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I recently asked a question about the transcendence degree of $\mathbb{C}$ over $\mathbb{Q}$. In a nice answer which I hope to understand better, two facts were given:

1: If $F$ is any infinite field and $K/F$ is an algebraic extension, then $\# K = \#F$.

2: For any infinite field $F$ and purely transcendental extension $F(X)$, we have $\# F(X) = \max (\#F, \# X)$.

For the first one, since $K$ is algebraic over $F$, then there is a surjection from $F[X]\to K$ mapping a polynomial $p(x)$ to one of its roots. Since $F$ is infinite, $\# F[X]=\# F$ so $\# F\geq\# K$. Since $K\supset F$, $\# K\geq \# F$, and equality follows. Is this correct?

I don't understand how the 2nd fact is true. Could some please explain it? Thank you.

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up vote 1 down vote accepted

The extension $F(X)$ is isomorphic to the field of fractions of the polynomial ring $F[X]$.

What is the cardinality of $F[X]$? Since for every $f\in F[X]$ there exists a finite $X_0\subseteq X$ such that $f\in F[X_0]$, and $\#F[X_0]=\#F$, we have $$\begin{align*} \#F[X] &= \#\bigcup_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}F[X_0]\\ &\leq \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F[X_0]\\ &= \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F\\ &= \#\{X_0\subseteq X\mid \#X_0\lt\infty\}\#F\\ &=\left\{\begin{array}{ll} \#X\#F &\text{if }\ \#X=\infty\\ 2^{\#X}\#F &\text{if }\ \#X\lt\infty \end{array}\right.\\ &= \max\{\#X,\#F\}. \end{align*}$$ On the other hand, $F\subseteq F[X]$ and $X\subseteq F[X]$, so $\max\{\#X,\#F\}\leq\#F[X]$. This gives equality.

The field of fractions can be viewed as a quotient of $F[X]\times(F[X]-\{0\})$ modulo the equivalence relation $(f,g)\sim (h,k)\Longleftrightarrow fk=gh$, and since $F[X]$ is infinite, this means that the cardinality is at most that of $F[X]$; since it contains a subring isomorphic to $F[X]$, we conclude that $\#F(X)=\#F[X]$, giving equality.

(A proof that the cardinality of the set of all finite subsets of an infinite set $X$ is equal to that of $X$ can be found here )

On rereading your argument for the first part, it is incomplete, I think.

If your $X$ represents a set of indeterminates, then you need to explain why it cannot have cardinality larger than $F$; if it represents a single indeterminate, then it is false that an algebraic extension must be a surjective image of the polynomial ring in one variable (for example, the algebraic closure of $\mathbb{Q}$ is certainly algebraic over $\mathbb{Q}$, but it is not simple, so it is not the surjective image of $\mathbb{Q}[x]$).

So you need to explain why $\#X\leq\#F$.

A better argument would be to note that there is a map from $K$ to $F[x]$ (single indeterminate) that sends every element of $K$ to its monic irreducible polynomial over $F$. This map is not one-to-one, but each element $f$ of $F[x]$ has at most $\deg(f)$ preimages. As this is finite, it follows that each fiber of the map $K\to F[x]$ is finite. Therefore, $$\begin{align*} \#K &\leq \sum_{f\in F[x]}\deg{f}\\ &\leq \sum_{f\in F[x]} \aleph_0\\ & = \#F[x]\aleph_0\\ &=\#F[x]\\ &=\#F \end{align*}$$ (since $F$ is infinite). Then noting $F\subseteq K$ gives the other inequality.

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Thank you, this is a nice comprehensive answer. Also, for peace of mind, was my rough explanation of the first fact correct? – jain Jan 28 '12 at 23:33
@jain: not quite: if your $X$ has more than one element, why can't it have more elements than $F$ has? – Arturo Magidin Jan 28 '12 at 23:35
@Arturo: On my system, your use of \! causes the sum sign to overlap the summands. Perhaps you should consider using \substack (see AMS short-math-guide) instead? – Zhen Lin Jan 29 '12 at 0:14
Thanks for clarifying my mistakes. – jain Jan 29 '12 at 0:23

Theorem I: If $F$ is an infinite ring then $F[x]$ has the same cardinality as $F$.

Proof: Since every polynomial in $F[x]$ is uniquely determined by its coefficients we have that $F[x]\simeq\{\langle x_0,\ldots,x_n\rangle\mid n\in\mathbb N, x_i\in F\}$, that is all the finite sequences of elements in $F$.

Since $|F|$ is infinite, we have that $|F|=|F\times F|=|\underbrace{F\times\cdots\times F}_{n\text{ times}}|=\ldots$, therefore the cardinality of $F[x]$ is the countable union of all those finite sequences therefore $|F[x]|=|\bigcup F^n|=\aleph_0\times|F|=\max\{\aleph_0,|F|\}=|F|$.

(Of course this requires quite some axiom of choice, but we're alright with that.)

Note that every algebraic number has a minimal polynomial, therefore there are at most $|F|$ many algebraic elements over $F$. Therefore every algebraic extension cannot have added more than $|F|$ new elements, thus if $K$ is algebraic over $F$ then $|K|=|F|$.

By induction (and a bit more) we have that for a countable number of variables $x_0,x_1,\ldots$ the polynomial ring in those variables has the same cardinality as $F$.

Once we add "enough" we are bound to create transcendental elements, simply because there can only be so many polynomials over $F$. Transcendental extensions correspond to the field of fractions of $F[X]$ for some set of variables $X$ (which can be infinite, of course).

The field of fraction has cardinality of $|F[X]\times F[X]|=|F[X]|$. Every polynomial in $F[X]$ is a finite sequence from $X$ and a finite sequence from $F$. As we did in Theorem I we have that the finite sequences from both $X$ and $F$ have the cardinalities of $X$ and $F$ respectively, therefore $|F[X]|=|F|\cdot|X|=\max\{|F|,|X|\}$.

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Thank you, I appreciate it. – jain Jan 29 '12 at 0:24

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