[proofplan] We construct an infinite strictly ascending chain of ideals in $R$. For each positive integer $n$, let $I_n$ be the ideal generated by the first $n$ variables. The inclusions $I_n \subset I_{n+1}$ are immediate, and properness follows by evaluating $x_1, \ldots, x_n$ at $0$ while leaving $x_{n+1}$ unchanged. Since a [Noetherian ring](/page/Noetherian%20Ring) satisfies the ascending chain condition on ideals by definition, this strictly increasing chain proves that $R$ is not Noetherian. [/proofplan] [step:Construct the ascending chain generated by the first variables] Let \begin{align*} R := k[x_1, x_2, x_3, \ldots]. \end{align*} For each positive integer $n$, define the ideal \begin{align*} I_n := (x_1, \ldots, x_n) \trianglelefteq R. \end{align*} Equivalently, \begin{align*} I_n = \left\{\sum_{i=1}^{n} x_i f_i : f_1, \ldots, f_n \in R\right\}. \end{align*} Since the generating set $\{x_1, \ldots, x_n\}$ is contained in $\{x_1, \ldots, x_n, x_{n+1}\}$, we have \begin{align*} I_n \subset I_{n+1} \end{align*} for every positive integer $n$. [/step] [step:Prove that each inclusion is proper by evaluating the earlier variables at zero] Fix a positive integer $n$. Define the $k$-algebra homomorphism \begin{align*} \pi_n: R \to k[x_{n+1}, x_{n+2}, x_{n+3}, \ldots], \qquad x_i \mapsto 0 \text{ for } 1 \le i \le n, \qquad x_i \mapsto x_i \text{ for } i \ge n+1. \end{align*} This map is well-defined because every element of $R$ is a polynomial involving only finitely many variables. If $f \in I_n$, then there exist $f_1, \ldots, f_n \in R$ such that \begin{align*} f = \sum_{i=1}^{n} x_i f_i. \end{align*} Applying $\pi_n$ gives \begin{align*} \pi_n(f) = \sum_{i=1}^{n} \pi_n(x_i)\pi_n(f_i) = 0. \end{align*} However, \begin{align*} \pi_n(x_{n+1}) = x_{n+1} \ne 0. \end{align*} Therefore $x_{n+1} \notin I_n$. Since $x_{n+1} \in I_{n+1}$, the inclusion $I_n \subset I_{n+1}$ is proper. [guided] The only point that needs proof is that the new variable $x_{n+1}$ is not already generated by the earlier variables $x_1, \ldots, x_n$. To test this, we use an evaluation map that kills precisely the first $n$ variables. Fix a positive integer $n$. Define a $k$-algebra homomorphism \begin{align*} \pi_n: R \to k[x_{n+1}, x_{n+2}, x_{n+3}, \ldots], \qquad x_i \mapsto 0 \text{ for } 1 \le i \le n, \qquad x_i \mapsto x_i \text{ for } i \ge n+1. \end{align*} This is a legitimate homomorphism because elements of $R = k[x_1, x_2, x_3, \ldots]$ are finite polynomials: each polynomial mentions only finitely many variables, so substituting $0$ for $x_1, \ldots, x_n$ produces another ordinary polynomial. Now take any $f \in I_n$. By the definition of the generated ideal $I_n = (x_1, \ldots, x_n)$, there exist polynomials $f_1, \ldots, f_n \in R$ such that \begin{align*} f = \sum_{i=1}^{n} x_i f_i. \end{align*} Applying $\pi_n$ to this expression gives \begin{align*} \pi_n(f) = \sum_{i=1}^{n} \pi_n(x_i)\pi_n(f_i). \end{align*} For every index $i$ with $1 \le i \le n$, we have $\pi_n(x_i)=0$, so the right-hand side is zero: \begin{align*} \pi_n(f) = 0. \end{align*} Thus every element of $I_n$ is killed by $\pi_n$. But the next variable is not killed: \begin{align*} \pi_n(x_{n+1}) = x_{n+1}. \end{align*} The polynomial $x_{n+1}$ is nonzero in the [polynomial ring](/page/Polynomial%20Ring) $k[x_{n+1}, x_{n+2}, x_{n+3}, \ldots]$. Hence $x_{n+1}$ cannot belong to $I_n$, because every element of $I_n$ maps to $0$ under $\pi_n$. Since $x_{n+1}$ is one of the generators of $I_{n+1}$, we have $x_{n+1} \in I_{n+1} \setminus I_n$, proving that $I_n \subset I_{n+1}$ is proper. [/guided] [/step] [step:Conclude that the ascending chain condition fails] The ideals constructed above form the chain \begin{align*} I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \cdots. \end{align*} Thus there is an ascending chain of ideals in $R$ that never stabilizes. A Noetherian ring is precisely a ring satisfying the ascending chain condition on ideals. Therefore $R = k[x_1, x_2, x_3, \ldots]$ is not Noetherian. [/step]