras that are central simple algebras at every point, though. These are known as sheaves of Azumaya algebras; I mention them since they were brought up in th …

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 …

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 …

Let $A$ be the coordinate ring of a smooth affine curve $X$ over $\mathbb C$, and let $p$ be a point of infinite order in the class group of $A$. Let $B$ be the coordinate ring of $X \smallsetminus \{ …

$\mathcal{E}xt^q(F, I)$ is 0 for $q > 0$. On the other hand it follows easily by considering extensions by 0 that $\mathcal{H}om(F, I)$ is flabby, hence acyclic.

The dimension of $M$ is the maximum of the dimensions of $M'$ and $M''$, so what you are asking for cannot happen (at least not in the noetherian case). To Dylan: the dimension of a module is the dime …