# Splitting fields & degrees of extensions

In the last post, We’ve seen ways to extend an existing field. In this post, I would like to do two things:

1. Discuss about the degrees of extensions
2. Bring the term of a splitting field and prove it’s uniqueness (up to isomorphism).

Let’s begin

## Multiplicative Property

Suppose that $F\subseteq L \subseteq K$ are three fields. I’ve already proved that $L,K$ are vector spaces over $F$ and $K$ is also a vector space over $L$. Then we can look at their dimentions: $[K:F],[K:L],[L:F]$.

Let’s try to find a relation between these 3 numbers.

Suppose that:

• $\{u_1,u_2,\dots,u_n\}\sub K$ is a basis for $K/L$.
• $\{v_1,v_2,\dots,v_m\}\sub L$ is a basis for $L/F$

Consider some element $\alpha\in K$, we can write $\alpha$ as a linear combination of $u_1,u_2,\dots,u_n$ with saclars from $L$:

\alpha=a_1u_1+a_2u_2+\cdots+a_nu_n=\sum_{i=1}^na_iu_i

However, $a_i\in L$, so we can express each one of them as a linear combination of $v_1,v_2,\dots,v_k$ with scalars from $F$:

a_i=b^i_1v_1+b^i_2v_2+\cdots+b^i_mv_m=\sum_{j=1}^mb_j^iv_j

We now have:

\alpha=\sum_{i=1}^na_iu_i=\sum_{i=1}^n(\sum_{j=1}^mb_j^iv_j)u_i=\sum_{i=1}^n\sum_{j=1}^mb_j^iv_ju_i

So $\alpha$ is a linear combination of elements from the set $\{u_iv_j\}_{1\leq j\leq n,1\leq i\leq m}$. In other words, this set is spanning $K$ as a vector space over $F$, that is: $\text{span}\{v_ju_i\}_{1\leq j\leq n,1\leq i\leq m} = K$.

Great, we now have a spanning set for $K$ as a vector space over $F$. Is that set also linearly independent? Let’s figure it out: Consider the linear combination:

\sum_{i=1}^n\sum_{j=1}^mb_j^iv_ju_i=0

Where $b_j^i\in F$. Note that $b_j^iv_ח$ is an element in the field $L$. Moreover:

\sum_{i=1}^n\sum_{j=1}^mb_j^iv_ju_i=\sum_{i=1}^n\underbrace{(\sum_{j=1}^mb_j^iv_j)}_{\in L}u_i=0

Therefore, we found a linear combination of the $u_i$‘s with scalars from $L$. Since $\{u_i\}_{i=1}^n$ is a basis for $K$ over $L$, we conclude that the scalars $\sum_{j=1}^nb_j^iv_j$ must be zero:

\sum_{j=1}^mb_j^iv_j=0

However, this is a linear combination of the $v_j$‘s with scalars in $F$, again, since $\{v_j\}_{j=1}^m$ is a basis for $L$ over $F$, the scalars must be zero: $b_j^i=0$. Hence, the initial linear combination is ‘trivial’, so the set is linearly indepenet as we thought.

Cool, we’ve found a basis for $K/F$, which is $\{u_iv_j\}$. It contains $n\cdot m$ element, therefore:

[K:F]=n\cdot m = [K:L]\cdot[L:F]

And that’s a really neat connection. This property is called the Multiplicative Property of field extentions.

##### Quick exmaple

Consider the fields:

\begin{array}{c}
\mathbb{Q}[\sqrt{2},\sqrt{3}]\\
|\\
\mathbb{Q}[\sqrt{2}]\\
|\\
\mathbb{Q}
\end{array}

We would like to know what is the degree of the extension $\mathbb{Q}[\sqrt{2},\sqrt{3}]/\mathbb{Q}$.

The multiplicative Property of field extentions tells us that:

[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}[\sqrt{2}]]\cdot[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]

We know that $[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}] = 2$ since the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $m_{\sqrt{2}}(x)=x^2-2$.

