By Quillen D.

Unpublished MIT lecture notes

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Unpublished MIT lecture notes

**Rings, Extensions, and Cohomology**

"Presenting the court cases of a convention held lately at Northwestern collage, Evanston, Illinois, at the party of the retirement of famous mathematician Daniel Zelinsky, this novel reference offers updated insurance of themes in commutative and noncommutative ring extensions, specially these related to problems with separability, Galois thought, and cohomology.

On the center of this brief advent to type conception is the assumption of a common estate, vital all through arithmetic. After an introductory bankruptcy giving the fundamental definitions, separate chapters clarify 3 ways of expressing common houses: through adjoint functors, representable functors, and bounds.

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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.