Right ideal $eS$ in a simple semigroup $S$ without zero, with idempotent $e$ and minimal left ideal $L$ is a minimal right ideal of $S$.

Suppose that $S$ is like in the title, a simple semigroup without zero, with a minimal left ideal $L$ and idempotent $e$. We can always assume that $L = Se$. I've already proved few facts about $S$. I know that $S$ is a sum of it's minimal left ideals, that $Se$ is a left group, and that $H_e = eSe$ is a group. This was also a part of the exercise.How do I prove that $eS$ is a minimal right ideal? If possible, I'd like a hint before a full solution. It's an exercise from a subsection '$0$-minimal ideals and $0$-simple semigroups' if that helps....Read more

lie superalgebras - left ideals in Lie super algebras

Let $\mathfrak g$ be a Lie superalgebra. If $\mathfrak a$ is not a grade subspace of $\mathfrak g$, then why does $[\mathfrak g, \mathfrak a]$ and $[\mathfrak a, \mathfrak g]$ are not same?For me as sets they are linear span of $[a,x]$ and $[x,a]$ and hence they are same. But in book it is given they are different and the author has defined left and right ideal separately.I am reading the book "Lie superalgebras and enveloping algebras by Ian M.Musson" Proposition 1.2.2.Kindly help me with this.Thank you....Read more

Radical Ideal for algebra

Show the following:a) $\text{rad}(IJ)=\text{rad}(I\cap J)=\text{rad}(I) \cap \text{rad}(J)$b) $\text{rad}(I)=R$ if and only if $I=R$c) if $P$ is prime $\text{rad}(P^n)= P$ for all $n$d) Let $F$ be a field and $T$ a subset of $F^n$ . Show that the ideal$I(T) = \{f ∈F[x_1,\ldots,x_n] : f(a_1,\ldots,a_n)=0$ for all $(a_1,\ldots,a_n) \in T\}$ is a radical ideal.My answers so far:a)Let $\text{rad}(I)=\{r \in R,r^n \in I$,for some positive $n\}$ and $\text{rad}(J)=\{r \in R,r^m \in J$, for some positive $m\}$. Then $r^nr^m = r^{n+m} \in IJ$. By hom...Read more

ideals - Computing the radical of $\mathfrak{gl}(2,\mathbb{C})$ without using the semisimplicity of $\mathfrak{sl}(2,\mathbb{C})$.

I have been trying to show that the radical of $\mathfrak{gl}(2,\mathbb{C})$ is its center, i.e. scalar matrices, however all the proofs I have encountered (e.g. Radical of $\mathfrak{gl}_n$) have used the fact that $\mathfrak{sl}(2,\mathbb{C})$ is semisimple. Instead, I am aiming to find the radical of $\mathfrak{gl}(2,\mathbb{C})$ first, and then use the fact that $\mathfrak{gl}(2,\mathbb{C}) \cong \mathfrak{sl}(2,\mathbb{C}) \bigoplus \mathbb{C}$ along with the proposition:If $L$ is a Lie algebra, then $L/Rad(L)$ is semisimple,to conclude th...Read more

ideals - Restricting a filter in a Boolean algebra to a generating set and have it generate a filter

Let $B$ be a Boolean algebra and $S \subseteq B$ be a subset that generates $B$. Is it the case that every filter $x$ of $B$ is equal to the filter generated by $x \cap S$? What if $S$ itself is a lattice with respect to the induced partial order?Of course, every element of $x$ is $t(\bar a)$ where $\bar a \in S$ and $t(\bar x)$ is a term in the signature of Boolean algebra, and $t$ can be put in terms of nice normal forms, but I can't apply those facts to this problem....Read more

Associative algebra without nilpotent ideals is direct sum of minimal left ideals

In the book 'Spinors, Clifford and Cayley Algebras' Hermann states that any (finite dimensional) semisimple associative algebra is the direct sum of minimal left ideals. Here, 'semisimple' is defined as 'has no nilpotent ideals except $0$'. However, in the proof he seems to use an unit, which was not in the assumptions.Interestingly enough, later in the book it is mentioned as an exercise that semisimple algebras have an unit, but I think that exercise uses the earlier theorems.Therefore, my questions are:Is Hermann's statement generally true w...Read more

Is some One Dimensional subalgebra an Ideal of the 2 Dimensional Non-Abelian Lie Algebra?

