Why should the real numbers interact with number fields and Galois theory?

Question

When I wanted to show $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not Galois, my argument was it's a real field, while the normal closure clearly contains complex numbers. However I'm unsatisfied with this argument, suppose someone who never met $\mathbb{R}$ asks this, he just innocently takes $x^3-2$ and wonders if adjoining one root gives a Galois extension. How would he come up with the $\mathbb{R}$ argument? It feels unmotivated- it seems he'd need to say something like the algebraic closure of $\mathbb{Q}$ has an order two automorphism, and understand its interaction with the roots of $x^2 - 3$.

** There is an argument that avoids $\mathbb{R}$, namely that $\phi(3)=2$ so this divides the splitting field index, but it's a different argument and requires $\phi(n) \nmid n$.

Thus my question is:

1. Are there more arguments for why $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not Galois

2. Is there a better motivation to how and why $\mathbb{R}$ is helpful, and how one naturally arrives at it?

3. Are there more striking examples of $\mathbb{R}$ playing an 'algebraic' role? (I'm familiar with the geometry of numbers, but that's not the type of examples I'm looking for, there it plays an analytic role).

Answer 1

I'm not quite sure what you want to see, but here are a number of special "algebraic" properties of the field $\Bbb{R}$. They are highly interdependent and, of course, do rely on the topological properties of the real numbers (completeness mostly).

1. Every odd degree polynomial with real coefficients has a real zero, and hence $\Bbb{R}$ has no odd degree extensions.
2. The field $\Bbb{R}$ has no non-trivial automorphisms.
3. The field $\Bbb{R}(i), i^2+1=0$, happens to be algebraically closed.
4. The (associative) division algebras over $\Bbb{R}$ are few, and have degrees $1,2,4$ only.
5. In class field theory (algebraic number theory on steroids) we look at behavior of primes of a number field $K$ in its finite extensions. To get a complete and coherent picture of the situation we need to include "the infinite prime", which means that we also look at real/complex embeddings of $K$ (and the extension).

Items 1-3 are obviously connected, but don't really need the topological completeness of $\Bbb{R}$. In fact, they also hold for (define, even) so called real closed fields, see here for a local explanation of item 3. Furthermore, a theorem due to Artin and Schreier tells us that if an algebraically closed field $\overline{K}$ is a finite extension of its subfield $L$, then we are in characteristic zero and $\overline{K}=L(i)$.

Items 4 and 5 are also connected, but I don't know if the topological completeness of $\Bbb{R}$ is necessary or not. Some approaches/proofs rely on analytical techniques, so ostensibly need completeness. But I am too ignorant to say whether completeness is absolutely necessary there.

Answer 2

I think there should be two kind of separate questions here: can we avoid arguments based on the order on $\mathbb{Q}$, and can we avoid the use of the "non-algebraic" real numbers $\mathbb{R}$.

The answer to the second question is definitely yes. In fact, for basically any algebraic/arithmetic application, you can completely safely replace $\mathbb{R}$ by the real closure of $\mathbb{Q}$, which is an algebraic extension. You can describe it as the intersection of $\mathbb{R}$ and the algebraic numbers in $\mathbb{C}$, but it can be constructed with no mention whatsoever to $\mathbb{R}$ or $\mathbb{C}$. Jyrki's answer highlights some properties of real closed fields that can be useful. In particular, the algebraic closure of $\mathbb{Q}$ is obtained from this real closed field by adjoining a square root of $-1$, so you can mimic all arguments based on the complex conjugation.

The first question is more delicate. The first answer that comes to my mind is "but why would you want to, though?". The fact that $\mathbb{Q}$ is (uniquely) ordered is an interesting algebraic property, and there is no reason not to use it. There is a whole theory of field orderings, with connections with quadratic forms, valuation theory, algebraic geometry, etc. Using it as an argument to show that (for instance) some extension is or is not Galois should not be seen as "abandoning" the use of "purely arithmetic" arguments (whatever that means). There is no universal catch-all method for dealing with Galois extensions of $\mathbb{Q}$, and using the fact that $\mathbb{Q}$ is an ordered field, with a real closure, and that some number fields can be embedded in this closure while some cannot should be part of the standard toolkit.

Answer 3

We can show $K = \mathbb{Q}(\sqrt[3]{2})$ is not Galois using only computations over $\mathbb{Q}$, as follows:

We want to show the only root of $f=T^3 - 2$ in $K$ is $\alpha=\sqrt[3]{2}$. Suppose that $\beta \in K$ is a root of $f$, and write $\beta = x + y\alpha + z\alpha^2$. Compute the coefficients of $\beta^3 - 2$ w.r.t the basis $(1,\alpha,\alpha^2)$, by hand or like this in SageMath:

sage: K.<a> = NumberField(x^3 - 2)
sage: A.<x,y,z> = PolynomialRing(K)
sage: v = sum(vector(c)*m for (c,m) in (x+y*a+z*a^2)^3 - 2)
sage: show(v)

$$\left(x^{3} + 2 y^{3} + 12 x y z + 4 z^{3} - 2,\,3 x^{2} y + 6 y^{2} z + 6 x z^{2},\,3 x y^{2} + 3 x^{2} z + 6 y z^{2}\right)$$

These three polynomials define an ideal $I$ in the ring $\mathbb{Q}[x,y,z]$:

sage: B.<x,y,z> = PolynomialRing(QQ, order='lex')
sage: I = B.ideal(list(v))

Computing a Gröbner basis of $I$ w.r.t lexicographic ordering shows that $V(I)$ is finite:

sage: I.groebner_basis()
[x^3 + 2*y^3 + 3240/7*z^9 - 216*z^6 - 26/7*z^3 - 2,
 x*y - 270/7*z^8 + 18*z^5 + 9/14*z^2,
 x*z - 54*y^2*z^6 + 9*y^2*z^3,
 y^4 - 108*y*z^6 - 2*y*z^3 - y,
 y^3*z + 27/7*z^7 - 2*z^4 + 1/28*z,
 y*z^7 + 1/108*y*z,
 z^10 - 1/2*z^7 + 1/108*z^4 - 1/216*z]

The last element gives the equation $216z^{10} - 108z^7 + 2z^4 - z = 0$ which, by the rational root theorem, has $z=0$ as its only root in $\mathbb{Q}$. The fourth element then yields $y^4 - y = 0$, hence for $y \in \mathbb{Q}$ we get $y=0$ or $y=1$. If $y=0$ then the first element yields $x^3 - 2 = 0$ which is a contradiction for $x \in \mathbb{Q}$. So $y=1$, and the second element shows $x=0$. In summary $(x,y,z) = (0,1,0)$, which means $\beta=\alpha$, so we are done.

In general there are algorithms to compute a Gröbner basis, the radical of a zero-dimensional ideal, the intersections with the univariate polynomial rings, and rational roots of univariate polynomials. This may be not the most efficient way, but it works and uses only arithmetic in $\mathbb{Q}$.

Answer 4

Your question is equivalent to showing that $x^3 -2$ does not split into linear factors over $\mathbb{Q}(\sqrt[3]{2})$. You can show that it splits into a linear and a quadratic factor over it, and all you need to do is to show that quadratic factor is irreducible over $\mathbb{Q}(\sqrt[3]{2})$. From this point of view, it's quite natural to consider $\mathbb{R}$, as $\mathbb{Q}(\sqrt[3]{2})$ embeds into it, and we have a simple condition for irreducibility of degree two polynomials over $\mathbb{R}$. I'm not sure if it answers your question though.