By Irena Swanson
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Extra resources for Abstract Algebra [Lecture notes]
In particular, we want to finish finding the stable configuration when the graph is overloaded by one grain of sand in the center. By the lemma, we can topple overfull vertices one at a time, or all at the same time, so for speed we proceed with all overfull vertices at the same time. 2 1 0 3 3 3 3 4 3 3 3 3 0 1 2 3 4 3 1 → 4 0 4 4 3 0 2 3 0 1 2 5 0 5 1 → 0 4 0 0 5 0 2 5 0 1 2 1 3 1 1 → 3 0 3 0 2 1 0 3 1 1 2 This is one possible stable configuration, so an element of G. By dropping extra allowable grains of sand on the center or corner positions, this gives us actually 4 · 34 elements of G (4 possibilities for the center, 3 for each of the corners).
There exists M ∈ S such that for all I ∈ S, either I and M are incomparable, or else I ≤ M . We claim that M is a maximal element in R. First of all, since M ∈ S, M = R. Secondly, if I is an ideal in R such that M ⊆ I, then either I ∈ S or I ∈ S. In the latter case, necessarily I = R. In the former case, M ⊆ I implies M ≤ I, but by the definition of M , I ≤ M , whence I = M . This proves that M is a maximal ideal in R. * Zorn’s lemma is, contrary to its name, not something to prove: either you assume it for your set theory or you don’t, and no contradictions arise.
2 The only prime ideals in Z are (0) and (p), where p varies over the prime integers. 3 If R is a field, the only prime ideal is (0). 4 If X is a variable over R, then (X 2 + 1), X − r, as r varies over R are all prime ideals. 5 Let R = Q[X]/(X 2). Let P be a prime ideal in R. Certainly 0 = X 2 ∈ P , so by the definition of prime ideals, X ∈ P . Now verify that XR is a prime ideal. 7 Let R be a commutative ring with unity and let I be an ideal of R. Then I is a prime ideal if and only if R/I is a domain.
Abstract Algebra [Lecture notes] by Irena Swanson