How to show that $\mathbb Q[\sqrt 2]=\{a+b\sqrt 2:a,b\in\mathbb Q\}$ is the smallest subfield of $\mathbb R$ containing $\mathbb Q$ and $\sqrt 2?$
I've shown that $\mathbb Q[\sqrt 2]$ is a subfield of $\mathbb R.$ What about the smallest case?
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Is it clear to you what smallest means in this context? This kind of statement means that any other subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt2$ contains $\mathbb{Q}[\sqrt2]$. Can you show this? (It's fairly obvious and certainly the easier containment of the two to show...).
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$\begingroup$ Oh yes! BTW then any subring of $\mathbb R$ containning $\mathbb Q$ and $\sqrt 2$ contains $\mathbb Q[\sqrt 2].$ Isn't it? $\endgroup$ Apr 2, 2014 at 15:28
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$\begingroup$ It is actually the smallest subring containing $\mathbb Q$ and $\sqrt 2?$ $\endgroup$ Apr 2, 2014 at 15:29
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$\begingroup$ It is, yes, but be careful; in general a subring of a field isn't necessarily a field. $\endgroup$– ah11950Apr 2, 2014 at 15:30
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$\begingroup$ Such as $\mathbb Z[\sqrt 2]$ of $\mathbb R.$ right? $\endgroup$ Apr 2, 2014 at 15:31
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