4 votes
Accepted

isomorphism $\mathfrak{m}\otimes_A (A/\mathfrak{m})\cong\mathfrak{m}_\mathfrak{m}\otimes_{A_\mathfrak{m}}(A_\mathfrak{m} / \mathfrak{m}_\mathfrak{m})$

You can do this in a couple smaller, bite-sized pieces: $\newcommand\m{\mathfrak{m}}$ $\m_\m\otimes_{A_\m} (A_\m/\m_\m)\cong (\m \otimes_A A_\m)\otimes_{A_\m} (A_\m/\m_\m)$ since $S^{-1}M\cong S^{-1}...
  • 2,209
3 votes
Accepted

When is there a one to one correspondence between closed sets in the Zariski topology of the prime spectrum of a ring and radical ideals of that ring?

You're correct, there is a bijective correspondence between closed subsets of $\operatorname{Spec} A$ and radical ideals $I\subsetneq A$ and this always holds. This correspondence gets discussed a lot ...
  • 52.9k
2 votes
Accepted

Proving Proposition 1.1 in Atiyah and MacDonald

Hint Use the "canonical" projection $\rho:A\twoheadrightarrow A/\mathfrak a$, which sends each $x\in A$ to the coset $x+\mathfrak a$ in $A/\mathfrak a$. The bijection is between the sets $\...
  • 4,281
2 votes
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
2 votes

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
2 votes

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 $2$ ...
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

why is $\bar u \circ \bar v =v \circ u \circ f?$

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 ...
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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