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I'm looking for a field $F$ such that for subfields $K,E$ of $F$ the structure $K\cup E$ is not a field.

Clearly both $K$ and $E$ must be proper subfields, and while looking for the answer I've come across the primitive element theorem, which might have an answer in general to the question of a union of subfields being a field, but I don't know enough mathematics to understand that theorem - currently I'm just trying to find an example.

An explanation about the fundamental reasons why the general principle (that a union of proper subfields is not a field) holds would also be much appreciated!

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    $\begingroup$ This is one of those situations where almost anything is an example.... $\endgroup$ – Hurkyl Sep 25 '12 at 21:14
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    $\begingroup$ One can wonder how much you looked for examples! :-/ $\endgroup$ – Mariano Suárez-Álvarez Sep 25 '12 at 21:14
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    $\begingroup$ If $K\subseteq E$ or $K\supseteq E$, then the union is a field. Otherwise, it is not: Pick $r$ in $K$ but not $E$ and $s$ in $E$ but not $K$, and consider $r+s$. $\endgroup$ – Andrés E. Caicedo Sep 25 '12 at 21:15
  • $\begingroup$ Schweet, all of these answers are awesome. I dig it. $\endgroup$ – Jānis Lazovskis Sep 25 '12 at 21:21
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Take $K:=\Bbb C$, $E=\Bbb R$ and $F:=\{a+ib,a,\in \Bbb Q\}$. Then $e\in E$, $i\in F$ but $e+i\notin E\cup F$.

Actually, we need to have $E\subset F$ or $F\subset E$ , otherwise take $e\in E\setminus F$ and $f\in F\setminus E$; then $e+f\notin E\cup F$.

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Examples abound with finite fields. Let $\overline{\mathbb{F}}_{2}$ be an algebraic closure of $\mathbb{F}_{2}$. Consider the subfields $\mathbb{F}_{4}$ and $\mathbb{F}_{8}$. These fields only have $\mathbb{F}_{2}$ in common, so the order of their union is $4 + 8 - 2 = 10$. There is no finite field of order $10$.

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  • $\begingroup$ Very nice, I like the order argument :) $\endgroup$ – rschwieb Sep 25 '12 at 21:19
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For a field $F$, the rational function fields $F(x),F(y)\subseteq F(x,y,z)$,

but $F(x)\cup F(y)$ does not contain $xy$.

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Consider $\Bbb Q(\sqrt 2)\cup\Bbb Q(\sqrt 3)\subseteq \Bbb Q(\sqrt 2,\sqrt 3).$ The union does not contain $\sqrt 2+\sqrt 3.$

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Plenty of examples already, so I'll just add something in the line of intuition. Any field is an (abelian) group under addition. For arbitrary subgroups $H, K \subseteq G$ of a group $G$, $H \cup K$ is not a group unless $H \subseteq K$ or $K \subseteq H$ (I leave this as an exercise). Hence, the for the union of two fields to be a field, one must contain the other. This statement is also sufficient trivially.

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  • $\begingroup$ Indeed a great observation. $\endgroup$ – user41442 Sep 26 '12 at 4:06

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