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This is from Serge Lang's Algebra, in Chapter 5.

If $K$ is a field, and $K\subset E$ is a field extension, then for $\alpha\in E$, then $K(\alpha)$ is defined to be the smallest subfield of $E$ which contains $K$ and $\alpha$, which is given by the field of quotients $$K(\alpha) = \left\{\frac{f(\alpha)}{g(\alpha)}:f(x),g(x)\in K[x]\right\}$$.

I tried to prove this remark but am stuck on $(\subset)$.

The right hand side clearly contains $K$ by taking $f(x) = k$, and $g(x) = 1$ for any $k\in K$. But how can I represent $\alpha$ as such a quotient?

Edit: I think (my memory is a bit rusty here) that if $K$ is a field and $\alpha$ is some object not in $K$, we construct the field $K[x]$ which contains $K$ as a subfield, and identify $\alpha$ with the indeterminant $x$. Do I have to examine this embedding more closely or is this just going down the wrong train of thought?

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$f(x)=x$, $g(x)=1$. –  Gerry Myerson Nov 18 '12 at 0:25

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up vote 2 down vote accepted

The equality should be clear if you note that $K(\alpha)$ is the field of fractions of $K[\alpha]$ and recall how the field of fractions is constructed.

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ahh, i was writing an edit just as you typed this answer, I think I see what you are getting at –  Kyle Schlitt Nov 18 '12 at 0:27
    
actually when I think about it like that it seems like there is nothing to show. thanks! (will accept in 9 minutes) –  Kyle Schlitt Nov 18 '12 at 0:29

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