0
$\begingroup$
  • Let $S$ be a monoid with zero and $I_{\lambda}$ be an indexed set, then $B_{\lambda}(S) = \{ (\alpha, s , \beta ) : \alpha , \beta \in I_{\lambda}, s\in S \} \cup \{0\}$ is a semigroup and $J = \{ (\alpha, 0_s , \beta ) : \alpha , \beta \in I_{\lambda}\} \cup \{0\}$ , where $0_s$ is a zero of $S$, is an ideal of $B_{\lambda}(S)$. Define $$B_{\lambda}^o(S) = B_{\lambda} \backslash J$$

is called Brandt $\lambda^o$-extension of the semigroup $S$ with zero.

Let $S$ be an monoid with zero, $\lambda \geq 1$ any cardinal and $B_{\lambda}^o(S)$ the Brandt $\lambda^o$-extension of $S$. Then every non-trivial homomorphic image of $B_{\lambda}^o$ is the Brandt $\lambda^o$-extension of some monoid with zero.

I have tried:

when $\lambda = 1$ the proof is trivial. Therefore we assume that $\lambda \geq 2.$ Let $T$ be a semigroup and $h : B_{\lambda}^o(S) \rightarrow T$ is a homomorphism . Without loss of generality we can assume that the homomorphism $h$ is surjective map. suppose $(0_s)h = 0_T$ and $k$ be in $T$, then $$ k (0_s)h = (x)h . (0_s)h = (x.0_s)h = (0_s)h$$ thus $0_T$ is zero of $T$. Also we know that the semigroup $B_{\lambda}^o(1_s)$ is a congruence free, where $1_s$ is the identity element of $S$. So any homomorphism $g : B_{\lambda}^o(1_s) \rightarrow T$ is either trivial or isomorphism.

we fix $\alpha_o \in I_{\lambda}$, for every $\alpha , \beta \in I_{\lambda}$, we denote $1_{\alpha,\beta} = (\alpha, 1_s , \beta)h$ and $T^*_{\alpha,\beta} = \{ (\alpha, s , \beta)h : s \in S\backslash \{0\} \} \backslash \{0_T\}$ and $T_o= T^*_{\alpha_o , \alpha_o}$. Firstly we shall show that for any $\alpha, \beta , \gamma , \delta \in I_{\lambda}$, we have $|T^*_{\alpha,\beta}| = |T^*_{\gamma,\delta}|$, by defining the maps $$\phi_{(\alpha,\beta)}^{(\gamma, \delta)} : T^*_{\alpha,\beta} \rightarrow T^*_{\gamma,\delta}$$ by $$(x) \phi_{(\alpha,\beta)}^{(\gamma, \delta)} = 1_{\gamma ,\alpha}.x.1_{\beta,\delta}$$

similarly we can define the maping $$\phi_{(\gamma,\delta)}^{(\alpha, \beta)} : T^*_{\gamma,\delta} \rightarrow T^*_{\alpha,\beta}$$

Composition of both map is identity map, so both are mutually invertible maps and hence we have $|T^*_{\alpha,\beta}| = |T^*_{\gamma,\delta}| = |T_o|$.

Next I want to show that $T = I_{\lambda} \times T_o \times I_{\lambda} \cup \{0_T\}$, but I am unable how to prove.

$\endgroup$
7
  • $\begingroup$ This is not a difficult generalization of the proof aperiodic Brandt is congruence-free. If you identify triples whose outer elements are different you force those triples to become 0. $\endgroup$ Jan 3, 2017 at 13:29
  • $\begingroup$ You mean to say that $B_{\lambda}^o(S)$ is a free congruence. $\endgroup$
    – user120386
    Jan 3, 2017 at 18:31
  • $\begingroup$ No it has congruences. You can take lambda Brandt over over a quotient of S. $\endgroup$ Jan 3, 2017 at 18:58
  • $\begingroup$ on way for the proof is to show $T$ is isomorphic $ I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\}$ and other way is to show $T \subseteq I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\}$ and $ I_{\lambda} \times T_o \times \ I_{\lambda} \cup \{0_T\} \subset eq $. $\endgroup$
    – user120386
    Jan 3, 2017 at 19:20
  • $\begingroup$ The same question was posted on math.stackexchange.com $\endgroup$
    – J.-E. Pin
    Jan 4, 2017 at 13:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.