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The way I learned it was given a field extension $F \subset E$, and an element $\alpha \in E$

$$F(\alpha) := \{p(\alpha)/q(\alpha) : p(x), q(x) \in F[x] ,q(\alpha) \not = 0\} $$

Is there an easier way to think about the field $F(\alpha)$

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You can think of it as smallest subfield of $E$ that contains $F$ and $\alpha$ – Thomas Andrews Jul 9 '12 at 19:50
up vote 10 down vote accepted

The definition you've given yields the form of an arbitrary element of the extension. However, it is not the intuitive definition. The intuitive definition is simple: $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$.

In fact, your definition can also be somewhat simplified if $\alpha$ is algebraic over $F$. In this case $F(\alpha)=F[\alpha]$, that is, the smallest ring containing $F$ and $\alpha$, so $F(\alpha)=F[\alpha]=\lbrace p(\alpha)\vert p\in F[x]\rbrace$. Furthermore, algebraic elements have finite degree, so we only need to consider $p$ of degree smaller than the degree of $\alpha$ over $F$.

If $\alpha$ is transcendental over $F$, then $F(\alpha)$ is actually isomorphic to the field of rational functions over $F$, $F(x)$, in which case it is pretty much what you've written, but of course there's no restriction on $q$ in this case.

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+1 great answer – Belgi Jul 9 '12 at 20:07
also, if $\alpha$ is transcendental over $F$ then $F[\alpha]\cong F[x]$ – Belgi Jul 9 '12 at 20:08

Ok so the intuition is that you start with the field $F$ and throw into the pot this new "thing" $\alpha$ to make a field $F(\alpha)$, defined to be the smallest field containing $F$ and $\alpha$.

Let's try an example and discover what the field $\mathbb{Q}(\sqrt{2})$ looks like. Ok so this field has to contain all rational numbers. It also contains $\sqrt{2}$ and all rational multiples of this. Thus in fact $\mathbb{Q}(\sqrt{2})$ must contain all numbers of the form $a+b\sqrt{2}$ for rational $a,b$. It must also contain $\sqrt{2}^2, \sqrt{2}^3, ... $ and all rational multiples of these. But these powers of $\sqrt{2}$ are redundant in this counting (they can all be written in terms of numbers of the form $a+b\sqrt{2}$ and so are already accounted for).

Now $\{a+b\sqrt{2}\,|\,a,b\in\mathbb{Q}\}$ is actually a field itself (easily checked), so by minimality it must be equal to $\mathbb{Q}(\sqrt{2})$.

Now once you try to work out a few more of these you start to realise that the above process worked smoothly purely because of the fact that $\sqrt{2}$ is algebraic over $\mathbb{Q}$.

In general if $\alpha$ is algebraic over your initial field $F$ then this "redundancy" occurs...and that you only have to use the powers of $\alpha$ upto $d-1$, where $d$ is the degree of the "minimal polynomial" of $\alpha$ over $F$ (this being the smallest polynomial with $\alpha$ as a root over $F$).

We had to stop at the first power of $\sqrt{2}$ above because $\sqrt{2}$ satisfies a quadratic polynomial $x^2 - 2$ over $\mathbb{Q}$ and it satisfies no polynomial of smaller degree.

When $\alpha$ is transcendental over $F$ then all powers of $\alpha$ (positive and negative) and all sums, multiples, quotients of these must be thrown into the pot.

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For teaching purposes, I think you should also add the inverse to your second paragraph, i.e. "[..] contain $\sqrt{2}^2, \sqrt{2}^3, \ldots, \sqrt{2}^{-1}$ and all rational multiples [..]." Then it should be apparent to OP, that inverse $\alpha^{-1}$ is expressible in terms of lower powers of $\alpha = \sqrt{2},$ since $\alpha$ is algebraic over $\mathbb Q,$ namely $\alpha^{-1} = \frac{1}{2} \alpha.$ – user2468 Jul 9 '12 at 22:48
Yes this would probably have been better, although when checking that the required set is a field we do check this fact. – fretty Jul 10 '12 at 8:25

For an extension $E/F$ and $\alpha\in E$, if $\alpha$ is algebraic over $F$, then the extension $F(\alpha)\subseteq E$ can also be described in a way which is intrinsic to $F$ (i.e. which involves no mention of $E$).

To say that $\alpha$ is algebraic means there exists a non-zero (hence non-constant) polynomial $f\in F[X]$ with $f(\alpha)=0$. There is a unique $F$-algebra map $\varphi:F[X]\rightarrow E$ satisfying $X\mapsto(\alpha)$. The kernel is non-zero (because $f$ is in it) and proper (since this is a map of non-zero rings), so is of the form $(g)$ for a unique monic $g$, say of degree $d\geq 1$. One can characterize $g$ as the monic polynomial of minimal degree having $\alpha$ as a root. The homomorphism $\varphi$ gives rise to an isomorphism $F[X]/(g)\cong\mathrm{im}\varphi$. The image of $\varphi$ is $F[\alpha]\subseteq E$, the $F$-subalgebra of $E$ generated by $\alpha$, which can concretely be described as the set of polynomials in $\alpha$ with coefficients in $F$. This is a domain, being a subring of a field. One can also show that the elements $1+(g), X+(g),\ldots,X^{d-1}+(g)$ form a basis for $F[X]/(g)\cong F[\alpha]$ as an $F$-vector space. In particular, $F[\alpha]$ is of dimension $d$ as an $F$-vector space. From this it follows that $F[\alpha]$ is a field. Indeed, consider any non-zero element $\beta$ of $F[\alpha]$ and the $F$-endomorphism $x\mapsto\beta x:F[\alpha]\rightarrow F[\alpha]$. This is injective, hence surjective, since source and target have the same finite dimension, so there is $x$ with $x\beta=1$, and $\beta$ is a unit. Since, by definition, $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$, and $F[\alpha]$ is such a field, we must have $F[\alpha]=F(\alpha)$.

In summary, we have constructed an isomorphism (of $F$-algebras) $F[X]/(g)\cong F(\alpha)$ (along the way we showed that $F(\alpha)$ consists of all polynomials in $\alpha$ with coefficients in $F$). Since $F[X]/(g)$ is a field, $(g)$ must be a maximal ideal, and $g$ is irreducible. It follows that $g$ is the unique monic irreducible polynomial in $F[X]$ having $\alpha$ as a root. In this way you get a description of $F(\alpha)$ as $F[X]/(g)$ which doesn't directly involve $E$.

Conversely, if $g\in F[X]$ is a monic irreducible polynomial, then $(g)$ is maximal, so $E:=F[X]/(g)$ is a field containing $F$, and if $\alpha:=X+(g)$, then $\alpha$ is a root of $g$ in $E$, and $E=F(\alpha)$.

Note that this is only valid for $\alpha$ algebraic over $F$. As pointed out in one of the other answers, if $\alpha$ is transcendental over $F$, then the map $\varphi:F[X]\rightarrow E$ given by $X\mapsto\alpha$ is injective (this is exactly what it means for $\alpha$ to be transcendental over $F$), so one has $F[X]\cong F[\alpha]$ as $F$-algebras. This isomorphism of domains then extends uniquely to an $F$-algebra isomorphism of fraction fields $F(X)\cong F(\alpha)$.

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