# smallest subfield containing transcendental or algebraic element

Consider the extension $\mathbb R/ \mathbb Q$. what is the smallest subfield $\mathbb Q(\pi)$ of $\mathbb R$ containing $\mathbb Q$ and $\pi$. I think it is $\mathbb R$. A more general question: for all transcendental elements $a$ in an extension $L/K$, we have that the smallest subfield $K(a)$ of $L$ containing $K$ and $a$ is $L$ itself. But if the element $a$ is algebraic then the smallest subfield $K(a)$ of $L$ containing $K$ and $a$ is the field $K[a]\subset L$ of elements $p(a)$, $p\in K[X]$. Is that correct? Do we have an example in which the element $a$ is algebraic and still the smallest subfield $K(a)$ of $L$ containing $K$ and $a$ is $L$ itself?

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$\mathbb Q(\pi) = \{\frac{p(\pi)}{q(\pi)} \mid p, q\in \mathbb Q[X], q \ne 0 \}$. That doesn't equal $\mathbb R$, as it is countable. – martini Jun 25 '12 at 10:56
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If $a$ is transcendental over $K$ and lies in some extension $L$, then the smallest subfield of $L$ containing $K$ is... just $K(a)$! That is, it consists of rational functions in $a$ with coefficients in $K$. That's it. – Qiaochu Yuan Jun 25 '12 at 11:09
so for transcendental elements, the smallest subfield $K(a)$ containing $K$ and $a$ is the field of fractions, $p(a)/q(a)$, $p,q\in K[X]$ is that a general statement for all transcendental elements? – palio Jun 25 '12 at 11:10
Oh i see now that we have a definition confusion. actually i DEFINE $K(a)$ as the smallest subfield containing $K$ and $a$ and not as the field of rational functions in $a$ with coefficients in $K$ and if i well understand you these two notions coincide when $a$ is transcendantel? – palio Jun 25 '12 at 11:13

Let $K$ be a field and let $L$ be an overfield of $K$. For any $a\in L$, we define $K[a]$ to be the smallest subring of $L$ that contains $K$ and $a$, and we define $K(a)$ to be the smallest subfield of $L$ that contains $K$ and $a$. We have:

Theorem. Let $K\subseteq L$ be fields, and let $a\in L$.

1. $K[a] = \{p(a)\mid p(x)\in K[x]\}$.
2. $K(a) = \left.\left\{\frac{p(a)}{q(a)}\;\right|\; p(x),q(x)\in K[x], q(a)\neq 0\right\}$.
3. If $a$ is algebraic over $K$, then $K[a]=K(a)$, and $K(a)\cong K[x]/\langle f(x)\rangle$, where $f(x)$ is the minimal irreducible of $a$ over $K$.
4. If $a$ is transcendental over $K$, then $K(a)\cong K(x)$, the field of rational functions in one indeterminate over $K$.

In particular, note that for $K=\mathbb{Q}$, $L=\mathbb{R}$, and $a=\pi$, we have that $\mathbb{Q}(\pi)\cong\mathbb{Q}(x)$ is countable, and so cannot be isomorphic to $\mathbb{R}$.

Proof. 1. The homomorphism $K[x]\to K[a]$ induced by mapping $x\to a$ shows that $\{p(a)\mid p(x)\in K[x]\}$ is contained in $K[a]$. Conversely, the image of this map is a subring of $L$ that contains $K$ and $a$, and so $K[a]$ is contained in the right hand side; this proves 1.

1. $K[a]\subseteq K(a)$, so $K(a)$ contains the field of fractions of $K[a]$, which is the right hand side of 2. Conversely, the right hand side is easily seen to be a field that contains both $K$ and $a$, giving the other inclusion.

3 and 4. If $a$ is transcendental, then the map $K[x]\to K(a)$ is one-to-one, and so induces an embedding $K(x)\hookrightarrow K(a)$. By the description in 2 the map is onto, giving the isomorphism in 4. If $a$ is algebraic, then the kernel of the map $K[x]\to K[a]$ given by evaluation is the ideal generated by the minimal polynomial of $a$ over $K$. This is an irreducible, so $\langle f(x)\rangle$ is a maximal ideal of $K[x]$, hence the quotient $K[x]/\langle f(x)\rangle$ is a field; this field embeds into $K[a]$ and by 1 is onto, showing that $K[a]$ is actually a field (and hence equal to $K(a)$), and giving the isomorphism in 3. $\Box$

Your other assertions are incorrect: Say $L=\mathbb{Q}(x)$, $K=\mathbb{Q}$, and $a=x^2$. Then $a$ is transcendental over $K$, but $K(a)\neq L$; worse, take $L=\mathbb{Q}(x_1,\ldots,x_n)$ and $a=x_1$; then $L$ is a transcendental extension of $K(a)$.

As to the final question, note that by the Primitive Element Theorem, if $L$ is any finite separable extension of $K$, then there exists $a\in L$ such that $L=K(a)$; so for example every finite algebraic extension of $\mathbb{Q}$ will have one such $a$.

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Most of the question has been handled in the comments, all but the very last bit.

If $a$ is algebraic over $K$, and we take $L$ to be $K(a)$, then of course the smallest subfield of $L$ containing $K$ and $a$ is $L$ itself.

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