Similarly, $[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}[\sqrt{2}]]=[(\mathbb{Q}[\sqrt{2}])[\sqrt{3}]:\mathbb{Q}[\sqrt{2}]] = 2$ since the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}[\sqrt{2}]$ is $m_{\sqrt{3}}(x)=x^2-3$ (try to prove it!).

Finally, we get:

[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\sqrt{3}]:\mathbb{Q}[\sqrt{2}]]\cdot[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]=2\cdot 2=4

This property is a going to be quite handy in the future, so make sure that you understand it.

## Splitting fields

Let $K/F$ be a field extention, and $f\in F[x]$ a polynomial. We say that $f$ splits in $K$ if there are $\alpha_1,\alpha_2,\dots,\alpha_k\in K$. such that:

f(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)\in K[x]

and we say that $K$ splits $f$.

I think that the same ‘splits’ fits perfectly here, and makes the definition really intuitive.

As It turns out – every polynomial $f$ over some field $F$ splits in some field $K$ that extends $F$. Before I prove it, I want to prove a short statement first:

#### Just a small statement

\alpha\in K\text{ is a root of }f\iff (x-\alpha)|f(x)

Suppose that $(x-\alpha)|f(x)$, then $f(x)=(x-\alpha)g(x)$, hence: $f(\alpha)=(\alpha-\alpha)g(\alpha)=0$. So, $\alpha$ is a root of $f$.

Conversely, suppose that $\alpha$ is a root of $f$. Since $K[x]$ is an euclidian domain we can devide $f$ by $(x-\alpha)$ with a remainder:

f(x)=(x-\alpha)q(x)+r(x)

where $\deg(r(x)). Thus, $\deg(r(x))=0$ which implies that $r(x)=c$ for some $c\in K$. We can now plug in $\alpha$ in the equation to get:

f(\alpha)=(\alpha-\alpha)q(\alpha)+c

Since $\alpha$ is a root of $f$, $f(\alpha) = 0$ and we get: $0=c$. So, $f(x)=(x-\alpha)q(x)$, and that’s exactly the same as: $(x-\alpha) | f(x)$.

#### The proof

I am going to prove the statement by induction on the degree of $f$. The case $n=1$ is trivial (it is a linear polynomial and we don’t need to extend the field).

For the general case, there are two options:

• $f = f_1f_2$ is reducible: Then $\deg f_1,\deg f_2 < \deg f$. By induction, there is a field $K$ that extends $F$ and $f_1$ splits in it. Again, by induction, there is a field $E$ that extends $k$ where $f_2$ splits in. Now notice that $E$ also splits $f$.
\begin{array}{cc}
E & _{(f_{2})}\\
|\\
K & _{(f_{1})}\\
|\\
F
\end{array}
• $f$ is irreducible. Then we already proved that there exists a field $K$ that extends $F$, where $f$ has a root in ($K=F[x]/\langle f(x)\rangle$). Denote this root by $\alpha$. Then by the lemma, we get that $f(x)=(x-\alpha)g(x)$ in $K$ where $\deg g = \deg f -1$. By induction, there is a field $E$ that extends $K$ that splits $g$, and this field splits $f$ as well.

## Improving the statement

We just proved the existence of a field that splits $f$, however, we don’t really know anything about it except the fact that it splits $f$, and there might even be more than one field that splits $f$.

It turns out that if $f\in F[x]$ is a polynomial of degree $n$, then there is a field $k$ that splits $f$ such that $[K:f]\leq n!$.

