Intereting Posts

Strangest Notation?
Using $O(n)$ to determine limits of form $1^{\infty},\frac{0}{0},0\times\infty,{\infty}^0,0^0$?
Hensel Lifting and solving with mods
If $f(x)$ is uniformly continuous for $x \ge 0$, then it is bounded above by a straight line.
Extending the logical-or function to a low degree polynomial over a finite field
Demonstrate using determinant properties that the determinant of matrix “A” is equal to, 2abc(a+b+c)^3
Dedekind's cut and axioms
$ \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx$
Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
Integration of exponential and square root function
Definition of time-constructible function
How fast does the function $\displaystyle f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $ grow?
Reference for automorphic forms
Show that number of solutions satisfying $x^5=e$ is a multiple of 4?
algebraic element over a field

My definition of a number field is “a finite extension of $\mathbb Q$”. I want to prove that if $L$ is a finite field extension of $\mathbb Q$, then $L = \mathbb Q(\alpha)$ for some $\alpha$ algebraic over $\mathbb Q$.

I can prove things that look helpful. I know that if $L/K$ is a finite extension with $\mathrm{char}K = 0$ and with finitely many intermediate fields, then $L = K(\alpha)$ for some algebraic $\alpha$. I also know that if $\alpha$ is algebraic over $K$, then there are only finitely many intermediate fields between $K$ and $K(\alpha)$ (where those fields are determined by the factors of the minimal polynomial of $\alpha$ over $K$).

I can’t pull a proof together though. I’d obviously be done if I could prove that a finite extension of $\mathbb Q$ has finitely many intermediate extensions. Any hints would be greatly appreciated.

- Are there any good algebraic geometry books to recommend?
- Elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$.
- Beginner's text for Algebraic Number Theory
- When exactly is the splitting of a prime given by the factorization of a polynomial?
- Is there any trivial reason for $2$ is irreducible in $\mathbb{Z},\omega=e^{\frac{2\pi i}{23}}$?
- Eisenstein and Quadratic Reciprocity as a consequence of Artin Reciprocity, and Composition of Reciprocity Laws

- Other interesting consequences of $d=163$?
- On determining the ring of integers of a cubic number field generated by a root of $x^3-x+1$
- The uniqueness of a special maximal ideal factorization
- Showing that $\operatorname {Br}(\Bbb F_q)=0$
- Decomposition of a primitive regular ideal of a quadratic order
- What is the quotient $\mathbb Z/(1+2\sqrt{3})$?
- $\overline{\mathbb{Z}}$ is not a Dedekind domain.
- Proof of Stickelberger’s Theorem
- When is the sum of two squares the sum of two cubes
- Dedekind's theorem on the factorisation of rational primes

Here’s a proof that picks up with your observations (that it suffices to verify that there are finitely many intermediate subfields):

Let $L^{gc}$ denote the Galois closure of $L$. Then every subfield of $L$ is a subfield of $L^{gc}$, and subfields of $L^{gc}$ are, by Galois theory, in 1-1 correspondence with the subgroups of the Galois group $\operatorname{Gal}(L^{gc}/\mathbb{Q})$. Since the Galois group is finite, there are finitely many subgroups, hence finitely many subfields.

The result you’re looking for is more general. In fact, it holds for any finite separable extension of fields. In particular, it holds for any finite extension of a perfect field (e.g. of $\mathbf{Q}$). I will sketch a proof below.

Claim: If $L/K$ is a finite separable extension of fields, then there is an element $\gamma\in L$ such that $L=K(\gamma)$.

**Notation**: If $x\in L$ and if $F$ is a subfield of $L$, then I will let $m_{x,F}$ denote the minimal polynomial of $x$ over $F$. Note that the minimal polynomial satisfies the following property: if $b\in L$ and if $P\in F[X]$ with $P(b)=0$, then $m_{x,F}|P$ in $F[X]$.

**Proof**: Suppose that $L/K$ is a finite separable extension of an infinite field $K$ (if $K$ is finite, the claim is easy to show) and let $\overline{L}=\overline{K}$ be an algebraic closure (one can quickly show that every finite extension of $\mathbf{Q}$ is separable; see Keith Conrad’s notes on perfect fields). Since $L/K$ is finite, there are elements $a_1,\ldots, a_n\in L$ such that $L=K(a_1,\ldots, a_n)$. We can prove that $L=K(\alpha)$ for some $\alpha\in L$ by induction on the number of generators.

Let $F=K(a_1,\ldots, a_{n-2})$, so that $L=F(a_{n-1},a_n)$. Put $\alpha=a_{n-1}$ and $\beta=a_n$. Consider the finite set $$S=\{\lambda\in K | \lambda=\frac{\alpha-\alpha’}{\beta’-\beta}\text{ with } \alpha’, \beta’\in\overline{K} \text{ roots of }m_{\alpha,K}\text{ and }m_{\beta,K}\text{ resp., and }\beta’\neq\beta\}$$ Let $\lambda\in K\backslash S$ (which exists since $K$ is infinite). Consider the sub-extension $L/F(\gamma)/K$ where $\gamma=\alpha+\lambda\beta$. Suppose that $L\neq F(\gamma)$; we will search for a contradiction. Note that if $\beta\in F(\gamma)$, then so is $\alpha$ (since $S$ contains $0$, $\lambda\neq 0$). We therefore must have $\deg m_{\beta,F(\gamma)}(X)\geq 2$. Consider the polynomial $m_{\alpha,K}(\gamma – \lambda X)\in F(\gamma)[X]$, which has $\beta$ as a root. By the general fact on minimal polynomials that I stated at the top, we have that $m_{\beta,F(\gamma)}|m_{\alpha,K}(\gamma – \lambda X)$ in $F(\gamma)[X]$. In particular, if $\beta’\neq \beta$ is a root of $m_{\beta,F(\gamma)}$ (which exists since $m_{\beta,F(\gamma)}$ is of degree $>1$), then $\beta’$ is also a root of $m_{\alpha,K}(\gamma – \lambda X)$; that is to say, the element $\alpha’:=\gamma – \lambda\beta’$ is a root of $m_{\alpha,K}(X)$, and therefore $\lambda = \frac{\alpha – \alpha’}{\beta’-\beta}$ and $\lambda\in S$ (a contradiction).

Therefore, $L=F(\gamma)=K(a_1,\ldots,a_{n-2},\gamma)$ and we’ve reduced the number of generators of $L/K$ by $1$, go by induction now.

Use the primitive element theorem to get the existence of a generator. Then the fact that a number field forms a finite extension of $\mathbb{Q}$ tells you that any such generator has to be algebraic (since finite extensions are automatically algebraic).

- Operator norm of orthogonal projection
- Generators of a certain ideal
- Polar Decomposition: Adjoint
- Expression for the Maurer-Cartan form of a matrix group
- Unit and counit are close to being inverses
- show $\sum_{cyc}(1-x)^2\ge \sum_{cyc}\frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$
- Structure of $x^2 + xy + y^2 = z^2$ integer quadratic form
- Does $|x^G|$ divide the order $\langle x^G\rangle$?
- What is the practical difference between abstract index notation and “ordinary” index notation
- Let $\{K_i\}_{i=1}^{\infty}$ a decreasing sequence of compact and non-empty sets on $\mathbb{R}^n.$ Then $\cap_{i = 1}^{\infty} K_i \neq \emptyset.$
- Closed subspace of $l^\infty$
- Composing permutations in factorial notation
- Proof concerning Mersenne primes
- Cesàro operator is bounded for $1<p<\infty$
- Well-ordering theorem and second-order logic