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\}.

Advertisements

2 Responses to 305 -Fields (5)

  1. […] Last lecture we characterized subfields and used the characterization to provide many new examples of fields. Now we start to explore systematically which subfields of the complex numbers are suitable to study the question of which polynomial equations can be solved. […]

  2. […] knew that Corollary 8 holds in a few particular cases, for example for for See also lecture 4.5. The argument above is much more general, and reduces significantly the amount of computations […]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: