Homology of commutative rings by Quillen D.

By Quillen D.

Unpublished MIT lecture notes

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Homology of commutative rings

Unpublished MIT lecture notes

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For 1 ≤ i ≤ n, we have canonical projections πi : M n −→ M m = (mj )1≤j≤n −→ mi and canonical injections ιi : M −→ M n m −→ (0, . . , 0, m, 0, . . , 0). Let f : M −→ M be an endomorphism. For 1 ≤ i ≤ n, let fi = πi ◦ f . Then fi : M n −→ M is R-linear and for all m ∈ M n , we have n n f (m) = (f1 (m), . . , fn (m)). Now observe that fi (m) = fi (ι1 (m) + · · · + ιn (m)) = fi (ι1 (m)) + · · · + fi (ιn (m)). Putting things together, we finally get n n πi ◦ f ◦ ι1 , . . , f= i=1 πi ◦ f ◦ ιn . i=1 The idea of the proof is that, according to the formula above, f should be completely determined by the maps πi ◦ f ◦ ιj .

Let us show that Q = (i, 1 + 2i)Q(i) is a division algebra. Assume to the contrary that Q is split. 5, we have 1 + 2i = NQ(i)(ζ8 )/Q(i) (ξ) for some ξ ∈ Q(i)(ζ8 )× , where ζ8 denotes a primitive 8-th root of 1. As in the previous example, this implies that (1 + 2i)z 2 = x2 − iy 2 , for some x, y, z ∈ Z[i]. 1. PROPERTIES OF QUATERNION ALGEBRAS 25 Since Z[i] is a unique factorization domain, one may assume that x, y, z are coprime, and show as before that 1 + 2i y, using that 1 + 2i is an irreducible element of Z[i].

This contradicts the minimality of m. Hence m = 1, so I contains an element of the form 1 ⊗ b. Since B is simple, arguing as at the beginning of the proof shows that I contains 1 ⊗ 1, so I = A ⊗k B and we are done. 7. This result is not true if A is not central. For example, C is a simple R-algebra. However, we have ∼R C × C. C ⊗R C = Since C × {0} is a non-trivial ideal of C × C, it follows that C ⊗R C is not simple. 8. Let A and B be two k-algebras, and let L/k be a field extension. Then the following properties hold: (1) if A and B are central simple, so is A ⊗k B; (2) A is central simple over k if and only if A ⊗k L is central simple over L.

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