Semigroup Theory

**spoon737** · December 28th 2009, 08:36 AM

I'm trying to understand a particular proof, but it has a step which isn't really given any justification. Here is the part that's giving me trouble:

"Let $I$ be a semiprime ideal of a semigroup $S$ , and let $d \in S-I$ . Letting $M = \{d^n| n = 1, 2, ...\}$ , we obtain an m-system disjoint from $I$ ."

For those unfamiliar with the terminology, a semiprime ideal is an ideal of a semigroup $S$ with the property that for any $a \in S, aSa \subseteq I \Longrightarrow a \in I$ . An m-system is a non-empty subset $A$ of S such that $a,b \in A \Longrightarrow$ there exists $x \in S$ such that $axb \in A$ .

What I'm trying to understand is why $M$ should be disjoint from $I$ . I understand why it is an m-system, but I don't understand why, say, $d^2$ can't be in $I$ . Any insights would be appreciated.

**tonio** · December 28th 2009, 09:03 AM

Originally Posted by spoon737

I'm trying to understand a particular proof, but it has a step which isn't really given any justification. Here is the part that's giving me trouble:

"Let $I$ be a semiprime ideal of a semigroup $S$ , and let $d \in S-I$ . Letting $M = \{d^n| n = 1, 2, ...\}$ , we obtain an m-system disjoint from $I$ ."

For those unfamiliar with the terminology, a semiprime ideal is an ideal of a semigroup $S$ with the property that for any $a \in S, aSa \subseteq I \Longrightarrow a \in I$ . An m-system is a non-empty subset $A$ of S such that $a,b \in A \Longrightarrow$ there exists $x \in S$ such that $axb \in S$ .

What I'm trying to understand is why $M$ should be disjoint from $I$ . I understand why it is an m-system, but I don't understand why, say, $d^2$ can't be in $I$ . Any insights would be appreciated.

What book(s) are you using? The ones I have and some web sites define a semiprime ideal I as a non-empty subset of a semigroup S s.t. $s^2\in I\Longrightarrow s\in I\,,\,\,\forall x\in S$ , and then it is clear why M is disjoint from I ...

Tonio

**spoon737** · December 28th 2009, 11:11 AM

I'm using Introduction to Semigroups by Mario Petrich. It' pretty old (1973), so it's not entirely unlikely that the terminology is out of date.

The definition you gave is given in my book as the definition of a completely semiprime ideal. It makes a distinction between semiprime and completely semiprime. I agree, it's pretty trivial why M and I would be disjoint in that case, but my book only says that I is semiprime.

**Swlabr** · December 28th 2009, 11:25 AM

Originally Posted by tonio

What book(s) are you using? The ones I have and some web sites define a semiprime ideal I as a non-empty subset of a semigroup S s.t. $s^2\in I\Longrightarrow s\in I\,,\,\,\forall x\in S$ , and then it is clear why M is disjoint from I ...

Tonio

Are $S$ -systems defined on general semigroups, or on Monoids? The great John Howie's book only seems to define them for monoids (unless I am missing something). In monoids everything here works - the condition given by Tonio follows from spoon737's definition, and so the result holds...

Also, can you clarify your definition of an $M$ -system? I mean, do you mean to say $axb \in S$ instead of, say, $I$ ?

**spoon737** · December 28th 2009, 11:57 AM

My book's definition of an m-system is exactly as stated: given a non-empty subset $A$ of a semigroup $S$ , $A$ is an m-system if for any $a,b \in A$ there exists $x \in S$ such that $axb \in A$ . So yes, $axb$ is required to be in $A$ . I have yet to come across a definition for $S$ -system.

I think I've figured it out, though. After searching around a bit online, I found that my book's definition of a semiprime ideal is equivalent to the following:

An ideal $I$ is semiprime if for any ideal $J$ , $J^2 \subseteq I$ implies $J \subseteq I$ .

Technically, the result I found at PlanetMath was given for rings, but it was easy enough to see that this was the same for semigroups. Thinking of it this way made the result much more obvious. However, since the author makes no hint of this equivalent definition, yet he still states that $M$ and $I$ are disjoint in such a way that seems too trivial to require proof, I have to wonder if it really does follow immediately from his definition and I'm just missing something.

**NonCommAlg** · December 28th 2009, 03:15 PM

Originally Posted by spoon737

I think I've figured it out, though. After searching around a bit online, I found that my book's definition of a semiprime ideal is equivalent to the following:

i'd like to see how you "figured it out" because the claim, as you've given us, is basically false! here's a simple counter-example:
let $S = \mathbb{M}_2(\mathbb{R}),$ the set of $2 \times 2$ matrices wih real entries. $S,$ with matrix multiplication, is a semigroup and $I=(0)$ is a semiprime ideal of $S.$ let $d=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}.$ then $d \notin I$ but $d^2 = 0 \in I.$

**NonCommAlg** · January 5th 2010, 02:12 PM

this is for spoon737, the OP:

if anything in my answer is unclear for you, you should ask me in this website not asking it somewhere else! that's just not right!

you (almost) copied my answer in there and, even worse,

asked them to confirm it!! that's a big insult to me! haha ... ok, i forgive you. i just wanted to tell you that what you did was wrong.

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