The Algebraic Closure of the Rational Numbers

A while ago, I asserted that the field A is not a number field, since it has infinite dimension over Q. This is because the set {SQRT(2), SQRT(3), SQRT(5), SQRT(7), …} is linearly independent over Q. A commenter named Esther asked me for a proof, so here it is.

Suppose on the contrary that the set is linearly dependent. That is, that there exists a finite subset, say {SQRT(2), SQRT(3), …, SQRT(p)} such that a2*SQRT(2) + a3*SQRT(3) + … + a(p)*SQRT(p) = 0 and the a(p)’s are rational numbers not all zero.

There must be a minimal p that satisfies that condition, that is a p such that {SQRT(2), SQRT(3), …, SQRT(p)} is linearly dependent, but {SQRT(2), SQRT(3), …, SQRT(p‘)} is linearly independent, where p‘ is the largest prime less than p. We can write SQRT(p) as a rational combination of {SQRT(2), SQRT(3), …, SQRT(p‘)}, i.e. a2*SQRT(2) + … + a(p‘)*SQRT(p‘).

Now, let us look at the conjugates of the elements we have. SQRT(p) has minimal polynomial x^2 – p, so it has exactly two conjugates: itself, and -SQRT(p). The conjugates of a2*SQRT(2) + … + a(p‘)*SQRT(p‘) are of the form (+/-)a2*SQRT(2) (+/-) … (+/-) a(p‘)*SQRT(p‘).

Given any two primes between 2 and p‘, say q and r, we can look at a conjugate of a2*SQRT(2) + … + a(p‘)*SQRT(p‘) defined by SQRT(q) –> SQRT(q), SQRT(r) –> -SQRT(r), if it exists. If this conjugate is equal to SQRT(p), then by subtraction, we get that a linear equation on {SQRT(2), SQRT(3), …, SQRT(p‘)} whose SQRT(r) coefficient is 2a(r) is 0. By linear independence, then, a(r) = 0. Similarly, if this conjugate is equal to -SQRT(p), then a(q) = 0.

In other words, given any two different primes, at least one must have a zero coefficient. This means that at most one prime can have a nonzero coefficient, say q. Then a(q)*SQRT(q) = SQRT(p), i.e. (a(q))^2 * q = p in Q, which is false since p/q is never a perfect square for distinct primes p, q.

The final step, showing that the conjugate we want exists, requires working with Q(SQRT(q), SQRT(r)). It has at most four homomorphisms, defined by the images of SQRT(q) and SQRT(r). Showing it has exactly four, i.e. that it has degree 4, will force one of them to be what we want. I’ll do it in another post; here I’ll just show it has degree at least 3. This is enough because one homomorphism either sends SQRT(q) to SQRT(q) and SQRT(r) to -SQRT(r), or SQRT(q) to -SQRT(q) and SQRT(r) to SQRT(r); either will be sufficient for the proof.

But suppose that b1 + b2*SQRT(q) + b3*SQRT(r) = 0, where b(i) is in Q. Then b1 + b2*SQRT(q) = -b3*SQRT(r). Squaring both sides, we get b1^2 + 2b1b2*SQRT(q) + q*b2^2 = r*b3^2. Only SQRT(q) is irrational, so b1b2 = 0. If b1 = 0, then b2 and b3 are 0 since by assumption, SQRT(q) and SQRT(r) are contained in a linearly independent set. If b2 = 0, then b1 + b3*SQRT(r) = 0, so straightforwardly b1 = b3 = 0. Then b1 = b2 = b3 = 0, which means the set {1, SQRT(q), SQRT(r)} is linearly independent, and Q(SQRT(q), SQRT(r)) has degree at least 3.

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3 Responses to The Algebraic Closure of the Rational Numbers

  1. Uh oh maths..

    (tries to look intelligent, and then slinks off into obscurity)

  2. Alon Levy says:

    Hey, this I was actually asked to post.

  3. [...] Local fields are a lot more involved, because their algebraic closures have infinite degree over them. The same of course applies to global fields – recall that A has infinite degree over Q, for instance. Hensel’s lemma provides one local field analog of polynomial factorization in R, which more or less corresponds to the statement that every real polynomial of odd degree has a real root. [...]

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