Reconsideration of the universal mapping property for quotients: If R is a ring and I is a 2-sided ideal in R, an abstract quotient of R modulo I is a ring homomorphism pi : R -----> A with I <= Ker(pi) satisfying the condition that if f : R -----> S is any ring homomorphism with I <= Ker(f), then there is a unique ring homomorphism lambda : A -----> S such that lambda\circ pi = f .
If it is not assumed that I <= Ker(pi), but merely that each of the homomorphisms f factors uniquely through pi as above, then pi need not be surjective. Prove that this is the case when
R is the ring Z of integers.
I is the principal ideal pZ where p > 1 is prime.
A is the localization Z_{p} of Z at p:
pi is the unique ring homomorphism Z -----> Z_{p}.
In the ring Z[t] of polynomials with coefficients in the ring Z of integers, let I be the ideal of all f for which f(0) = 0, and let J be the ideal of all f for which f(0) is an even integer. Verify the following:
I is a principal ideal.
I is a prime ideal.
J is a maximal ideal.
J is not a principal ideal.
Let A be a commutative ring in which 2 is a unit. Show that the quotient ring A[t]/(t^{2}-1)A[t] is isomorphic to the ring-theoretic Cartesian product A \times A.
For a given field K determine up to equivalence all valuations of the field K(t) of ``rational functions'' with coefficients in K that are trivial on K (when embedded as the subfield of constants).