Transfinite Construction in Differential Fields proof 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:

*

*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?


*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.
 A: 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 $K^\omega$.
