Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice

Solution 1:

This answer builds on Qiaochu's and uses the same definition as Qiaochu, to wit: A ring $R$ is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$.

Theorem: If $R$ is noetherian, then $R[x]$ is noetherian.

This proof is basically taking the standard proof and rephrasing it to use Qiaochu's definition and be careful about choice.

I'm going to try to systematically use the following conventions: Ideals in $R[x]$ get capital letters; ideal in $R$ get overlined capital letters. Sets of ideals in $R[x]$ get calligraphic letters. I found that I could manage to write this without ever assigning a name to a set of ideals in $R$.

For any ideal $I \subseteq R[x]$, and any integer $j \geq 0$, define $$s_j(I) := \{ r \in R : \mbox{there is an element of $I$ of the form } r x^j + r_{j-1} x^{j-1} + \cdots +r_0 \}$$ Observe that $s_j(I)$ is an ideal and $s_j(I) \subseteq s_{j+1}(I)$.

Lemma: If $R$ is noetherian, and $I \subseteq R[x]$ is an ideal, then there is an index $j$ such that $s_k(I) = s_j(I)$ for $k \geq j$.

Proof: Let $s_j(I)$ be a maximal element in $\{ s_j(I) \}_{j \geq 0}$. Let $k \geq j$. Then we observed above that $s_k(I) \supseteq s_j(I)$. But, by the definition of a maximal element, we do not have $s_k(I) \supsetneq s_j(I)$, so $s_k(I) = s_j(I)$. $\square$.

We will denote the ideal $s_j(I)$ defined in the above lemma as $s_{\infty}(I)$. For $\mathcal{I}$ a collection of ideals in $R[x]$, we will write $s_j(\mathcal{I})$ or $s_{\infty}(\mathcal{I})$ for the result of applying $s_j$ or $s_{\infty}$ to each element of $\mathcal{I}$. So $s_j(\mathcal{I})$ is a set of ideals in $R$.

Let $\mathcal{I}$ be a collection of ideals in $R[x]$. Since $R$ is noetherian, there is a maximal element $\bar{J}$ in $s_{\infty}(\mathcal{I})$. Let $\mathcal{J}$ be the set of all $I \in \mathcal{I}$ with $s_{\infty}(I) = \bar{J}$.

Note that no element of $\mathcal{I} \setminus \mathcal{J}$ contains an element of $\mathcal{J}$, by the maximality of $\bar{J}$, so it is enough to show that $\mathcal{J}$ has a maximal element.

Choose an ideal $K \in \mathcal{J}$. (Making one choice does not use AC.) Let $m$ be an index such that $s_m(K) = \bar{J}$. Let $\mathcal{K}$ be the collection of ideals $I \in \mathcal{J}$ for which $s_m(I) = \bar{J}$. Note that no element of $\mathcal{J} \setminus \mathcal{K}$ can properly contain an element of $\mathcal{K}$, so it is enough to show that $\mathcal{K}$ has a maximal element.

We now make finitely many dependent choices. Choose a maximal element $\bar{J}^{m-1}$ in $s_{m-1}(\mathcal{K})$; let $\mathcal{K}_{m-1}$ be the set of $I \in \mathcal{K}$ with $s_{m-1}(I)=\bar{J}^{m-1}$; it is enough to show that $\mathcal{K}_{m-1}$ has a maximal element.

Choose a maximal element $\bar{J}^{m-2}$ in $s_{m-2}(\mathcal{K}_{m-1})$; let $\mathcal{K}_{m-2}$ be the set of $I \in \mathcal{K}^{m-1}$ with $s_{m-2}(I)=\bar{J}^{m-2}$; it is enough to show that $\mathcal{K}_{m-2}$ has a maximal element.

Continue in this manner to construct $\mathcal{K}_{m-3}$, $\mathcal{K}_{m-4}$, ..., $\mathcal{K}_0$. Since we are only making finitely many choices, we don't need AC; see my answer here.

At the end, we have a nonempty collection $\mathcal{K}_0$ of ideals such that, for any $I$ and $J \in \mathcal{K}_0$, and any $j \geq 0$, we have $s_j(I)= s_j(J)$. I claim that any element of $\mathcal{K}_0$ is maximal.

Let $I$ and $J \in \mathcal{K}_0$ and suppose that $I \supseteq J$. I will prove that $I=J$. This shows that every element of $\mathcal{K}_0$ is maximal.

Let $I_{\leq d}$ be the set of polynomials in $I$ of degree $\leq d$. I will show by induction on $d$ that $I_{\leq d} = J_{\leq d}$. The base case is $d=-1$, where both sides are $\{ 0 \}$. Since $I \supseteq J$, I just need to show that $I_{\leq d} \subseteq J_{\leq d}$.

Let $f \in I_{\leq d}$ and let the leading term of $f$ be $r x^d$. Then $r \in s_d(I) = s_d(J)$ so there is some $g \in J_{\leq d}$ with leading term $r$. Since $I \supseteq J$, we have $g \in I$ and hence $f-g \in I$. Since $\deg(f-g) < d$, by the induction hypothesis, we have $f-g \in J$. So $f = (f-g)+g \in J$. QED

Solution 2:

This problem reduces to proving Hilbert's basis theorem without choice, which I don't currently know how to do, but which I believe can be done. Let me explain the reduction. Below "ring" means "commutative unital ring."

Definition: A ring $R$ is Noetherian if any non-empty collection of ideals of $R$ has a maximal element.

(This definition is equivalent to the usual definitions in the presence of dependent choice, but in the absence of dependent choice I believe it is known that this definition is stronger. In any case it implies the other definitions.)

Fields are obviously Noetherian. Note that Noetherian rings contain maximal ideals by definition.

Proposition: Let $R$ be a Noetherian ring and $f : R \to S$ a surjective ring homomorphism. Then $S$ is Noetherian.

Proof. Let $I_i$ be a non-empty collection of ideals in $S$. Then $f^{-1}(I_i)$ is a non-empty collection of ideals in $R$ which by assumption has a maximal element $I_j$. Since $f$ is surjective, $I_j$ is also maximal in $S$.

Assumption: If $R$ is Noetherian, then $R[x]$ is Noetherian.

From this assumption it follows that any finitely-generated ring over a Noetherian ring is Noetherian, hence any ideal in such a ring is contained in a maximal ideal (in particular a prime ideal).

Let me explain the problem with the standard proof. This proof (as found in e.g. Atiyah-MacDonald) shows that if every ideal of $R$ is finitely-generated, then every ideal of $R[x]$ is finitely-generated (and I am not sure this works without dependent choice either). As far as I know, this condition is not equivalent to Noetherian (as defined above) in ZF. What we can prove in ZF is the following.

Proposition: If $R$ is Noetherian, then every ideal of $R$ is finitely-generated.

Proof. Let $I$ be an ideal of $R$. Then the collection of finitely-generated ideals contained in $I$ has a maximal element $J = (r_1, ... r_n)$. If $I$ is not all of $J$, then there exists $r_{n+1} \in I$ which is not in $J$, but then $(r_1, ... r_{n+1})$ is a finitely-generated ideal containing $J$, which contradicts the minimality of $J$.

However, as far as I can tell, we cannot prove the converse without dependent choice.

Solution 3:

To simplify the notations, we assume the characteristic of $k$ is $0$. The positive characteristic case can be proved similarly.

By Noether normalization lemma(this can be proved without AC), there exist algebraically independent elements $x_1, ..., x_n$ in $A$ such that $A$ is a finitely generated module over the polynomial ring $A' = k[x_1,..., x_n]$. Let $K$ and $K'$ be the fields of fractions of $A$ and $A'$ respectively. Let $L$ be the smallest Galois extension of $K'$ containing $K$. Let $G$ be the Galois group of $L/K'$. Let $B$ be the integral closure of $A$ in $L$. It is well known that $B$ is a finite $A'$-module. Let $g = \prod_{\sigma \in G} \sigma(f)$. Since $A'$ is integrally closed, $g \in A'$. Since $g$ is non-invertible, there exists an irreducible polynomial $h$ which divides $g$. Then $P = hA'$ is a prime ideal of $A'$ containing $g$. By this, there exists a prime ideal $Q$ of $B$ lying over $P$. Since $g \in Q$, there exists $\sigma \in G$ such that $\sigma(f) \in Q$. Then $f \in \sigma^{-1}(Q)$. Hence $f \in A \cap \sigma^{-1}(Q)$ QED