Part of your question is about the relationship between primary decomposition and localization. Here, I think the basic connection is that if $I$ is an ideal in a Noetherian ring $R$ whose minimal ...

Here's a variation on user2035's answer. Let $R=\mathbb Q[x, z, y_1, y_2, \ldots]/ \langle x-zy_1, x-z^2y_2, \ldots \rangle$. Then take $M=R$ and $I=\langle z \rangle$, and then $\cap_{i=1}^n I^n = \...

One algebraic version of this statement is that if $A$ is a local ring with embedding dimension $m$ and $A$ is the quotient of a regular local ring, then $A$ is the quotient of a regular local ring of ...

No, the cardinalities alone are not enough to identify a local commutative ring. For example, you can take $R = \mathbb F_p[x]/\langle x^3 \rangle$ and $S = \mathbb F_p[x,y] / \langle x^2, xy, y^2\...

No, this is not true. A counter example is the ideal $\langle xz, yw, xw+yz \rangle$, which has $3$ generators, but has an associated prime of codimension 4 at the origin. I'm not sure how much it ...

This can be done in a few steps in probably any computer algebra package. You take the generators of your original ideal $I$, and bi-homogenize them, as described in the question. Then saturate with ...

This is essentially the vertex enumeration problem in convex geometry. Let $A$ be an $n \times m$ matrix whose columns form a basis the vector space $W$. Then the vectors you're looking for are ...

Yes. Every proper, birational morphism between smooth algebraic surfaces can be obtained by the iterated blowup at reduced points. I don't have a reference at hand, but it's contained somewhere in ...

Not all finitely generated monoids $P$ will come from the cone construction. You need to assume that: $P^{gp}$ is torsion-free: If $x \in P^{gp}$ and $n\cdot x = 0$ then $x = 0$. $P$ is cancellative: ...

Yes this is always possible. What you do is you start with a simple normal crossing resolution $M$. Let $D_1, \ldots, D_k$ be the irreducible components of the exceptional divisor. Then, by the simple ...

Even with the additional assumption in your edit, the answer is still no. Take $X=Y=\mathbb A^1$ and let $f\colon X \rightarrow \mathbb P^2$ be given by sending $x$ to $(1:x:x^2)$. Let $(a:b:c)$ ...

Yes. See the beginning of section 3 of "Compactifications of subvarieties of tori" by Jenia Tevelev. He has a finitely generated integral domain $A$ (he calls it $\mathcal O(X)$) over an algebraically ...

ADDED: The following is based on my misinterpretation of irreducible as "cannot be nontrivially written as a direct sum." Moreover as Torsten Ekedahl points out, it is easy to generalize this example ...

The answer to the first part (about finding a linear combination which has full rank) is no. A counterexample with $n=3$ and $k=2$ is given by the quadratic forms $xy$ and $xz$. A general linear ...

I'm going to assume that $X^0$ is the same as $X_0$ and that the Picard group of $X$ is freely generated by the $D_i$, since without that the question doesn't make much sense. In this cases, the ...

No, it is not true. I'm going to describe everything in terms of commutative algebra, so take $X=Y=\operatorname{Spec} R$. Let $G$ be the group $\mathbb Z \times \mathbb Z$ with the lexicographic ...

I believe that the answer to the first question is no for the reason that a local complete intersection is pretty far from a complete intersection. Take $Z$ to be $3$ points in $\mathbb P^3$ which ...