Given a homomorphism of rings 0 RS and an element t S prove

Given a homomorphism of rings ?₀: R?S and an element t ? S, prove that there is a unique homomorphism ?: R[x]?S so that ?(r) = ?₀(r), for all r ? R, and ?(x) = t. Here R[x] denotes the ring of polynomials over R with indeterminate x. Hint: If ? exists, then ?(a0 + …+ aix i + …+ anxn ) = ?₀(a0) +…+ ?0(ai)ti +…+ ?0(an)tn .

6. Given a homomorphism of rings Po: R S and an element t E S, prove that there is a unique homomorphism op: RIx] S so that p(r) F (po (r) for all r E R, and p(x) t. Here RIx] denotes the ring of polynomials over R with indeterminate x. Hint: If p exists, then op(ao aix

Solution

Solution: First, if : R ! S is one-to-one, and ; : T ! R are ring homomorphisms
with = , then it follows immediately that (a) = (a) for all a 2 T, and so
= .
Conversely, suppose that : R ! S is a ring monomorphism. Consider the subset
T = f(a; b) j a; b 2 R R and (a) = (b)g of the direct sum R R. This subset
is closed under addition since if (a1) = (b1) and (a2) = (b2), then (a1 + a2) =
(a1) + (a2) = (b1) + (b2) = (b1 + b2). It is easily checked that it contains
(0;0) and is closed under taking additive inverses. Furthermore, if (a1) = (b1) and
(a2) = (b2), then (a1a2) = (a1)(a2) = (b1)(b2) = (b1b2), and so the subset is
closed under multiplication. The identity (1;1) belongs to the subset, so we conclude
that T is a subring of the direct sum R R.
De ne : T ! R by ((a; b)) = a, and de ne : T ! R by ((a; b)) = b. It can be
checked easily that and are ring homomorphism, and it follows from the de nition
of T that ((a; b)) = (a) = (b) = ((a; b)). Since is a ring monomorphism,
we must have = . It follows that if (a) = (b), then a = b, and therefore is
one-to-one.