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    Every nonzero homomorphism from a field is injective, and whether the same holds for domains

    P-Z4O2M

    Area
    Algebra
    Topics
    Fields, Homomorphisms, Integral Domains
    Status
    draft

    Let \(R\) and \(S\) be commutative rings with multiplicative identity.

    1. Prove that when \(R\) is a field, every non-zero ring homomorphism \(\phi: R\to S\) is injective.

    2. Does (a) still hold if we only assume that \(R\) is a domain? If so, prove it, and if not provide a counterexample.

    Solution

    Let \(R\) be a field and let \[ \phi:R\to S \] be a nonzero ring homomorphism. Its kernel is an ideal of \(R\), so \[ \ker\phi=(0)\quad\text{or}\quad R. \] The second possibility would make \(\phi\) the zero map. Hence \[ \ker\phi=(0), \] so \(\phi\) is injective. This argument does not require a convention that ring homomorphisms preserve \(1\).

    The statement fails for integral domains. For example, \[ \mathbb Z\to\mathbb Z/2\mathbb Z, \qquad n\mapsto n\bmod2, \] is a nonzero ring homomorphism, but \[ \ker\phi=2\mathbb Z\ne0. \] Thus a nonzero homomorphism from a domain need not be injective.

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