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Let $K\subset L$ be a field extension and let I be an ideal of the polynomial ring $K[X_1,...,X_n]$. I want to show that if $1\in J=(I)_{L[X_1,...,X_n]}$ then $1\in I$.$\quad$ (J is the ideal of $L[X_1,...,X_n]$ generated by the elements of I). For the case K is an algebraically closed field, i suppose that $1\notin I$ then there exist a maximal ideal m of $K[X_1,...,X_n]$ such that $I\subset m$ and we know that in this case $m=(X_1-a_1,...,X_n-a_n)$ for some $(a_1,...,a_n)\in K^n$ and we continue in this way we get a contradiction... Now i don't know what to do if K is not necessarily algebraically closed field. Thanks for your help.

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Hint:

$L[X_1,\dots,X_n]$ is a faithfully flat $K[X_1,\dots,X_n]$-algebra, hence $$IL[X_1,\dots,X_n]\cap K[X_1,\dots,X_n]=I.$$

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  • $\begingroup$ Thank you so much for this quick proof. I'm looking for a proof according to the best of my knowledge... but i see your proof hides a lot of important thing, so i ask if you can send some reference to see this in details. $\endgroup$ – SAM Jun 14 '17 at 2:47
  • $\begingroup$ Here are two: Matsumura, Commutative Ring Theory, Ch.3 Properties of Ring Extensions, theorem 7.5. Also Bourbaki, Commutative Algebra, ch. I, Flat Modules, §3 (Faithfully Flat Modules), n°5, proposition 8. $\endgroup$ – Bernard Jun 14 '17 at 8:55

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