UCLA Algebra Qualifying Exam Solutions
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چکیده
Problem 2. Let A be a central division algebra (of finite dimension) over a field k. Let [A, A] be the k-subspace of A spanned by the elements ab − ba with a, b ∈ A. Show that [A, A] = A. Solution. Let K be the algebraic closure of k, and consider B = A ⊗ k K. Then B ∼ = M n (K) for some n ∈ N, and thus we can understand [B, B] ∼ = [A, A] ⊗ k K. In this case, [B, B] contains only matrices of trace 0, since it is clear that tr(XY − Y X) = 0 for every X, Y ∈ B. Therefore [B, B] = B. As such, we could not have had [A, A] = A. Problem 3. Given ϕ : A → B a surjective morphism of rings, show that the image by ϕ of the Jacobson radical of A is contained in the Jacobson radical of B. Solution. We use the following characterisation of the Jacobson radical: J(R) = {x ∈ R : xy − 1 ∈
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UCLA ALGEBRA QUALIFYING EXAM Solutions
Proof. Let L/F be an algebraic extension. Let f : L −→ L be a homomorphism fixing F . Recall that field homomorphisms are always injective, it remains to show that it is surjective. Let a ∈ L. As L/F is algebraic, there exists a1, . . . , ad ∈ F such that a satisfy p(x) = x + a1xd−1 + . . .+ ad. Let S = {s ∈ L : p(s) = 0}. As f is a homomorphism fixing the coefficients of the polynomial p(x), i...
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