It is well known that there exist divisors (on a normal projective variety over say the complex numbers) that are big (= is a sum of an ample and an effective divisor) but not ample. However, if we strengthen ample to very ample, does the statement hold true?
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See the linked duplicate for an instance of a very ample divisor which when added to an effective divisor does not give a very ample divisor. – KReiser Dec 19 '23 at 03:14
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Sorry about that. I've changed the question slightly, haven't found an answer. – Calculus101 Dec 19 '23 at 10:39
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Now asked and answered on MO. – KReiser Dec 27 '23 at 17:50