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We call such modules projective. If you take $N\mapsto Hom(N,M)$ then you get injective modules. This is fairly basic, and covered in any homological algebra book, and mentioned on wikipedia.

Isn't that the only one of length two? And for length three, shouldn't $\mathbb{F}_q[x,y]/(x^2,xy,y^2)$ work? I believe (possibly incorrectly) that here length and dimension over the base field agree. …

In addition to all the papers, there's even a book! "Integral closure: rees algebras, multiplicities, algorithms" by Vasconcelos.

Let $R=\mathbb{C}[x,y]/(y^2-x^2(x-1))$. This is the nodal cubic in the plane. Look at the prime $\mathfrak{p}=(x,y)$, corresponding to the nodal point. The completion here is isomorphic to $\mathbb …

As for what "is" means here, it means "is naturally isomorphic to" that is, you should be able to find an isomorphism that doesn't depend on any choices you make (that is, it'll be functorial). As fo …

The homomorphisms in the category of sheaves are not sheaves themselves. The hom sheaves have the data of things that are only homomorphisms over open subsets. So if $Y,Z$ are coherent $\mathcal{O}_ …

No, but for somewhat trivial reasons. Let R be the polynomial ring in countably many variables, with no relations. This is a countably generated k-algebra, and it can't be the ring of functions on a …

It's easy to tell when a polynomial is squarefree (or not): that's just the question of the vanishing of the discriminant, which can be dealt with as the resultant of $f$ and $f'$. However, given a p …

I've been reading about D-modules, and have seen a proof that D_X, the ring of differential operators on a variety, is "almost commutative", that is, that its associated graded ring is commutative. N …

This is normalization. As for characterizing them, over C, in the complex topology on the variety, think of it as meaning (roughly, not exactly) locally irreducible. Most important is that normal va …

Here's an example stolen blatantly from wikipedia. Let $R=\mathbb{Z}[\sqrt{-3}]$, let $a=4=2*2=(1+\sqrt{-3})(1-\sqrt{-3})$ and $b=2(1+\sqrt{-3})$. Now, $2$ and $1+\sqrt{-3}$ are both maximal among d …

Ok, I might be missing something (I often am) but I believe that this does it. Scroll up to page 452, line 3.9. The book is Effective Methods in Algebraic Geometry by Rossi and Spangher.

Should be able to use Groebner bases and elimination theory to do it, I think. Though in the case of linear and quadratic equations, I'm not sure how much help it will be.

Is there a good software package for doing computations in the cohomology ring of Grassmannians? Things like, I can write down a polynomial in, in fact, special Schubert classes, but it's one where do …