# Hilbert's Basis Theorem

Hilbert's Basis Theorem is a result concerning Noetherian rings. It states that if $A$ is a (not necessarily commutative) Noetherian ring, then the ring of polynomials $A[x_1, x_2, \dotsc, x_n]$ is also a Noetherian ring. (The converse is evidently true as well.)

Note that $n$ must be finite; if we adjoin infinitely many variables, then the ideal generated by these variables is not finitely generated.

The theorem is named for David Hilbert, one of the great mathematicians of the late nineteenth and twentieth centuries. He first stated and proved the theorem in 1888, using a nonconstructive proof that led Paul Gordan to declare famously, "Das ist nicht Mathematik. Das ist Theologie. [This is not mathematics. This is theology.]" In time, though, the value of nonconstructive proofs was more widely recognized.

## Proof

By induction, it suffices to show that if $A$ is a Noetherian ring, then so is $A[x]$.

To this end, suppose that $\mathfrak{a}_0 \subset \mathfrak{a}_1 \subset \dotsb$ is an ascending chain of (two-sided) ideals of $A[x]$.

Let $\mathfrak{c}_{i,j}$ denote the set of elements $a$ of $A$ such that there is a polynomial in $\mathfrak{a}_i$ with degree at most $i$ and with $a$ as the coefficient of $x^j$. Then $\mathfrak{c}_{i,j}$ is a two-sided ideal of $A$; furthermore, for any $i' \ge i$, $j' \ge j$, $$\mathfrak{c}_{i,j} \subset \mathfrak{c}_{i',j}, \mathfrak{c}_ {i,j'} .$$ Since $A$ is Noetherian, it follows that for every $i \ge 0$, the chain $$\mathfrak{c}_{i,0} \subset \mathfrak{c}_{i,1} \subset \dotsb$$ stabilizes to some ideal $\mathfrak{m}_i$. Furthermore, the ascending chain $$\mathfrak{m}_0, \mathfrak{m}_1, \dotsc$$ also stabilizes to some ideal $\mathfrak{m}=\mathfrak{c}_{A,B}$. Then for any $i \ge A$ and any $j\ge 0$, $$\mathfrak{c}_{i,j} = \mathfrak{c}_{A,j} .$$ We claim that the chain $(\mathfrak{a}_k)_{k=0}^\infty$ stabilizes at $\mathfrak{a}_A$. For this, it suffices to show that for all $k \ge A$, $\mathfrak{a}_k \subset \mathfrak{a}_A$. We will thus prove that all polynomials of degree $n$ in $\mathfrak{a}_k$ are also elements of $\mathfrak{a}_A$, by induction on $n$.

For our base case, we note that $\mathfrak{c}_{k,0} = \mathfrak{c}_{M,0}$, and these ideals are the sets of degree-zero polynomials in $\mathfrak{a}_k$ and $\mathfrak{a}_M$, respectively.

Now, suppose that all of $\mathfrak{a}_k$'s elements of degree $n-1$ or lower are also elements of $\mathfrak{a}_M$. Let $p$ be an element of degree $n$ in $\mathfrak{a}_k$. Since $$\mathfrak{c}_{k,n} = \mathfrak{c}_{A,n},$$ there exists some element $q \in \mathfrak{a}_A$ with the same leading coefficient as $p$. Then by inductive hypothesis, $$p-q \in \mathfrak{a}_A ,$$ so $$p \in \mathfrak{a}_A,$$ as desired. $\blacksquare$