(February 2008) Let \(\varphi:R\to S\) be a ring homomorphism. Let \(I\subseteq R\) be an ideal of \(R\), and set \[ J=\{x\in I\mid \varphi(x)=0_S\}, \] where \(0_S\) denotes the zero element of \(S\). Prove that \(J\) is an ideal of \(R\).
Solution
Proof. (Nonempty) We have 0 R∈I and φ(0R) = 0S, so 0 R∈J. Thus, J⁄= ∅. (Closure under−) Given x,y∈J, we have x,y∈I and hence x−y∈I. Moreover, φ(x−y) =φ(x)−φ(y) = 0S− 0S = 0S, so x−y∈J. (Sticky) Given x∈J and r∈R, we have x∈I, and hence rx,xr ∈I. Moreover, φ(rx) =φ(r)φ(x) =φ(r)0S = 0S, and φ(xr) =φ(x)φ(r) = 0Sφ(r) = 0S, so rx,xr ∈J. The Fundamental Theorem of Ring Homomorphisms, Also called the First Homomorphism Theorem for Rings or the Basic Homomorphism Theorem for Rings , this theorem says that if φ : R→ S is a ring homomorphism, then there is a ring isomorphism ~φ :R/Ker(φ)≃ Im(φ) defined by ~φ(Ker(φ) +r) =φ(r). 12 Quotient Rings and Fields Criteria for R to be a Field. Know that a commutive ring with unit is a field if and only if its only ideals are{0} and the whole ring. Be prepared to prove this. For example, here is a problem asking to prove one direction of this statement: