9
$\begingroup$

Let $A$ be a commutative ring with an identity. Suppose that every non-empty set of ideals of $A$ has a maximal element. Let $A[[x]]$ be the formal power series ring over $A$. Can we prove that every non-empty set of ideals of $A[[x]]$ has a maximal element without Axiom of Choice?

Remark The same question was asked in MSE.

$\endgroup$
  • 2
    $\begingroup$ Is there a motivation for using this particular definition of a noetherian ring? I would have thought that in absence of choice, the more natural definition is that every ideal is finitely generated (or equivalently, every directed set of ideals has a largest element). I believe that the standard proof that $A$ noetherian implies $A[[x]]$ noetherian goes through without choice under the “finitely generated” definition. $\endgroup$ – Emil Jeřábek supports Monica Jan 14 '14 at 0:25
  • 6
    $\begingroup$ @EmilJeřábek For example, this definition implies that every ideal is contained in a maximal ideal. We've had a bunch of questions about this over the years, e.g. mathoverflow.net/questions/7025 and mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma and most of the logicians seem to agree that the OP's definition is better in the absence of choice. For example, see Francois Dorais' comment on the second question I linked. $\endgroup$ – David E Speyer Jan 14 '14 at 1:26
  • 3
    $\begingroup$ The dispute is similar to the difference between well-foundedness as every-nonempty-set-has-minimal-element vs. no-descending-sequences, but in the absence of AC, it is usually the former that is considered the more robust concept, and it is this that corresponds to the OP's definition here. $\endgroup$ – Joel David Hamkins Jan 14 '14 at 1:34
  • 7
    $\begingroup$ Side remark: all definitions proposed here are distinct as shown by Hodges, Six impossible rings, J. Algebra 31 (1974), 218–244; MR0347814. $\endgroup$ – François G. Dorais Jan 14 '14 at 2:52
  • 2
    $\begingroup$ In my comment David links to, I say that the maximal element definition is "the right definition" in the absence of choice. There are nuances here and context matters. The maximal element definition is arguably "the right definition" for ZF. Emil's proposal is also "the right definition" but in different settings, such as constructive mathematics. I'm not sure the ACC definition is "the right definition" in any context but it is a very useful one in ZFC. $\endgroup$ – François G. Dorais Jan 14 '14 at 3:00
4
$\begingroup$

I believe that the answer is yes.

Let $A$ be a commutative ring with $1$, such that any non-empty set of ideals has a maximal element. For each ideal $I\leq A[[x]]$, and each $n\in \mathbb{N}$, define $I^{(n)}$ to be the set

$$\{0\}\cup\{a\in A\ :\ a\text{ is the leading non-zero coefficient, in degree $n$, of some }f\in A[[x]]\}.$$

This is an ideal of $A$. Also define $I^{(\omega)}=\bigcup_{n\in \mathbb{N}}I^{(n)}$, which is also an ideal of $A$.

Now suppose $\{I_{\alpha}\}_{\alpha}$ is a set of ideals in $A[[x]]$. The set of ideals $\{I_{\alpha}^{(\omega)}\}_{\alpha}$ has a maximal element, say $I^{(\omega)}_{\beta}$. Further, it isn't hard to prove that the form of the noetherian hypothesis we are assuming implies finite generation of ideals (provable in ZF). Thus, we may fix a generating set $a_1,a_2,\ldots, a_k$ for $I^{(\omega)}_{\beta}$. This means that $I^{(\omega)}_{\beta}=I^{(n)}_{\beta}$ for some $n\in \mathbb{N}$, which we also fix. (For instance, we can take the integer $n$ to be the lowest degree where each of $a_1,a_2,\ldots, a_k$ occurs as a leading coefficient.)

Now, restrict the set of $\alpha$ to only those for which $I_{\alpha}^{(n)}=I_{\beta}^{(n)}$. (Note that if $I_{\alpha}^{(n)}\neq I_{\beta}^{(n)}$, then $I_{\alpha}$ does not contain $I_{\beta}$. So this restriction costs us nothing.) Next, we may as well also assume $\beta$ is chosen such that $I_{\beta}^{(n-1)}$ is a maximal element of $\{I_{\alpha}^{(n-1)}\}_{\alpha}$. We then may assume $I_{\beta}^{(n-2)}$ is maximal in $\{I_{\alpha}^{(n-2)}\}_{\alpha}$, after restricting $\alpha$ again. Recursively repeating this process (only finitely many times), the resulting $\beta$ is such that $I_{\beta}$ is maximal in $\{I_{\alpha}\}_{\alpha}$.

$\endgroup$
  • $\begingroup$ Thanks. I would appreciate if you would add an explanation why $I_{\beta}$ is maximal. $\endgroup$ – Makoto Kato Jan 22 '15 at 4:21
  • $\begingroup$ Assume, that we have $I_{\beta}\subseteq I_{\gamma}$. This implies $I_{\beta}^{(m)}\subseteq I_{\gamma}^{(m)}$ for every $m\in \mathbb{N}\cup\{\infty\}$. Then just use the maximality of $I_{\beta}^{(\omega)}$, followed successively by the maximality of $I_{\beta}^{(n)},I_{\beta}^{(n-1)},\ldots, I_{\beta}^{(0)}$. $\endgroup$ – Pace Nielsen Jan 22 '15 at 4:42
  • $\begingroup$ Could you explain why $I_{\beta}^{(m)}=I_{\gamma}^{(m)}$ for all $m$ implies $I_{\beta}=I_{\gamma}$? $\endgroup$ – Makoto Kato Jan 22 '15 at 5:08
  • $\begingroup$ Sure! Let $f(x)\in I_{\gamma}$. Since $I_{\beta}^{(m)}=I_{\gamma}^{(m)}$ for each $m\in \mathbb{N}$, after subtracting an element of $I_{\beta}$, we may as well assume the terms of $f(x)$ in degrees $< n$ are all zero. (Here, $n$ is an integer where $I_{\beta}^{(n)}=I_{\beta}^{(\omega)}$.) Letting $a_1,a_2,\ldots, a_k$ be a generating set for $I_{\beta}^{(n)}$ (as before), fix polynomials $g_1(x),g_2(x),\ldots, g_k(x)\in I_{\beta}$ where the leading term of $g_i(x)$ is $a_i$, and it occurs exactly in degree $n$. (to be continued...) $\endgroup$ – Pace Nielsen Jan 22 '15 at 5:18
  • $\begingroup$ We can now appeal to the axiom of dependent choice, and a standard argument shows that $f$ is a linear combination of $g_1,\ldots, g_k$. If you don't want even dependent choice (which I imagine is the case), I'm stuck at this point! (So I'm glad you pushed me on this issue.) My gut feeling is that if we start working with finite generation methods, we will need to appeal to dependent choice. I'll keep thinking about whether we can avoid going down to $g_1,\ldots, g_k$. $\endgroup$ – Pace Nielsen Jan 22 '15 at 6:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.