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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 4 / c / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 20 4 c
Created 2026-10-03 Updated 2026-10-06  0 By others on same topic  0 Discussions Create my own version
Over an affine open subscheme SpecA of the target, a closed immersion has the form Spec(A/I)→SpecA. Pulling it back along an affine scheme map SpecB→SpecA gives
Spec((A/I)⊗A​B)≃Spec(B/IB).
(1)
The identification follows directly by tensoring the quotient presentation; it does not require B to be flat over A. Since B→B/IB is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.
Being a closed immersion is stable under arbitrary base change of a morphism of schemes.

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