By Hamilton W.R.

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**Example text**

Suppose M and N are normal sub-orthocryptogroups of S. Take m ∈ M and n ∈ N . Note that S satisfies the Eqs. 3). Then we have mn = mn(mn)0 = mnm −1 mn 0 ∈ N M since mnm −1 ∈ N and n 0 ∈ M. Thus, M N ⊂ N M and vice versa. Hence, M N = N M. Then (M N )(M N ) = M M N N = M N and so M N is closed under multiplication. Next we take m ∈ M and n ∈ N . We have (mn)−1 = m 0 n −1 m −1 n 0 ∈ M N M N = M N . Hence, M N is closed under taking inverse. Since M and N are full, M ⊂ M E(S) ⊂ M N and N ⊂ E(S)N ⊂ M N .

On the other hand, two concepts are not equivalent in general as we give examples of external spined products that admit no internal spined product decomposition. Further, we examine internal spined product of orthocryptogroups. Using a lattice theoretic method, we obtain a unique decomposition theorem similar to the Krull–Schmidt theorem in group theory. We also study completely reducible orthocryptogroups in which any normal sub-orthocryptogroup is a spined factor. We show that such an orthocryptogroup is an internal spined product of simple sub-orthocryptogroups.

If deg(g0 (x)) ≥ 1, then there exists α ∈ F such that g0 (α) = 0. Thus γ(α, y) cannot be defined. On the other hand, we note that γ(α, y) = h0 (y)/k0 (y) + (h1 (y)/k1 (y))α + · · · + (hn (y)/kn (y))αn , which is a contradiction. Thus g0 (x) ∈ F. Similarly, g1 (x), . . , gm (x) ∈ F. Hence γ(x, y) ∈ F[x, y]. Therefore F(x)[y] ∩ F(y)[x] = F[x, y], and so QE (R) = Matk (F(x)[y] ∩ F(y)[x]) = Matk (F[x, y]). 21 because the commutative domain F[x, y] is not Prüfer. Therefore R = Matk (F[x, y]) has no right extending ring hull.