 By I. M. Isaacs

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Additional info for Algebra [Lecture notes]

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We also note that given a right R-module M , we may naively try to make it a left R-module by defining rm = mr. Yet this only actually works if our multiplication is commutative, so we should not be fooled. Let M be a right R-module. Then N ⊆ M is a submodule if N is an additive subgroup of M and for n ∈ N and for all r ∈ R, we have that nr ∈ N . We continue with our definitions. 3. Suppose that U, V are right R-modules and ϕ : U → V is an additive homomorphism. Then ϕ is an R-module homomorphism if (ur)ϕ = (uϕ)r for all u ∈ U and r ∈ R.

Therefore I ∩ ann(x) = 0 and e − e2 is equal to zero. Hence e2 = e, and e is idempotent. Now e ∈ I and eR ⊆ I. We again use the minimality of I to conclude that eR is either 0 or I. If eR = 0, then e = 0, and x = xe is zero, which is contrary to our selection of x, e. Thus eR = I, as desired. We note that if e ∈ R is idempotent, then so is (1 − e). This is true as (1 − e)(1 − e) = 1 − 2e + e2 , which is 1 − 2e + e = 1 − e. This fact will be useful when proving the next lemma. 3. Let e ∈ R be idempotent and eR ⊆ ˙ where V is a right ideal.

Also, j ∈ J(R) ⊆ M and 1−j ∈ M . So j + (1 − j) = 1 ∈ M . Yet this contradicts the fact that M is maximal in R, so we must have (1 − j)R = R, and hence j is rqr. 2. Let I ⊆ R be a right ideal, and assume that all elements of I are rqr. Then I ⊆ J(R). Proof. It suffices to show that I ⊆ M for all maximal right ideals M of R. So fix M a maximal right ideal, and assume that I M . Then as I + M is a right ideal of R, we must have that I + M = R and hence we have elements i ∈ I and m ∈ M such that i + m = 1, or (1 − i) = m.