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alg. geometry

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alg. geometry

how can we show that
"ideal of an algebraic variety is radical"


hint: if f^n lies in the ideal I of the variety V, then f^n(x)=0 for every x in V. If we're working over a field, f(x) = 0 for all x in V, so f lies in I too. Therefore the radical of I is contained in I.

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