Is there any one dimensional subalgebra which is an Ideal of the two dimensional non-abelian Lie Algebra?i.e. is it invariant as a subalgebra of the 2D non-abelian algebra I read that "all the subalgebras of an abelian algebra are automatically invariant"~Mathematical Methods for Physicists - George B. ArfkenI would assume this includes the trivial case of the group itself.However, it then occurred to me that this would only hold if I consider the algebra as a subalgebra of itself and not necessarily in its own right.Furthermore, can you pro...Read more

Origin of ideals in order theory

I am trying to clarify in my head the different meanings of "ideals" in mathematics. We have ideals in Number Theory, as in Dedekind (derived from 'Ideal Complex Numbers' in Kummer), in Abstract Algebra (Ring Theory), as in Dedekind and, mostly, in Noether's development, and we also have, it seems, a notion of ideals in Order Theory, in which they are a special subset of a partially ordered set (poset).I have read the meaning in order theory developed from the second meaning (that in abstract algebra and ring theory).I would like to know:1) How...Read more

Localization of the polynomial ring at a prime ideal modulo maximal ideal is isomorphic to polynomial ring modulo prime ideal.

Let $p \in K[T]$ irreducible, s.t. $\text{LC}(p) = 1$. Then$$ K[T]/(p) \cong K[T]_{(p)}/pK[T]_{(p)}.$$What I have is:\begin{align*}&K[T] \hookrightarrow K[T]_{(p)} \text{ and } K[T]_{(p)} \twoheadrightarrow K[T]_{(p)}/pK[T]_{(p)}\\& \implies K[T] \overset{f}\longrightarrow K[T]_{(p)}/pK[T]_{(p)}\end{align*}The kernel of $f$ is $(p)$ so$$ K[T]/(p) \cong \text{im} f = \left\{g + pK[T]_{(p)} \mid g \in K[T] \right\} \overset{?} = K[T]_{(p)}/pK[T]_{(p)}.$$...Read more

semigroups - example of right / left ideals

I searched and searched for examples of right / left ideals, but could find none. I read that a right ideal of $S$ is a subset of $R$ of $S$ such that $RS \subseteq R$, and that symetrically, a left ideal is a subset $L$ of $S$ such that $SL \subseteq L$. What concrete examples of right / left ideals are there?...Read more

ideals - Noether polynomial ring over $2\Bbb Z$

Given a definition of $R=2\Bbb Z$ prove that ring $R[x]$ is not noether.I assume that the proof should be based on the fact that a ring is noether if and only if any ideal is finitely generated. However, I'm struggling with constructing a non-finitely generated ideal, because only ideals I can think of are "root-based" (i.e. polynomials that have certain roots), that are, of cource, finitely generated.I would be very grateful for a description of such ideal or a piece of work that can explain the structure of polynomial ring ideals in detail....Read more

Assuming that $I,J\subset S$ are monomial ideals, is it true that $I+J$ is monomial? What about $IJ$ and $I\cap J$?

Let $\mathbb{k}$ be a field and $S=\mathbb{k}[x_1,\dots,x_k]$ be a ring of polynomials over $\mathbb{k}$.Assuming that ideals $I,J\subset S$ are monomial, is it true that $I+J$ is monomial? What about $IJ$ and $I\cap J$?I do not know how to approach this problem. If we take for example $I=(x_1,x_2)$ and $I=(x_2,x_3)$ then how to find $I+J$ and so on?...Read more