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Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module.

I have no idea what can I do here, I mean, suppose $I$ is finitely generated, then there is a subset $S \subset I$ with $S=\{x_{i_1},...,x_{i_n}\}$ such that $\langle\{x_{i_1},...,x_{i_n}\}\rangle=I$. How can I arrive to a contradiction? Any suggestions would be appreciated.

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To start with, let me correct your claim: suppose $I$ is finitely generated, then there is a subset $S \subset I$ with $S=\{f_{1},...,f_{n}\}$ such that $\langle S\rangle=I$, where $f_i$ are polynomials in $I$.

These polynomials need only finitely many generators, say $x_1,\dots,x_N$. That is, we can write $f_i=x_1g_{i1}+\cdots+x_Ng_{iN}$ for all $i=1,\dots,n$. Since we know that $x_M\in I$ for $M>N$ and $I=\langle f_1,\dots,f_n\rangle$ we get $x_M=f_1h_1+\cdots+f_nh_n$. Now just evaluate this by sending $x_j$ to $0$, for all $j=1,\dots,N$ and get a contradiction.

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  • $\begingroup$ why u can just send $x_j$ to 0. Since $h_1, ..., h_n \in A$(so i can have a $h_i$ with$ x_M $in it), why i can be sure that $h_i f_j $ won't give me a polynomial with $x_M$ in it? $\endgroup$ – user10024395 Apr 14 '15 at 13:43
  • $\begingroup$ @user136266 Sure, but the $f_i$s are all zero, a contradiction. $\endgroup$ – user26857 Apr 14 '15 at 13:48
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A finite subset of $A$ contains words, which are made of a finite number of the $x_i$s. Given such a finite set $S$, there is $i$ such that $x_i$ does not appear in any element of $S$. It follows that $x_i\not\in\langle S\rangle$, unless $S$ contains some non-zero element of $K$, in which case $\langle S\rangle=A$.

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