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Fall 2016 - Problem 9

category theory

Let FF be a field and a∈Fa \in F. Prove that the functor from the category of commutative FF-algebras to Set\Set taking an algebra RR to the set of invertible elements of the ring R[X]/⟨X2−a⟩R\br{X}/\gen{X^2 - a} is representable.
Solution.

Denote the class of XX in the quotient via xx. Note that cx+dcx + d is invertible if and only if there exist c′,d′∈Rc', d' \in R such that

1=(cx+c)(c′x+d′)=cc′x2+(cd′+c′d)x+dd′=(cd′+c′d)x+(acc′+dd′).\begin{aligned} 1 &= \p{cx + c}\p{c'x + d'} \\ &= cc'x^2 + \p{cd' + c'd}x + dd' \\ &= \p{cd' + c'd}x + \p{acc' + dd'}. \end{aligned}

Consider the FF-algebra A=F[Z,W,Z′,W′]/⟨Z′W+Z′W,WW′+ZZ′a−1⟩A = F\br{Z, W, Z', W'}/\gen{Z'W + Z'W, WW' + ZZ'a - 1}. Any homomorphism φ ⁣:A→R[X]/⟨X2−a⟩\func{\phi}{A}{R\br{X}/\gen{X^2 - a}} is uniquely determined by φ(z),φ(z′),φ(w),φ(w′)\phi\p{z}, \phi\p{z'}, \phi\p{w}, \phi\p{w'}, and note that φ(z)x+φ(w)\phi\p{z}x + \phi\p{w} is invertible by construction. In fact, because inverses are unique, φ\phi is determined uniquely by φ(z),φ(w)\phi\p{z}, \phi\p{w}.

Conversely, given an invertible cx+dcx + d with inverse c′x+d′c'x + d', define f ⁣:F[Z,W,Z′,W′]→R[X]/⟨X2−a⟩\func{f}{F\br{Z, W, Z', W'}}{R\br{X}/\gen{X^2 - a}} via (Z,W,Z′,W′)↦(c,d,c′,d′)\p{Z, W, Z', W'} \mapsto \p{c, d, c', d'}. By construction, f(⟨Z′W+Z′W,WW′+ZZ′a−1⟩)=0f\p{\gen{Z'W + Z'W, WW' + ZZ'a - 1}} = 0, so ff descends to a function φ ⁣:A→R[X]/⟨X2−a⟩\func{\phi}{A}{R\br{X}/\gen{X^2 - a}}.

Thus, we obtain a bijection between (R[X]/⟨X2−a⟩)×\p{R\br{X}/\gen{X^2 - a}}^\times and homomorphisms φ ⁣:A→R[X]/⟨X2−a⟩\func{\phi}{A}{R\br{X}/\gen{X^2 - a}}, i.e., if F\mathcal{F} is the functor in question, then

F(−)≃Hom⁡F(A,−),\mathcal{F}\p{-} \simeq \Hom_F\p{A, -},

so AA represents F\mathcal{F}.