Ideals of a polynomial ring in infinitely many variables which are not finitely generated I'm doing this exercise in the Dummit and Foote textbook and got no clue for it. Hope some one can help me solve this. Thanks

Prove that a polynomial ring in a infinitely many variables with coefficients in any commutative ring contains ideals that are not finitely generated.

 A: Finding an infinitely generated ideal in particular means picking an infinite number of elements to do the generating. What is the most obvious infinite set to pick? The fact that there are a specifically infinite number of variables suggests an immediate answer! This is your "clue."

Here's a hint for how to prove this ideal is not finitely generated. Suppose $(x_1,x_2,\cdots)$ were finitely generated by a finite number of multivariable polynomials $\{\pi_1,\cdots,\pi_m\}$. How many variables appear in this finite generating set of polynomials? A finite number, say $\{x_s:s\in S\}$ with $S$ a finite subset of $\Bbb N$. So is $x_r\in(\pi_1,\cdots,\pi_m)$ possible if $r\not\in S$?
A: Hint: Consider the ideal $I=\langle x_1,x_2,x_3... \rangle$. 
Following what anon said. You can apply the same trick one below to prove what annon is asking you. 
Alternate: Let us prove that $R'=R[x_1,...]$ is not noetherian. Take the chain $(x_1)\subset (x_1,x_2)\subset\cdots$. If this stabilizes, then for some $n$ we have $(x_1,...,x_n)=(x_1,..,x_{n+1})$. Then $x_{n+1}=h_1x_1+\cdots+h_nx_n=f$. Evaluate $f$ at the point that is $0$ in every coordinate except $n+1$, here say it is $1$. Then you get $1=0$. A contradiction. 
Remember that a different characterization of Noetherian is that every ideal is finitely generated. Then by above since it is not noetherian, we have that there must be an ideal that is not finitely generated.
