## 305 -Fields (5)

At the end of last lecture we stated a theorem giving an easy characterization of subfields of a given field ${\mathbb F}.$ We begin by proving this result.

Theorem 18. Suppose ${\mathbb F}$ is a field and $S\subseteq{\mathbb F}.$ If $S$ satisfies the following 5 conditions, then $S$ s a subfield of ${\mathbb F}:$

1. $S$ is closed under addition.
2. $S$ is closed under multiplication.
3. $-a\in S$ whenever $a\in S.$
4. $a^{-1}\in S$ whenever $a\in S$ and $a\ne0.$
5. $S$ has at least two elements.

Proof. Notice that $0\in S:$ If $a\in S$ (and there is some such $a,$ by condition 5) then $-a\in S$ by 3, and therefore $0=a+(-a)\in S$ by 1.

Similarly, $1\in S:$ First, there is some nonzero element in $S,$ since $S$ has at least two elements by 5, and only one of them at the most can be zero. Let $a\in S,$ $a\ne0.$ Then $a^{-1}\in S$ by 4, and therefore $1=aa^{-1}\in S$ by 2.

Once we have that $0,1\in S,$ the verification of the field axioms (see Definition 1 in lecture 4.1) is straightforward since $S\subseteq {\mathbb F},$ and ${\mathbb F}$ is a field. ${\sf QED}$

Examples. 1. ${\mathbb Q}\subseteq{\mathbb R}\subseteq{\mathbb C},$ so ${\mathbb Q}$ is a subfield of ${\mathbb R}$ and ${\mathbb C},$ and ${\mathbb R}$ is a subfield of ${\mathbb C}.$

2. ${\mathbb Z}_2$ is a subfield of both the field of 4 elements and the field of 8 elements. However, the field of 4 elements is not a subfield of the field of 8 elements. This is because any nonzero element of the field of 4 elements satisfies $x^4=x$ while any nonzero element of the field of 8 elements satisfies $x^8=x.$ This means that we would have at least one $x\ne0,1$ such that $x^4=x^8,$ but $x=x^4=x^8(x^4)^{-1}=1,$ contradiction.

3. $S=\{a+b\sqrt2 : a,b\in{\mathbb Q}\}$ is a subfield of ${\mathbb R}.$ To see this, one verifies conditions 1–5 of Theorem 18. Conditions 1–3 and 5 are straightforward. To check condition 4, notice that if $x\in S$ and $x\ne0,$ say $x=a+b\sqrt2,$ with $a,b\in{\mathbb Q},$ then it cannot be that both $a$ and $b$ are zero. Then $\displaystyle \frac1x=\frac{a-b\sqrt2}{a^2-2b^2},$ because $a-b\sqrt2\ne0.$ This is because $a-b\sqrt2=0$ implies that $b\ne0,$ since otherwise from $a-b\sqrt2=0$ we also have $a=0,$ and we cannot have both $a,b$ being zero simultaneously since $x\ne0.$ But then $a-b\sqrt2=0$ implies that $\sqrt2=a/b\in{\mathbb Q},$ which we know is not the case.

But then $\displaystyle\frac 1x=\left(\frac a{a^2-2b^2}\right)+\left(\frac {-b}{a^2-2b^2}\right)\sqrt2 \in S,$ and condition 4 follows.

Definition 19. If $S$ is a subfield of ${\mathbb F}$ and $a\in {\mathbb F},$ the smallest subfield of ${\mathbb F}$ that contains $S$ and has $a$ among its elements is denoted by $S(a).$

Next lecture we will show that there is indeed such a smallest subfield. Notice that in Example 3 above, we in fact have $S={\mathbb Q}(\sqrt2).$ This is because if ${\mathbb F}\subseteq{\mathbb R}$ is a field, ${\mathbb Q}\subseteq {\mathbb F},$ and $\sqrt2\in{\mathbb F},$ then $S\subseteq {\mathbb F},$ since ${\mathbb F}$ must be closed under addition and multiplication. But, since $S$ is already a field, there can be no smaller subfield of ${\mathbb R}$ that contains ${\mathbb Q}$ and has $\sqrt2$ as an element.

4. Similarly, ${\mathbb Q}(\sqrt n)=\{a+b\sqrt n : a,b\in{\mathbb Q}\}$ for any integer (in fact, any rational) $n.$ Here, we see ${\mathbb Q}$ and $\sqrt n$ as being in ${\mathbb C}$ rather than ${\mathbb R}$ in case $n<0.$ Note that if $n$ is already the square of a rational, then ${\mathbb Q}(\sqrt n)={\mathbb Q}.$

5. ${\mathbb Q}(\root 3\of 2)=\{a+b\root 3\of 2+c\root 3\of 4 : a,b,c\in{\mathbb Q}\}.$ Again, we use Theorem 18, and only condition 4 requires special care. Suppose that $a,b,c\in{\mathbb Q}$ and that $a+b\root 3\of2 + c\root 3 \of 4\ne0.$ We claim that $a^3+2b^3+4c^3-6abc\ne0.$ This is because of the following identity, closely related to the inequality between the arithmetic and the geometric means, see this post for further discussion:

$\displaystyle a^3 +2 b^3 + 4 c^3 -6abc = \frac 12 (a + \root 3\of{2} b + \root 3\of{4} c)\times$ $((a-\root 3\of 2 b)^2+(a-\root 3\of 4 c)^2+(\root 3\of 2 b-\root 3\of 4 c)^2).$

Notice that $a^3+2b^3+4c^3-6abc$ is the determinant of the matrix $\displaystyle\left(\begin{array}{ccc}a&2c&2b\\ b&a&2c\\ c&b&a\end{array}\right),$ so the system of equations $\displaystyle\begin{array}{rcrcrcl} ad&+&2ce&+&2bf&=&1\\ bd&+&ae&+&2cf&=&0\\ cd&+&be&+&af&=&0\end{array}$ has a unique solution $(d,e,f),$ and $d,e,f$ must be rational since they are obtained from $a,b,c$ by means of the elementary operations ($+,-,\times,\div$). But this means that $(a+b\root 3\of 2+c\root 3 \of 4)(d+e\root 3\of 2+f\root 3\of 4)=1,$ or $(a+b\root 3\of 2+c\root 3 \of 4)^{-1}=d+e\root 3\of 2+f\root 3\of 4.$

Definition 20. A number $r\in{\mathbb C}$ is transcendental iff it is not a root of any polynomial with rational coefficients.

6. The number $\pi$ is known to be transcendental. This is a deep theorem of Ferdinand von Lindemann from 1882. This means that we have no “concrete” description of ${\mathbb Q}(\pi)$ as in the examples above, since (for example) if $a+b\pi+c\pi^2=1/\pi$ for some rational $a,b,c$ then $\pi$ would be a root of the cubic polynomial $cx^3+bx^2+ax-1=0.$ On the other hand, it turns out that one describe ${\mathbb Q}(\pi)$ in easy terms: ${\mathbb Q}(\pi)=\{p(\pi)/q(\pi) : p,q$ are polynomials with rational coefficients, $q(x)\not\equiv0\}.$