---
tag: BFN4MBMZ1E
kind: proof
name: Proof of Hilbert's basis theorem by leading coefficients
msc: ["13E05"]
proves: 0YKA9J3XEE
uses: [0MVH4Y1CRZ, 0ZCAXY9563, 0Q9EZS90GY, 0GYGHKX834, 0MYSA6MYNA, 030TZ0DQWT, 0P05RJC5XA, 0DPF6WRZW6]
methods: [0J0XBKG64S]
sources:
  - url: https://en.wikipedia.org/wiki/Hilbert%27s_basis_theorem
    locator: Proof, First proof
---

## Proof

We argue by contradiction, with the leading coefficients of a badly chosen sequence of polynomials.

1. 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]].
2. First we show: if every [[ideal|0MYSA6MYNA]] of $R[x]$ is finitely generated, then $R[x]$ is Noetherian.
3. 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.
4. 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 $U \subseteq I_N$.
5. 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]].
6. So it is enough to show that every ideal $I$ of $R[x]$ is finitely generated. Suppose that some ideal $I$ is not.
7. 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 $B_0 = \set{0}$.
8. 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 $0 \in B_k$.
9. 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.
10. 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 $J_0 \subseteq J_1 \subseteq \dots$.
11. 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 $a_N = \sum_{i<N} u_i a_i$ with $u_i \in R$.
12. Let $g = \sum_{i<N} u_i \, x^{\deg f_N - \deg f_i} f_i$. The exponents are nonnegative by Step 9. So $g \in B_N \subseteq I$.
13. 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 $a_N$.
14. Hence $f_N - g$ is $0$ or has degree less than $\deg f_N$. Also $f_N - g \in I$.
15. 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 $f_N - g \ne 0$.
16. 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.
17. So every ideal of $R[x]$ is finitely generated. By Steps 2 to 5, $R[x]$ is Noetherian.

## Idea

If 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.