By Berstel J., Perrin D.

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1. For m ∈ I − {0} and n ∈ M such that mn = 0, we have mRmn. 2. For m ∈ I − {0} and n ∈ M such that nm = 0, we have mLnm. 3. For any H class H ⊂ I − {0} we have H 2 = H or H 2 = {0}. Proof. 9. Let us prove 1. We have mnM ⊂ mM . Since mM is a 0-minimal right ideal and mn = 0, this forces the equality mnM = mM . The proof of 2 is symmetrical. Finally, to prove 3, let us suppose H 2 = {0}. Let h, h ∈ H be such that hh = 0. Then, by 1 and 2, hRhh and h Lhh . Since hLh and h Lhh , we have hLhh . 10 for minimal ideals instead of 0-minimal ideals.

For m ∈ K P ×Q and n ∈ K Q×R , the product mn is defined as the following element of K P ×R . Its value on (p, r) ∈ P × R is (mn)p,r = mp,q nq,r q∈Q When P = Q = R, we thus obtain an associative multiplication which turns K Q×Q into a monoid. Its identity is denoted IQ . A monoid of K-relations over Q is a submonoid of K Q×Q . It contains in particular the identity IQ . Version 19 juin 2006 J. Berstel and D. Perrin 7. Formal series 7 29 Formal series Let A be an alphabet and let K be a semiring.

Berstel and D. Perrin 6. Semirings and matrices 27 where ∞ ∈ K. The operations of K are extended to K by setting for x ∈ K, (i) x + ∞ = ∞ + x = ∞ (ii) if x = 0, then x∞ = ∞x = ∞ (iii) ∞∞ = ∞, 0∞ = ∞0 = 0 Extending the order of K to K by x ≤ ∞ for all x ∈ K, the set K becomes a totally ordered semiring. It is a complete semiring because any subset has an upper bound and therefore also a least upper bound. We define N = N ∪ ∞, R + = R+ ∪ ∞ to be the complete semirings obtained by applying this construction to N and R+ .