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Keith Kearnes

 Room for Mary Star and Keith Kearnes

Keith Kearnes
Jan 25, 2017 06:04
I am going to stop, since it looks like all of my lines are running together.
Keith Kearnes
Jan 25, 2017 06:04
.
Keith Kearnes
Jan 25, 2017 06:04
We do not have this, since $E_1/F(B_1\cup B_2)$ need not be an extension.
Keith Kearnes
Jan 25, 2017 06:03
Since the extensions $E/F(B_1)$ and $E/F(B_2)$ are algebraic, we have that the extensions $E_1/F(B_1\cup B_2)$ and $E_2/F(B_1\cup B_2)$ are also algebraic, or not?
Keith Kearnes
Jan 25, 2017 06:02
You are correct in your suspicion. We need not have $B_2\sibseteq E_1$. To see an example, let $\mathbb F = \mathbb Q$, $B_1 = \{e^{\sqrt{2}\}$, $B_2 = \{e^{\sqrt{3}\}$. Then $B_2\not\subseteq \mathbb E_1 = \mathbb Q(e^{\sqrt{2}})$.
Keith Kearnes
Jan 25, 2017 05:56
<p>I have done the following: </p>

<p>Let $B_1$ be a transcendental basis of $E_1/F$ and let $B_2$ be a transcendental basis of $E_2/F$. </p>

<p>So, we have that the extensions $E_1/F(B_1)$ and $E_2/F(B_2)$ are algebraic. </p>

<p>We have that $B_1\subset B_1\cup B_2$ and $B_2\subset B_1\cup B_2$. </p>

<p>Do we get the extensions $F(B_1)\leq F(B_1\cup B_2)\leq E_1$ and $F(B_2)\leq F(B_1\cup B_2)\leq E_2$ ? Or do we not know is $B_2\subseteq E_1$ and $B_1\subseteq E_2$ ? </p>
Keith Kearnes
Jan 25, 2017 05:55
Now I cut and paste from your question:
Keith Kearnes
Jan 25, 2017 05:53
I will make a few comments on your work.
Keith Kearnes
Jan 24, 2017 22:03
I have never used chat before, so I might make some mistakes.
Keith Kearnes
Jan 24, 2017 22:03
Yes.
Keith Kearnes
Jan 24, 2017 22:03
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