Skip to content
Part CCSIR NET June 2025injective-and-surjective-part-company-once-the-ring-is-infinite-dimensional

Injective and surjective part company once the ring is infinite dimensional

Let f:Q[X]→Q[X]f : \mathbb{Q}[X] \to \mathbb{Q}[X] be a ring homomorphism with f(1) = 1. For n ≥ 1, let fn=ff^{n} = f ∘ ⋯ ∘ f (n times). Which of the following statements are true?

  1. A.If f is onto, then so is fnf^{n} for all n ≥ 1.✓ Correct
  2. B.ker fn+1f^{n+1} =ker⁡fn= \ker f^{n} for some n ≥ 1.✓ Correct
  3. C.If f is onto, then f is one-to-one.✓ Correct
  4. D.If f is one-to-one, then f is onto.

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, and why this one tests finite-dimensional intuition.

See pricing

50 are analysed free — try those first.

The trap it tests

Finite-dimensional intuition

Something true in ℝⁿ, or in a nice space, assumed in general.

Drill statements like this

Related counterexample: Irreducible over ℤ ⇒ irreducible mod every prime

More on this topic

The chapter behind this: Irreducibility tests — free to read

From Rings and Fields › Polynomial rings and irreducibility tests

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram