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Part CCSIR NET June 2024invertible-in-the-target-is-not-invertible-in-the-image

Invertible in the target is not invertible in the image

Let R and S be non-zero commutative rings with multiplicative identities 1_R, 1_S, respectively. Let f : R → S be a ring homomorphism with f(1_R) = 1_S. Which of the following statements are true?

  1. A.If f(a) is a unit in S for every non-zero element a ∈ R, then S is a field
  2. B.If f(a) is a unit in S for every non-zero element a ∈ R, then f(R) is a field
  3. C.If R is a field, then f(a) is a unit in S for every non-zero element a ∈ R✓
  4. D.If a is a unit in R, then f(a) is a unit in S✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: Every prime ideal is maximal

More on this topic

The chapter behind this: Ideals, quotients and the Chinese remainder theorem — free to read

From Rings and Fields › Ideals, quotient rings, prime & maximal ideals, CRT

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