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I'm learning differential fields theory and given my background in model theory I found this book. On p. 203 I find this:

LEMMA 4.7.6. Let $K$ be a differential field, let $P \in K\{Y\} \neq$ be irreducible of order $r$, and let $Q \in K\{Y\}^{\neq}$have order $<r$. Then there is an element y of a differential field extension of $K$ with $P(y)=0, Q(y) \neq 0$, such that $K\langle y\rangle$ embeds over $K$ into any differentially closed differential field extension of $K$.

After which the authors mention:

Every differential field has a differential closure: this follows by a straightforward transfinite construction using Lemma 4.7.6

My questions are:

  1. what is exactly meant here by “transfinite construction”? The term is not used previously (or at any other point) in the book. Is it just another name for the transfinite induction?

  2. even if that is the case, I cannot see how that induction/construction follows? (I feel especially intimidated because it says “straightforward” and I don't know how to prodcue it).

Any help would be greatly appreciated.

Jordie Vincent
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  • Seems like it would mimic the proof of the existence of algebraic closures. http://www.cs.bsu.edu/~hfischer/math412/Closure.pdf – Robert Shore Feb 03 '23 at 00:42
  • @RobertShore how so? the proof you referenced seems rather different that Lemma 4.7.6 from above, doesn't use noetherianity nor does it employ embeddings and morphisms in a way similar to the proof in the book – Jordie Vincent Feb 03 '23 at 14:59
  • The point is that in the algebraic closure setting, you can always add one more algebraic element until you reach a maximal element in a chain, guaranteed by Zorn's Lemma. That seems to be what's guaranteed in this setting by Lemma 4.7.6. – Robert Shore Feb 03 '23 at 17:43
  • A transfinite construction is one which is done by transfinite recursion. This is similar, but not quite the same as transfinite induction (this is included in the article you link to), just the same as induction and recursion (over natural numbers) are not exactly the same thing. – tomasz Feb 04 '23 at 12:21
  • Regarding your second question, I'm not sure what the $\neq$ means in this notation, but I would guess that what they mean is that you can just ignore the $Q$ (take it to be $1$, for instance) and the embeddability statement and just recursively add zeroes for each irreducible polynomial $P$, then repeat that for each polynomial in the extension, repeating this $\omega$ many times. – tomasz Feb 04 '23 at 12:24
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    @tomasz I think the $\neq$ just means that $P$ and $Q$ should be non-zero. You can't just ignore the $Q$, because the definition of "differentially closed" says not just that you have solutions to differential polynomials, but that you have solutions to differential polynomials which are not solutions to differential polynomials of lower degrees. But Lemma 4.7.6 is exactly what you need to adjoin the necessary solutions. – Alex Kruckman Feb 05 '23 at 02:32

1 Answers1

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As tomasz says in the comments, "transfinite construction" here just means a construction by transfinite recursion. And the idea is to construct the differential closure by repeatedly (transfinitely) applying Lemma 4.7.6.

Let me spell out some details for you. Let $K$ be a differential field.

  • Let's define a differential problem in $K$ to be a pair $(P,Q)$, where for some $r$, $P\in K[Y,\dots,Y^{(r)}]$, $Q\in K[Y,\dots,Y^{(r-1)}]$, both are non-zero, and $Y^{(r)}$ occurs in $P$.
  • A solution to this problem is an element $y\in K$ such that $P(y) = 0$ and $Q(y)\neq 0$.
  • Then $K$ is differentially closed if every differential problem in $K$ has a solution in $K$.
  • Let's say that a differential field extension $K'$ of $K$ is small over $K$ (my terminology) if every embedding of $K$ into a differentially closed field extends to an embedding of $K'$.
  • Lemma 4.7.6 says that every differential problem in a differential field $K$ has a solution in a differential field extension $K'$ which is small over $K$.
  • Finally, a differential closure of $K$ is a differentially closed differential field extension of $K$ which is small over $K$.

Let's construct a differential closure of $K$. Enumerate the differential problems in $K$ as $(P_\alpha,Q_\alpha)_{\alpha<\kappa}$ for some cardinal $\kappa$. We build a chain $(K_\alpha)_{\alpha\leq\kappa}$ of differential field extensions of $K$.

Base case: $K_0 = K$.

Limit step: For $\alpha$ a limit ordinal, define $K_\alpha$ to be the union (directed colimit) of the chain $(K_\beta)_{\beta<\alpha}$.

Successor step: Apply Lemma 4.7.6 to find a differential field extension $K_{\alpha+1}$ which is small over $K_\alpha$ and contains a solution to the differential problem $(P_\alpha,Q_\alpha)$.

Now $K_\kappa$ is a differential field extension of $K$ which is small over $K$ (prove smallness by a straightforward transfinite induction) and contains solutions to all differential problems in $K$ (by construction, $(P_\alpha,Q_\alpha)$ has a solution in $K_{\alpha+1}$ and hence in $K_\kappa$). But $K_\kappa$ may not be differentially closed, since there are new differential problems in $K_\kappa$ that weren't in $K$.

To solve this, define $K^0 = K$ and $K^1 = K_\kappa$, and for each $K^i$, repeat the construction above to find $K^{i+1}$ which is small over $K$ and contains solutions to all differential problems in $K^{i}$. Let $K^\omega$ be the union (directed colimit) of $(K^{i})_{i\in \omega}$. Then $K^\omega$ is a differential closure of $K$, since every differential problem in $K^\omega$ is already in $K^i$ for some $i$ and hence has a solution in $K^{i+1}$, and hence in $K^\omega$.

Alex Kruckman
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