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Suppose $\Bbb F \subset \Bbb K_1 , \Bbb K_2 \subset \Omega$ all fields.

Denote by $\Bbb K_1\Bbb K_2$ the minimal field containing both $\Bbb K_1$ and $\Bbb K_2$

I need to prove that if $[\Bbb K_1:\Bbb F]$ and $[\Bbb K_2:\Bbb F]$ both finite then $[\Bbb K_1 \Bbb K_2 : \Bbb F]\le [\Bbb K_1:\Bbb F][\Bbb K_2:\Bbb F]$.

Not sure how to do it, hints/partial solution would be great.

Thanks !

1 Answers1

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Hint:

If $\;\{v_1,...,v_n\}\,,\,\,\{w_1,...,w_m\}\;$ are basis of $\;\Bbb K_1/\Bbb F\,,\,\,\Bbb K_2/\Bbb F\;$ resp., then

$$\;\Bbb K_1\Bbb K_2=\text{Span}_{\Bbb F}\,\{\,v_iw_j\}_{1\le i\le n,\,1\le j\le m}\;$$

DonAntonio
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  • Thanks! I tried to do exactly that, I can't see why the product would span $\Bbb K_1 \Bbb K_2$ –  May 15 '18 at 12:18
  • This works. Assuming that there exists a bigger field $\Omega$ that contains both $\Bbb{K}_1$ and $\Bbb{K}_2$. For otherwise the products $v_iw_j$ are not necessarily defined... – Jyrki Lahtonen May 15 '18 at 12:21
  • @Liad Take a product of elements $;xy,,,,x\in\Bbb K_1,,,,y\in\Bbb K_2;$ . Then clearly $;xy\in\text{Span},{v_1w_j};$ . Take now any finite sum with coefficients in $;\Bbb F;$ and products as above: it clearly is in the given span... – DonAntonio May 15 '18 at 12:21
  • $\Bbb K_1 \Bbb K_2$ is the smallest field containing both $\Bbb K_1 $ and $\Bbb K_2$ , why are you looking at products? @DonAntonio –  May 15 '18 at 12:22
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    @Liad you can show that the span described by DonAntonio is closed under products. If that span is taken inside some integral domain, it follows that the span is a field. No smaller field can possibly contain both the fields, so you are done. – Jyrki Lahtonen May 15 '18 at 12:23
  • And Jyrki is right: the composition of fields requires both fields contained in a bigger one – DonAntonio May 15 '18 at 12:23