Question 1. Consider the following argument:
Let U be the set of all sets. Define a partial ordering on U by inclusion: A ≤ B iff A ⊆ B for A,B ∈ U. Consider a chain C of U under this partial ordering: C : A1 ≤ A2 ≤ A3 ≤ ···. Define B = ∪i≥1Ai. Clearly, B ∈ U and it is an upper bound of the chain C. Hence, Zorn’s Lemma implies that U has a maximal element, say M.
The argument is clearly wrong since M is not a maximal element: M ⊂ {M,{M}} ∈ U. Identify which step in the argument is wrong and why.
Question 2. Let (G,·) be a finite group with the property that there exists only one element a2 ∈ G such that a2 6= e and . Define a bipartite graph
H = (G,G,E) as follows.
Edge (a,b) ∈ E if a 6= e and b = ak for k > 1, or a = e and b = a2.
Prove that the graph H has a perfect matching.
Question 3. Let R be a ring and a ∈ R. Define (a) = {b·a | b ∈ R}. Prove that (a) is an ideal of R.
Let polynomial C(x,y) = (x2 +y2 −1)·x. The curve C(x,y) = 0 is unit circle plus y-axis on the plane. (C) = {Q(x,y) · C(x,y) | Q(x,y) ∈ R[x,y]} is an ideal of the ring R[x,y], the ring of polynomials in two variables with coefficients in R.
Define R = R[x,y]/(C). For any point P ∈ R × R on the plane, define
R and g(P) 6= 0} and and g(P) 6= 0 and f(P) = 0}. Prove that
RP is a ring.
IP is a maximal ideal of RP.
For point P = (1,0), IP = (y).
For point P = (0,1), (x) ⊆ IP.
It can be shown that IP 6= (x). Therefore, ring RP contains information about whether curve C is degenerate at point P.
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