Theorem: The forgetful functor is representable. In particular .
Proof: First, we establish the following fact. For a commutative ring and element , define the evaluation map by , the map that evaluates the given polynomial at inside . Note that . This is indeed the unique homomorphism that sends . This is straightforward to see: Suppose were a homomorphism with . Let . Then,
and thus .
The preceding fact is most of the content of the proof. Indeed, define the natural transformation with components taking an element to the map . We verify naturality: Let be a ring homomorphism. We must have . Again is just left-composition by , and is just . So, let . Applying the right side to yields . Applying the left side to yields . But of course,
and . So is natural and .