Give an example of fields $k\subseteq K\subseteq L $, and $l\subseteq L$, for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has already been posted here , but has not yet had an answer, the truth is that I can not find such an example, could someone please help me? Thank you very much.
This is problem 3.18 from Patrick Morandi's Field and Galois Theory.     Give an example of fields $k \subseteq K \subseteq L$ and $l \subseteq L$ for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has been asked twice, namely  and , though neither had any answers, and only some inconclusive comments. Normally I wouldn't ask the question again, but after struggling with it for weeks at this point, I emailed Morandi, and he gave me some hints.  He told me that he remembers finding an example, but that he isn't sure. He told me that it would need to be over characteristic $p$ and that the extensions (at least some of them I suppose) could not be separable. Moreover, he told me that $K = k(x)(a)$, where $a$ is a root of an irreducible polynomial $f(x,y) \in k(x,y)$ (I'm not 100% what the root of a two variable polynomial is). Finally, he said that $l$ is an algebraic extension of $k$, where $f$ becomes reducible over $l$. Still however, I can't come up with anything. I was hoping that with these extra hints, someone would be able to figure it out.  He did point out that there is a chance that there is an error in the question, but I haven't managed to prove that such an example cannot exist.
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But I want:  1 This is my section For further information see subsection **3** within this section.   Is that possible?  [Edit] To make it more clear: I am looking for a command that works like \ref, but prints (the arabic number of) the value of the subsection counter at the position of the \label.  ======================= == Solution in LaTeX ==  There is another solution below. This is a different solution I came up with  (I cannot post an answer to my own question, so I put it here):  \documentclass{article}  \newcommand{\labelsubseccounter}[1]{     \renewcommand\thesubsection{\arabic{subsection}}     \addtocounter{subsection}{-1}     \refstepcounter{subsection}     \label{#1}     \renewcommand\thesubsection{\thesection.\arabic{subsection}} }  \begin{document}  \section{This is my section} For further information see subsection \ref{subsecthree} within this section,  which is called \ref{onepointthree}, not \ref{onepointfour}. \subsection{One} \subsection{Two} \subsection{Three}\label{onepointthree}\labelsubseccounter{subsecthree} \subsection{Four}\label{onepointfour}  \end{document}   (I would like to insert the compiled code, but don't know how, sorry) For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$ I don't understand the proof to the a/m claim. 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Here is an excerpt of the proof:  For an infinite ordinal $\kappa,$ the canonical well-ordering of $\kappa \times \kappa,$ denoted by $&lt;_{cw}$ is defined as follows: $(\alpha_1, \beta_1)&lt;_{cw} &lt; (\alpha_2, \beta_2)$ iff either one of the following holds.  $(1): \max\{\alpha_1,\beta_1\} &lt; \max\{\alpha_2,\beta_2\}$  $(2): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1&lt;\alpha_2$   $(3): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1 = \alpha_2$ and $\beta_1 &lt; \beta_2.$  Claim: For all infinite cardinals $\kappa, (\kappa \times \kappa, &lt;_{cw}) \cong (\kappa, \in).$  Proof: (It has been show that it is true for $\kappa = \omega$) Assume there is a counterexample  $\kappa.$ Let $A= \{\lambda \in \kappa + 1| \lambda\geq \omega \ is \ a \ cardinal \ and \ (\lambda \times \lambda, &lt;_{cw}) \not\cong (\lambda, \in)\}.$ Then, $\kappa \in A.$ Let $\mu$ be least element of $A.$ Thus, $\mu &gt; \omega$ and for all $\lambda,$ if $\omega \leq \lambda &lt; \mu$ and $\lambda$ is a cardinal, then $(\lambda \times \lambda, &lt;_{cw}) \cong (\lambda, \in).$  By Comparability theorem, $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, &lt;_{cw})$ since $|\mu \times \mu| \geq \mu$ and $\mu$ is cardinal. 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