I have stumbled upon a remarkable similarity between the proof of Baer's criterion and an extension theorem in field theory. Here are the statements:

Baer's criterion: Let $R$ be a ring. A left $R$-module $E$ is injective iff every $R$-map $f:I \to E$, where $I$ is a left ideal of $R$, can be extended to a map $R\to E$.

Extension theorem: let $F\subset K$ be an algebraic field extension, and $L$ an algebraically closed field. Then every field homomorphism $\sigma: F \to L$ can be extended to a field homomorphism $K\to L$.

Now, these two theorems don't seem to have anything to do with each other, but the proofs are strikingly similar. Here's how they go:

For Baer's criterion (the non-trivial statement, i.e. the sufficiency of the ideal condition): to prove injectivity, we let $A\subset B$ be a submodule and $f:A\to E$ an $R$-map, and define $X=\{(A',g'): A\subset A'\subset B,\, g': A' \to E,\, g'|_A=f\}$.

Let $\leq$ be the partial order in $X$ given by $(A',g')\leq (A'',g'') \iff A'\subset A'' \text{ and } g''|_{A'}=g'$.

By Zorn's lemma there's $(A_0,g_0)$ a maximal element. If $A_0=B$ we're done, if not, let $b\in B\setminus A_0$, define $I=\{r\in R: rb\in A_0\}$ and $h:I\to E$, $r \mapsto g_0(rb)$. Apply the hypothesis to find an extension $\tilde{h}$ of $h$ to $R$. Let $A_1=A_0 + Rb$ and $g_1:A_1 \to E$, $a_0+rb \mapsto g_0(a_0)+r\tilde{h}(1)$, and $(A_1,g_1)$ contradict the maximality of $(A_0,g_0)$.

For the extension theorem: define $M=\{(A,\tau):F\subset A \subset K,\, \tau:A\to L,\, \tau|_F=\sigma\}$. Let $\leq$ be the partial order on $M$ given by $(A,\tau)\leq (A',\tau') \iff A\subset A' \text{ and } \tau'|_A=\tau$.

By Zorn's lemma there's $(A,\tau)$ a maximal element. If $A=K$ we're done, if not let $\alpha \in K\setminus A$, I claim $\tau$ extends to $A(\alpha)\to L$, contradiction.

Let $p$ be the minimal polynomial for $\alpha$ over $A$. The polynomial $\tau p\in L[X]$ has a root $r\in L$ by hypothesis. Since $A(\alpha)= \frac{A[X]}{\langle p \rangle}$, define

$\tilde{\tau}: \frac{A[X]}{\langle p \rangle} \to L$ as $\tilde{\tau}|_A=\tau$, $\tilde{\tau}(X)=r$, and $(A(\alpha),\tilde{\tau})$ contradicts the maximality of $(A,\tau)$.

The resemblance of both proofs shouldn't come as such a surprise since both are about extending morphisms, so maybe the only observation to make is "it's a useful technique, remember it's useful for extending morphisms". If that's it, then this question is useless. But I'm intrigued. Is this technique used to prove other extending theorems? Is it possible to generalize it and write a single categorical proof for situations of this kind? Any other observations are appreciated (or perhaps it's just a dumb observation that doesn't serve any purpose).

  • $\begingroup$ It is a very common pattern. $\endgroup$ – Mariano Suárez-Álvarez Feb 17 '11 at 16:37
  • $\begingroup$ It is the very standard application of Zorn's Lemma. I used it in a paper to construct amalgams of nilpotent groups, by constructing "partial amalgams" and then applying Zorn's Lemma to extend them. The general idea is that if you can do a "single-step extension", then you can do an arbitrary extension by transfinite induction "one-step-at-a-time". Don't know if it can be categorified, but I doubt it since it fails in many situations when you try to go from the finite to the infinite case. $\endgroup$ – Arturo Magidin Feb 17 '11 at 16:48
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    $\begingroup$ Every proof using Zorn's lemma looks the same. (Imre Leader said in a lecture once something like "once you've seen one application of Zorn's lemma, you've seen them all," but he gave three anyway.) $\endgroup$ – Qiaochu Yuan Feb 17 '11 at 16:53
  • $\begingroup$ If you want to crystallize a strategy out of this similarity, here it is: If you want to define a map from some arbitrarily-infinite space (like $R$ in Baer's criterion, or $K$ in the extension theorem), but you can only do it on small finite parts of it (such as finitely generated submodules of $R$ in Baer's criterion, or finite extensions of $F$ in the extension theorem), then you can use the Zorn lemma. Here, "arbitrarily-infinite" means that it MAY be very large (infinite, not finitely generated, etc.), although it needs not be - we just don't know how large it is. $\endgroup$ – darij grinberg Feb 17 '11 at 19:07

I do see some nontrivial commonality here, but it is not precisely between the two theorems you suggest.

Namely, just as the injective modules are (obviously!) the injective objects in the category of $R$-modules, the algebraically closed fields are the injective objects in the category of fields. Indeed, the latter statement is precisely your Extension Theorem.

Thus your Extension Theorem doesn't strike me as a "Baer criterion" per se. Perhaps it is better the other way around: if we define an injective field as being a field which satisfies the conclusion of the Extension Theorem, then maybe "Baer's Criterion" is that a field is injective iff it is algebraically closed? (To be clear, this is definitely a true fact; it's not clear whether it is worthy of this name.)

Let me try this: to check whether a field $K$ is injective, it suffices to check that for every maximal ideal $I$ of $K[t]$ there exists a $K$-algebra map $K[t]/I \rightarrow K$. This is perhaps "Baeresque"?

Note that I came to the analogy here by trying to understand why the injective envelope $M \rightarrow E(M)$ is not a natural [in the categorical sense] construction: it is unique up to nonunique isomorphism over $M$. This is just the same situation as the algebraic closure of a field. And then I found this paper by Adamek, Herrlich, Rosicky and Tholen exploring this issue in a more general categorical context.

  • $\begingroup$ On the other hand, this makes me think one can define the "absolute Galois group" (well-defined up to an inner automorphism!) of an $R$-module as the group of automorphisms of "the" extension $M \hookrightarrow E(M)$. Anyone know anything about the Inverse Galois Problem in this context? $\endgroup$ – Pete L. Clark Feb 18 '11 at 6:09
  • $\begingroup$ your answer is great food for thought. Thank you. $\endgroup$ – Bruno Stonek Feb 20 '11 at 22:07

I suspect you'll find highly interesting the following paper on related topics, which I mentioned in some prior answers that you may find of interest. Below is the introduction of said paper. enter image description here enter image description here enter image description here

  • $\begingroup$ Somewhat related is Schmidt, J: "Uber die Rolle der transfiniten Schlussweisen in einer allgemeinen Idealtheorie." Math. Nachr. 7 (1952) 165--182. $\endgroup$ – Willie Wong Jan 17 '12 at 10:23
  • $\begingroup$ I've just read this introduction. It's really interesting. Thank you very much! I think I see the link between Theorem 1 and the proof of the characterization of semisimple modules: "a module is direct sum of simple modules iff every submodule has a complement". I will download & print the paper, and I'll try to read and think it through when I have more time. Thank you again, your intuition was right, I do find this highly interesting! $\endgroup$ – Bruno Stonek Jan 17 '12 at 13:31

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