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Write the right-hand side as $Hom_B(P/P^2,B)$. If the map you are interested in is surjective, then the preimage of the trace ideal of $P/P^2$ in $B$ must be contained in the the trace ideal of $P$ in …

Such extension is called "cyclically pure". An extension is called pure if the induced map $A\otimes_A M\to B\otimes_A M$ is injective for any $A$ module $M$. If the map $A\to B$ splits as map of $A$- …

It should be noted that the answer is yes if $R$ is normal and $M$ is torsion-free. That is because of the: Fact: if a map $f:A \to B$ of reflexive modules is locally an isomorphism in codimension on …

The authors reduce to the case of $R$ complete with infinite residue field and use them implicitly at a couple of places in the proof. This is a fairly standard practice. For instance, to assert that …

Take $I=(a^3,b^3)$ and $J=(ac^2-bd^2)$. Then according to Macaulay2, $I\cap J$ has generators in degrees $7,8,9$, for instance $a^3c^6-b^3d^6$. So the answers to Question 3 and 1 are no.

We use the fact that in an Artinian Gorenstein ring, any ideal contains the socle. The assumption tells us that the socle of $A$ is $\mathfrak m^2$, which is principal. Let $I\neq (0)$ be a non-maxima …

This new wonderful note, F-singularities: a commutative algebra approach, written by Linquan Ma and Thomas Polstra, two card-carrying commutative algebraists, is perhaps what you need. From the Introd …

This holds for $n\leq 3$ but may fail for $n=4$ and higher. See Proposition 2.6 and Example 2.9 in the paper "On monomial Golod ideals" (but probably known to experts before).

When $d\geq 3$, these are isolated hypersurface singularities of dimension at least $4$, so are UFD by the Grothendieck's local Lefschetz Theorem. When $d=2$ and the field has characteristic $0$, the …

Since, $X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)((X-Y)^2+(Y-Z)^2+(Z-X)^2)$, taking $X,Y,Z$ close to each other give some non-trivial and cheap solutions. For instance $(k+1,k,k)$ for $N=3k$, $(k+1,k+1,k)$ …

No. Let $(R,\mathfrak m)$ be commutative local Artin ring, then the radical of $M$ is $\mathfrak mM$ and your condition is equivalent to $\mathfrak m^2M=0$. One can not bound the length of such indeco …

I followed the reference suggested by KConrad in the comments and found perhaps the answer to Question 1: Annie MacKinnon, who got her PhD from Cornell in 1894 with the thesis "Concomitant Binary Form …

Recently I dug up some biographical details of Lindsay Burch, of Hilbert-Burch Theorem fame, whose few papers have had quite an impact on commutative algebra. This made me curious about the first wome …

Interestingly, the equality you seek holds in one important special case. If $I$ is any monomial ideal and $J$ is the radical of $I$, then $pd_S(I)\leq pd_S(J)$. See the proof of Theorem 2.6 in this p …

As I proved in the answer to this question, the following holds: Proposition: For an Artinian ring $A$, the following are equivalent: $A$ is Gorenstein. Any finitely generated faithful module $M$ ov …