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• 4,281
Accepted

### Is $R[X_1,X_2,\cdots]/\langle X_1,X_2,\cdots\rangle$ finitely presented?

No: over $P$. $P=P^1$. Not as a $P$-module.
• 3,838

### Why $\operatorname{dim}\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,y}} \kappa(y)=0$ in certain situation?

use $\mathcal{O}_{X,x}\otimes \kappa(y)=\mathcal{O}_{f^{-1}y, \:x}$ . Then $dim(f^{-1}y)=dim(\overline{\{x\}})=dim(Z)$ as all components of the fibre $f^{-1}y$ have the same dimension, hence (the ...
• 1,382

### Is $[A,[A,B]]=0$ generally true in the ring theory of operator?

You can find $2\times 2$-matrices $A,B\in M_2(K)$ with $[A,B]=AB-BA=B$, either by a direct computation, or by the adjoint matrices of the nonabelian Lie algebra $\mathfrak{r}_2(K)$ of dimension $2$ ...
• 120k
1 vote

### Counterexamples to Theorems/Corollary related to going down theorems

In $\mathbb Z\subset\mathbb Z[i]$, $(2\mathbb Z[i])\cap\mathbb Z=2\mathbb Z$ is maximal, but $2\mathbb Z[i]$ is not even prime and definitely not maximal. In general, $p\mathcal O_K$ is not ...
• 9,026
1 vote
Accepted

There is a typo, it should be Next, we have $\bar u \circ \bar v = 0$, that is $f \circ v \circ u = 0$ for all $f \colon M’’ \to N$. Why? Well, if $\bar u \circ \bar v \colon \text{Hom}(M’’,N) \to \... • 18k 1 vote Accepted ### Examples of non-constant polynomials$f(x)^3+g(x)^2=c$for some non-zero constant$c$This is possible in characteristic$2$or$3$. For instance, in characteristic$2$, you could have$c=1$,$f(x)=x^2$, and$g(x)=x^3+1$in characteristic$2$. More generally, if$f(x)$is a ... • 302k 1 vote ### Examples of non-constant polynomials$f(x)^3+g(x)^2=c$for some non-zero constant$c$Another way to see that this is possible in characteristic 2 or 3 but nowhere else is by using a little more algebraic geometry. If we have polynomials$f(x),g(x)$satisfying$f^3+g^2=c$for$c\neq 0$,... • 52.9k 1 vote Accepted ### Determinant of surjective endomorphism of$R^{\oplus n}$Call$e^1,\cdots, e^n$the canonical basis of$R^n$, call$F$the matrix that represents$f$in that basis and call$F^1,\cdots, F^n$the colums of$F$. Saying that$\operatorname{col}F=R^n\$ means ...

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