By Bernard R. McDonald

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1 0 0 0 0 f0 0 0 = Step 7. 0 2 0 0 0 f0 0 0 = 0. Step 9. 0 0 0 1 0 f0 0 0 = = 0 0 2 0 0 0 0 f0 Step 8. 2 f0 f0 0 . Step 6. 0 0 f0 0 0 0 0 1 . Step 4. 0 f0 0 0 0 0 0 f0 0 0 0 0 0 0 = 0. = 0. = 0. Thus by Step 1 through Step 9, there is a multiplication •3 on V such that •3 extends the R-module multiplication of V over R: a1 + f0 · r1 f0 · s1 2b1 c1 a2 + f0 · r2 f0 · s2 •3 2b2 c2 x y z w = , where x = a1 a2 + 2s1 r2 + 2a1 s2 + 2c1 s2 + f0 · r1 a2 + f0 · a1 r2 + f0 · r1 r2 , y = 2a1 b2 + 2r1 b2 + 2b1 c2 , z = f0 · s1 a2 + f0 · c1 s2 + f0 · s1 r2 , w = 2s1 b2 + c1 c2 .

The multiplications •1 , •2 , •3 , and •4 are well defined and they extend the R-module multiplication of V over R. Thus (V, +, •1 ), (V, +, •2 ), (V, +, •3 ), and (V, +, •4 ) are all possible compatible ring structures on V . AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 25 13 Define θ2 : (V, +, •2 ) → (V, +, •1 ) by θ2 a + f0 · r f0 · s 2b c = a + 2r + f0 · r f0 · s 2b c . Then we see that θ2 is a ring isomorphism. Also define θ3 : (V, +, •3 ) → (V, +, •1 ) and θ4 : (V, +, •4 ) → (V, +, •1 ) by θ3 a + f0 · r f0 · s 2b c = a + 2s + f0 · r f0 · s 2b c , and a + 2r + 2s + f0 · r 2b a + f0 · r 2b = .

Note that if v = g r 0 0 because v = ve1 . By Claim 1, and noting that V = e1 V e1 +e1 V e2 +e2 V e1 +e2 V e2 , A 0 Thus e1 V e1 = f0 0 0 0 0 0 or e1 V e1 = a 0 = 0 0 0 2b 0 0 + p+f g + 0 0 + 0 0 0 c a + p + f 2b f0 0 , = g c 0 0 hence g = 0, c = 0, 2b = 0, and f0 = a + p + f . Thus a + p = 0 and f = f0 . Hence with a, b, p, c ∈ A and f, g ∈ Hom (2AA , AA ). So f0 0 0 0 = a 0 0 0 Since RR ≤ess VR , there is 0= −a + f0 0 −a + f0 0 + 0 0 0 0 −a + f0 0 with x 0 2y z ∈ R such that x 0 2y z = −ax + f0 · x 0 0 0 −2ay + 2y 0 ∈ e2 V e1 .

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