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Liapin wrote: Two-sided ideals are also all the possible unions of principal two-sided ideals. I get this as:

  1. If $I$ is an ideal of $S$ then for $a_1...a_k\in S$ we have $I= \cup(a_i)$.

  2. Also, I think $a_i\in S$ can be chosen from $D$-classes' representatives of $S$ because they form a partition of $S$. Am I right?

J.-E. Pin
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Gozal
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    @hardmath: I edited it. – Gozal Jul 10 '19 at 17:17
  • You have not told your Readers what $S$ is, although $I$ is described as a (two-sided?) "ideal of $S$". While you seem to be asking about such ideals, you introduce a further undefined notation for $D$-classes. For these reasons it is difficult to reply to "Am I right?" – hardmath Jul 11 '19 at 00:54
  • @hardmath: Accept my apology. I am new. Sorry man. Thank you for your consideration. – Gozal Jul 11 '19 at 18:30
  • No apology needed! I'm just trying to get the content polished for everyone's benefit. – hardmath Jul 11 '19 at 18:36

1 Answers1

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Let $S$ be a semigroup and let $S^1$ be the semigroup equal to $S$ if $S$ is a monoid and equal to $S \cup \{1\}$, where $1$ is a new identity element, otherwise. This notation is useful in the context of two-sided ideals because if $X$ is a subset of $S$, then the two-sided ideal generated by $X$ is $S^1XS^1$.

In particular, a subset $I$ of $S$ is a two-sided ideal if and only if $I = S^1IS^1$. Thus, if $I$ is a two-sided ideal, then $$ I = S^1IS^1 = S^1(\bigcup_{s \in I}\{s\})S^1 = \bigcup_{s \in I} S^1sS^1 $$ and hence $I$ is the union of the principal two-sided ideals $S^1sS^1$, where $s \in I$. Conversely, let $(S^1s_jS^1)_{j \in J}$ be a family of principal two-sided ideals and let $X = \{s_j \mid j \in J\}$. Then $$ \bigcup_{j \in J} S^1s_jS^1 = S^1XS^1 $$ is a two-sided ideal. This is the meaning of Ljapin's statement.

I am not sure to follow your first remark but for the second one, yes: if $I$ is a two-sided ideal, then $$ I = \bigcup_{s \in I/\mathcal{D}} S^1sS^1 $$ where the notation $s \in I/\mathcal{D}$ means that you select only one representative of each $\mathcal{D}$-class contained in $I$.

J.-E. Pin
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  • Indeed, this is what I wanted to know. Sorry for bad English. Excuse me. – Gozal Jul 11 '19 at 18:29
  • So you last words mean that if we have, for example, 3 D-classes so we have 3 ideals? Regards. – Gozal Jul 11 '19 at 18:32
  • Dear friend, I am just reading the Clifford's neat book and I have some questions. I know that you are not here to answer JUST me. and this is a kind of you. I have some up to 4 or five questions about ideals in mind. Since I am reading it by myself so, some of them may seem ridiculous. How can I ask them? Can I ask them in a question? Regards. – Gozal Jul 23 '19 at 19:23