If $F ⊂ \mathbb{C}$ and $α_1,\ldots,α_n$ ∈ $\mathbb{C}$ are algebraic over $F$, prove that $F(α_1, \ldots, α_n)$ is a simple extension of $F$.

Question

The statement is true for $n = 1$: $F(α_1)/F$ is finite (since $α_1$ is algebraic over $F$) and is simple, so we need to show that it holds for $n ≥ 2$. By induction on $n$; suppose the theorem is true for the base case $n = 2$. a) How do I prove it for $n + 1$?

Now we need to prove that it holds for $n = 2$. If $α$, $β ∈ \mathbb{C}$ are algebraic over $F$, we need to show that there is $γ$ such that $F(α,β) = F(γ)$. Let $f(x)$, $g(x)$ be the minimal polynomials of $α$, $β$ (respectively) over $F$. Let $k = \deg(f(x))$ and $l = \deg(g(x))$. Let $α_1 = α, α_2, \ldots,α_k$ be the roots of $f(x)$ in $\mathbb{C}$, and $β_1 = β, β_2, \ldots, β_l$ be the roots of $g(x)$ in $\mathbb{C}$. Let $a ∈ F$ be such that $a ≠ (α_i − α)/(β − β_j)$. b) Why does $a$ exist?

Let $γ = α + aβ$. We need to show that $F(α,β) = F(γ)$. c) How do we know that $F(γ) ⊂ F(α,β)$?. We need to show that $F(α,β) ⊂ F(γ)$. It is enough to show that $α,β ∈ F(γ)$. Let $h(x) = f(γ − ax)$. d) Why is $h(x) ∈ (F(γ))[x]$? e) How do we know that $β$ a root of $h(x)$? f) How do we show that none of $β_2, ···, β_l$ is a root of $h(x)$?

Let $g_0(x)$ be the minimal polynomial of $β$ over $F(γ)$. g) How do we know that $g_0(x)\mid g(x)$, and $g_0(x)\mid h(x)$? and that $h$) $β$ is the only root of $g_0(x)$? If this is true then $g_0(x) = a(x − β)b$ for some $b ≥ 1$ and $a ∈ F(γ)$. Now $g_0(x)$ is monic, $a = 1$, and $g_0(x) = (x − β)b$. Moreover, $g_0(x)$ is irreducible over $F(γ)$, so it has no repeated roots so $b = 1$ and $g_0(x) = x − β$. i) How do we know that $β ∈ F(γ)$? and that j) $α ∈ F(γ)$?

Answer 1

You may want to see Lubin's beautiful proof of the primitive element theorem here:

Geometric interpretation of primitive element theorem?

Answer 2

So many question...but alright here we go:

The statement is true for $n = 1$: $F(α_1)$/$F$ is finite (since $α_1$ is algebraic over $F$) and is simple, so we need to show that it holds for $n ≥ 2$. By induction on $n$; suppose the theorem is true for the base case $n$ = 2. a) How do I prove it for $n + 1$?

Once you know it for $n=2$ you then say $F(\alpha_1, \dots, \alpha_n,\alpha_{n+1})= F(\alpha_1, \dots, \alpha_n)(\alpha_{n+1}) = F(\gamma')(\alpha_{n+1})= F(\gamma)$, where first you use induction hypothesis and then the case $n=2$ for $\gamma'$ and $\alpha_{n+1}$.

Now we need to prove that it holds for $n$ = 2. If $α$, $β$ ∈ $\mathbb{C}$ are algebraic over $F$, we need to show that there is $γ$ such that $F(α,β)$ = $F(γ)$. Let $f(x)$, $g(x)$ be the minimal polynomials of $α$, $β$ (respectively) over $F$. Let $k = deg(f(x))$ and $l = deg(g(x))$. Let $α_1 = α, α_2, ···,α_k$ be the roots of $f(x)$ in $\mathbb{C}$, and $β_1 = β, β_2, ···, β_l$ be the roots of $g(x)$ in $\mathbb{C}$. Let $a ∈ F$ be such that $a ≠ (α_i − α)/(β − β_j)$. b) Why does $a$ exists?

It's just $F$ is infinite (note it is characteristic $0$ and contains the integers for instance) and only finitely many values are excluded.

Let $γ = α + cβ$.

That $c$ should be an $a$.

We need to show that $F(α,β) = F(γ)$. c) How do we know that $F(γ) ⊂ F(α,β)$?.

The element $α + aβ$ is in $F(α,β)$, but that's $\gamma $ and the claim follows.

We need to show that $F(α,β) ⊂ F(γ)$. It is enough to show that $α,β ∈ F(γ)$. let $h(x) = f(γ − ax)$. **d) Why is $h(x) ∈ (F(γ))[x]$?

You have a polynomial $f \in F[X]$, you plug in $γ − ax$ in, you expand, and collect the variables together, the coefficients will be made up of elements from $F$ and $\gamma$, so they are in $F(\gamma)$

e) How do we know that $β$ a root of $h(x)$?

Note that $h(\beta)= f(\gamma - a \beta)= f(\alpha)=0$. Recall $f$ was the minimal polynomial of $\alpha$.

f) How do we show that none of $β_2, ···, β_l$ is a root of $h(x)$?**

Because $\beta_i$ a root of $h(x)$ would mean $f(\gamma - a \beta_i)=0$ so $\gamma - a \beta_i$ would be a root of $f$, so one of the $\alpha_j$. But $a$ was chosen exactly so that this cannot be.

Let $g_0(x)$ be the minimal polynomial of $β$ over $F(γ)$. g) How do we know that $g_0(x)|g(x)$, and $g_0(x)|h(x)$? and that h) $β$ is the only root of $g_0(x)$?

Since $g$ is a polynomial over $F$ it is also a polynomial over $F(\gamma)$ and it has $\beta$ as a root. The minimal polynomial of $\beta$ over $F(\gamma)$ divides every polynomial over $F(\gamma)$ that has $\beta$ as root, whence in particular $g(x)$.

Also $h$ is a polynomial over $F(\gamma)$ that has $\beta$ as a root.

If this is true then $g_0(x) = a(x − β)b$ for some $b ≥ 1$ and $a ∈ F(γ)$. Now $g_0(x)$ is monic, $a = 1$, and $g_0(x) = (x − β)b$. Moreover, $g_0(x)$ is irreducible over $F(γ)$, so it has no repeated roots so $b = 1$ and $g_0(x) = x − β$. i) How do we know that $β ∈ F(γ)$? and that j) $α ∈ F(γ)$?

Because the $\beta$ arises as a coefficient of $g_0$ and this is polynomial over $F(\gamma)$ by construction. Then you get $\alpha$ from $\gamma$ and $\beta$ as $\alpha = \gamma - a \beta$, recall the $a$ was in $F$ so this is in $F(\gamma)$.

Hope I didn't miss any.