# How to think of the field $F(\alpha)$

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

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

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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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