Show that TFAE:
\(A\in \Field\)
\(A\) is a simple ring, so \(\Id(A) = \ts{ 0, A }\).
If \(B\in \Field\) is nonzero then every ring morphism \(A\to B\) is injective.
Solution
Let \(A\) be a commutative ring with \(1\neq0\); ring homomorphisms preserve \(1\). Number the three conditions (1), (2), (3) in the order stated.
1. (1) implies (2).
Let \(I\) be a nonzero ideal of the field \(A\) and \(0\neq x\in I\). Then \(1=x^{-1}x\in I\), so \(a=a\cdot1\in I\) for every \(a\in A\) and \(I=A\).
2. (2) implies (3).
Let \(\varphi\colon A\to B\) be a ring homomorphism to a field. \(\ker\varphi\) is an ideal, and \(\varphi(1)=1\neq0\) gives \(\ker\varphi\neq A\). By (2), \(\ker\varphi=0\), so \(\varphi\) is injective.
3. (3) implies (1).
Let \(0\neq x\in A\) and suppose \((x)\neq A\). The proper ideal \((x)\) lies in a maximal ideal \(\mathfrak m\) (Zorn’s lemma), and \(\pi\colon A\to A/\mathfrak m\) is a homomorphism to a field. By (3), \(\mathfrak m=\ker\pi=0\), so \(x\in\mathfrak m=0\), a contradiction. Hence \((x)=A\), so \(x\) is a unit, and \(A\) is a field.
4. Q.E.D.
Steps 1, 2 and 3 give \((1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)\).