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  "body": "\n## Proof\n\nWe argue by contradiction, with the leading coefficients of a badly chosen sequence of polynomials.\n\n1. Let $R$ be a [[Noetherian|0MVH4Y1CRZ]] [[commutative ring|0GYGHKX834]]. Let $R[x]$ be its [[polynomial ring|0Q9EZS90GY]]. For $S \\subseteq R[x]$, write $(S)$ for the [[ideal generated by|030TZ0DQWT]] $S$. For a nonzero polynomial, $\\deg$ is its [[degree|0DPF6WRZW6]].\n2. First we show: if every [[ideal|0MYSA6MYNA]] of $R[x]$ is finitely generated, then $R[x]$ is Noetherian.\n3. Let $I_1 \\subseteq I_2 \\subseteq \\dots$ be ideals of $R[x]$, and let $U$ be their union. Two elements of $U$ lie in a common $I_k$, since the $I_k$ form a chain. So $U$ is an ideal.\n4. Let $U = (g_1, \\dots, g_m)$. Each $g_i$ lies in some $I_{k_i}$. Let $N$ be the largest $k_i$. Then all $g_i$ lie in $I_N$, so $U \\subseteq I_N$.\n5. For $n \\ge N$ we get $I_N \\subseteq I_n \\subseteq U \\subseteq I_N$, so $I_n = I_N$. This is the [[ascending chain condition|0ZCAXY9563]].\n6. So it is enough to show that every ideal $I$ of $R[x]$ is finitely generated. Suppose that some ideal $I$ is not.\n7. We choose $f_0, f_1, f_2, \\dots \\in I$ recursively. Suppose $f_0, \\dots, f_{k-1}$ are chosen, and let $B_k = (f_0, \\dots, f_{k-1})$, with $B_0 = \\set{0}$.\n8. Since $I$ is not finitely generated, $B_k \\ne I$. Choose $f_k \\in I \\setminus B_k$ of least degree. Note $f_k \\ne 0$, because $0 \\in B_k$.\n9. The degrees satisfy $\\deg f_k \\le \\deg f_{k+1}$. Indeed, $f_{k+1} \\notin B_{k+1}$ and $B_k \\subseteq B_{k+1}$, so $f_{k+1} \\in I \\setminus B_k$. The minimality of $\\deg f_k$ applies.\n10. Let $a_k \\in R$ be the [[leading coefficient|0P05RJC5XA]] of $f_k$. Let $J_k$ be the ideal of $R$ generated by $a_0, \\dots, a_k$. Then $J_0 \\subseteq J_1 \\subseteq \\dots$.\n11. Since $R$ is Noetherian, there is $M$ with $J_k = J_M$ for all $k \\ge M$. Let $N = M + 1$. Then $a_N \\in J_{N-1}$, so $a_N = \\sum_{i\u003cN} u_i a_i$ with $u_i \\in R$.\n12. Let $g = \\sum_{i\u003cN} u_i \\, x^{\\deg f_N - \\deg f_i} f_i$. The exponents are nonnegative by Step 9. So $g \\in B_N \\subseteq I$.\n13. Each term $u_i x^{\\deg f_N - \\deg f_i} f_i$ has degree at most $\\deg f_N$. Its coefficient of $x^{\\deg f_N}$ is $u_i a_i$. So the coefficient of $x^{\\deg f_N}$ in $g$ is $a_N$.\n14. Hence $f_N - g$ is $0$ or has degree less than $\\deg f_N$. Also $f_N - g \\in I$.\n15. If $f_N - g \\in B_N$, then $f_N = (f_N - g) + g \\in B_N$, which is false. So $f_N - g \\in I \\setminus B_N$, and in particular $f_N - g \\ne 0$.\n16. So $f_N - g$ is an element of $I \\setminus B_N$ of smaller degree than $f_N$. This contradicts the choice of $f_N$ in Step 8.\n17. So every ideal of $R[x]$ is finitely generated. By Steps 2 to 5, $R[x]$ is Noetherian.\n\n## Idea\n\nIf an ideal of $R[x]$ needed infinitely many generators, we could pick them one by one with the smallest possible degree. Their leading coefficients generate a chain of ideals in $R$, which must stop growing. Then one new leading coefficient is a combination of earlier ones, and subtracting a matching combination of earlier polynomials lowers the degree. This contradicts the minimal choice.\n",
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