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Results tagged with ra.rings-and-algebras
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user 10503
Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
4
votes
Accepted
Annihilator of minimal prime ideal in a commutative Noetherian ring
Put $P=(N:M)$.
Because $N$ is a maximal submodule, $P$ is a maximal ideal.
Because $M$ has finite length, there is some minimal $k$ such that $MP^k=MP^{k+1}$. By Nakayama, there exists $s\in 1 …
- 23k
0
votes
Polynomial Rings
Let me write $R[x]=S[[y]]$ to avoid confusion.
I. Note that 1+y is a unit. Therefore (thinking of y as an element of $R[x]$), we have $y\in R$.
II. Now mod out $y$ on both sides: $\overline{R}[x …
- 23k
0
votes
Accepted
Another question about primitive central idempotents in associative unital rings (yes, again!)
I have no idea what an infinite sum means in an arbitrary associative unital ring.
Your question is equivalent to asking whether it is possible to write 1 as a sum of primitive central idempotents. …
- 23k
10
votes
Accepted
chain of prime ideals in polynomial ring $S=\Bbb{R}[x_1,x_2,...,x_n,...]$
Pick your favorite bijection $\phi$ from the natural numbers to the rational numbers. For each real number $\alpha$, let $I_\alpha$ be the ideal generated by all $x_n$ with $\phi(n)<\alpha$.
Then $I …
- 23k
4
votes
Accepted
Depth of polynomial ring $S=\Bbb{R}[x_1,x_2,x_3,...,x_n,...]$
In the same spirit as the answer to your earlier question, pick a bijection $\phi$ from the natural numbers to the rationals. For each real number $\alpha$, let $I_\alpha$ be the ideal generated by a …
- 23k
0
votes
When is this diagram of tensor powers an equalizer?
I've not ground this out, but:
For $(x,y)\in A\times A$, let [x,y] be the corresponding generator in the free $B$-module over the set $A\times A$.
Then if $x\otimes1-1\otimes x=0$, there must be a …
- 23k
3
votes
Accepted
ring with prescribed K group
Every abelian group $G$ is the class group of some Dedekind domain $R$ (theorem of Luther Claborn), so we have $K_0^{red}(R)= G$.
- 23k
5
votes
Can we deduce that two rings $R_1$ and $R_2$ are isomorphic if their polynomial ring are iso...
Let X be an affine variety with two non-isomorphic vector bundles V and W that become isomorphic after adding a trivial line bundle to each. Then the coordinate rings of the total spaces of V and W s …
- 23k
12
votes
Accepted
Ring with vanishing $K_0$
Question 1:
Does $K_0(R)=0$ imply that $R^m\approx R^n$ for some $m\neq n$?
Answer to Question 1:
Yes.
Choose any distinct natural numbers $m$ and $n$. Using square brackets to denote $K_0$ classes …
- 23k
1
vote
Accepted
Finite extension of a field
Suppose $k\subset Quot(A/p)$ is a finite extension of fields. Then every nonzero element $x$ of $A/p$ is algebraic over $k$ and so satisfies a minimal polynomial with non-zero constant term $a_0\in k …
- 23k
5
votes
Accepted
Algebra structure $Tor(A,A)$
More generally, if $A,B,C,D $ are $k$-algebras, there's a multiplication
$$Tor_m^k(A,B)\otimes_k Tor_n^k(C,D)\rightarrow Tor^k_{m+n}(A\otimes_k C,B\otimes_k D)\qquad(1)$$
If you take $C=A$ and $B=D$, …
- 23k
1
vote
Tensoring with descending chain of modules
Let $A=k[X]$, let $M_i$ be the $A$-ideal generated by $X^i$, and let $B=k(X)$. Then $\cap M_i=0$ is certainly finite and free, but
$$0=(\cap M_i)\otimes B\neq \cap(M_i\otimes B)=B$$
which is a coun …
- 23k
0
votes
zeros of a homogeneous polynomial
Over the field of three elements, you can take $\lambda=-1$. Over the field of five or seven elements, you can take $\lambda=1$. Over the field of eleven or thirteen elements, you can take $\lambda …
- 23k
3
votes
Accepted
Second summand to make projective module free
If you want something completely general, LSpice's comment is the answer.
For the special case of an ideal $P$ in the ring $R$ of integers of a number field (or more generally if $R$ is a Dedekind dom …
- 23k
7
votes
Accepted
When are two projective modules of equal rank isomorphic?
If $R$ is noetherian of dimension d, then we have:
The Bass Cancellation Theorem: If $M$ has rank $\ge d+1$ after localizing at any prime, and if there exists a finitely generated projective module …
- 23k