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Results tagged with ra.rings-and-algebras
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user 18060
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.
2
votes
Accepted
Morphism of algebras
No. Consider the matrix corresponding to $i$. It is semisimple with eigenvalues $i,i,-i,-i$ . So it's centralizer is $8$-dimensional - clearly, it is $M_2(\mathbb Q(i))$. Since this algebra is ramifie …
- 148k
15
votes
Accepted
rings in which every element is a sum of two commuting idempotents
If $x$ and $y$ are two commuting idempotents, then $(x+y)^3 = x+y+6xy$, $(x+y)^2=x+y+2xy$, so $z=x+y$ satisfies the equation $z^3-3z^2+2z=0$. Plugging $z=3$ into the equation, we obtain $6=0$. So the …
- 148k
2
votes
Accepted
Annihilators in algebras
It is false. The ring of two-by-two matrices has such a basis:
$$a_1=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},a_2=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},a_3=\begin{pmatrix} 0 & 1 \\ 1 & 0 …
- 148k
4
votes
Accepted
Rings of Quaternions
Assume $xy=0$ but $yx\neq 0$. We can take $x$ and $y$ to be $2$-adic quaternions, and then we have $xy=0$ modulo $2^s$ but $yx\neq 0$ modulo $2^s$. We can factor out the highest possible power of $2$ …
- 148k
10
votes
$\Lambda$-Ring Structures on $\mathbb A^2$
This answer contains a list of all the $\Lambda$-ring structures that I have found. If anyone can find any others, they should edit this list.
The two $\Lambda$-ring structures on $\mathbb A^1$ give …
5
votes
Maximal centralizer in full matrix ring
Let $x \in M_n(F)$. If the characteristic polynomial of $x$ has distinct prime factors in the ring $F[t]$, then there is some idempotent that commutes with everything that commutes with $x$, hence, if …
- 148k
6
votes
Does every finite free R-algebra have a basis starting with 1?
If $R$ has a finite-dimensional vector bundle that is nontrivial but stably trivial, then obviously not. If $M$ is the corresponding module, then $R[M]/M^2$ is a counterexample.
If $R$ has no such bu …
- 148k
0
votes
Vector "product" diagonalization
Let $k$ be your base field, then consider the ring $k[\epsilon]/\epsilon^{n+1}$. The elements of the ring such that $v^2=1$ are just $v=\pm1$. These elements do not span the ring as a vector space ove …
- 148k
13
votes
Connection between determinant and quotient rule
If $f$ is $g$ times a constant $c$ then the quotient is $c$ and has derivative zero and the two columns of the Wronskian are linearly dependent, the left column equalling the right column times $c$, a …
- 148k
17
votes
Accepted
Rings with group of units cyclic of prime order
For $p=2$, $\mathbb F_3$ is an example.
Otherwise, $p$ is odd, so $-1$ is a unit of order $2$ unless it is equal to $1$, so $1=-1$, so $2=0$. We can form the ring:
$\mathbb Z[x]/(2,x^p-1)= \mathbb F …
- 148k
4
votes
Description of p-adics tensor the reals
Since $\mathbb R$ is a $\mathbb Q$-vector space, we have $\mathbb Z_p\otimes_{\mathbb Z} \mathbb R= \mathbb Q_p \otimes_{\mathbb Q} \mathbb R$. So this is the special case of the general phenomenon of …
- 148k
7
votes
Accepted
$\lambda$-ring endomorphisms of ${\mathbb Z}[x]$
No, it is not cyclic.
For each of the rings, the semigroup of endomorphisms consists of one element of degree $n$ for each positive integer $n$ plus one or two elements of degree $0$. For a positive i …
- 148k
13
votes
Accepted
Topology on a module over a topological ring
I think the right definition to use in general is "The finest topology making multiplication $R \times M \to M$ and addition $M \times M \to M$ continuous". This has the obvious advantage of making $M …
- 148k
3
votes
Principal ideal subrings of formal power series rings
This happens if and only if the coefficients $a_i$ satisfy a simple Frobenius-linear recurrence $a_i = \sum_{j=1}^n a_{i-j}^{p^j} c_j$ for all $i \geq n$.
First assume that $R$ has dimension $1$.
$R …
- 148k
3
votes
In the group ring $\mathbb{Z}_p [G]$, what elements satisfy $(\sum a_g g)^p = \sum a_g g^p$?
If $G$ has no $p$-torsion, then the Jacobian matrix of this system of equations is invertible modulo $p$ (the derivative of the left side vanishes mod $p$, and the derivative of the right side is a pe …
- 148k