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I am trying to prove the following statement:

Let R be a ring and $I=\{\sum_{i=1}^n a_i x_i : a_i\in R\}$ the ideal generated by $S=\{x_1,\ldots, x_n\}$. Then $I$ is the intersection of ideals $J$ in R containing $S$.

Let $a=\sum_{i=1}^n a_i x_i\in I$ then $a$ is in every J as $x_i\in J$, so $a_ix_i\in J$, so also their sum. So $I$ is a subset of the intersection of all J. On the other hand $I$ is an ideal containing S, so the intersection is a subset of I.

Is this proof valid?

Phil-ZXX
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1 Answers1

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Yes, it is perfect. $\ \ \ \ \ \ \ \ \ \ $

Berci
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    Well, provided that the ring is commutative and has unit element. Else, the generated ideal looks a bit different... Can you also write that down? – Berci Mar 24 '13 at 13:40