The proof for that is done by induction on the degree of $n$, suppose that $p|f$ is an irreducible divisor of $f$. Therefore, in the field $F_1=F[x]/\langle p\rangle$ $p$ has a root $\alpha$. In addition, we also know that $[F_1:F]=\deg p\leq \deg f$. Notice that $\alpha$ is a root of $f$ as well, therefore, in the field $F_1$ : $f(x)=(x-\alpha)f_1$ where $\deg f_1 = \deg f -1$. By induction, there exists a field $E$ that extends $F_1$ that splits $f_1$ with a degree bounded by $(n-1)!$. The field $E$ splits $f$ as well, we can now use the multiplicative property to get:

[E:F]=[E:F_1]\cdot[F_1:F]\leq(n-1)!\cdot n=n!

As we wanted.

## The splitting field

Suppose that $f\in F[x]$. We say that $(F\sub)E$ is a splitting field if $E$ splits $f$ while also being the minimal field with this property. i.e. there is no proper subfield $E^\prime \sub E$ where $f$ splits in.

How can we create such a filed? Let’s try to find out: Suppose that $E$ is a splitting filed, therefore, in $E$ we can write $f$ as a product of linear polynomials:

f(x)=(x-\alpha_1)\cdots (c-\alpha_n)

From this fact we can conclude that $F\subseteq F[\alpha_1,\dots,\alpha_k]\subseteq E$. However, by the minimality of $E$, we conclude that $F[\alpha_1,\dots,\alpha_k]=E$.

Great, we found exactly what $E$ is – It is the field over $F$ that is generated by all the roots of $f$.

For example, Consider the polynomial

f(x)=(x^2-2)(x^2-3)\in\mathbb{Q}[x]

We know that $\mathbb{R}$ splits the polynomial since:

f(x)=(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})\in \mathbb{R}[x]

The roots of $f$ are $\pm \sqrt{2}, \pm \sqrt{3}$, therefore, the splitting field should be $\mathbb{Q}[\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}]$. However, since $-\sqrt{2} = -1\cdot \sqrt{2}, -\sqrt{3} = -1\cdot\sqrt{3}$, we can represent this splitting field as $\mathbb{Q}[\sqrt{2},\sqrt{3}]$, and this is a splitting field for this polynomial.

As you may noticed already, we can create the splitting field in so many ways: If $\alpha_1,\dots,\alpha_k$ are the roots of a polynomial $F$, we can adjoin $\alpha_1$ then to new field we can adjoin $\alpha_2$ and so on until we reach $\alpha_k$, on the other hand, we can start by adjoining $\alpha_k$ first, and then $\alpha_4$ and then another root until all of the roots are in the field. Those differenet ways of constructing the field leads us to the question: Is a splitting field uniqiue? Is there a way to construct more that one splitting field?

As it turns out, a splitting field is indeed unique (up to isomorphism). This fact allows us to refer a splitting field as The splitting field.

In order to prove the uniqueness, we are going to work pretty hard now, however, we will also develop tools that will make our life easier in the future, in fact – we are now going to present what I consider as the core of galois theory – You may not see it right away, but we’ll use what I am going to present here a lot, and I mean it, a lot.

## Extending field’s embedding

The main thing we are going to discuss about here is how many ways are there to embed some field in another field. First of all, notice that every ring-homomorphism between fields must be an embedding! Why? suppose that

\varphi:F\rightarrow E

Is a ring homomorphism. By the first isomorphism theorem, we know that:

F/\ker\varphi\cong\text{Im}\varphi\sub E

However, we also know that $\ker\varphi$ is an ideal in $F$. But $F$ is a field, it has no non-trivial ideals, therefore $\ker\varphi=(0)$ – so $\varphi$ is an embedding.

Now our initial state is an embedding $\varphi:E\to F$, and a field $K$ that extends $F$.

\begin{array}{ccc}
K\\
\cup\\
F & \overset{\varphi}{\longrightarrow} & E
\end{array}

Our goal is to find how many embeddings $\varphi_1:K\to E$ are there such that $\varphi_1|_F=\varphi$.

\begin{array}{ccc}
K & \overset{\varphi_{1}}{\longrightarrow} & E\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E
\end{array}

In other words, we want to find how many ways are there to extend $\varphi$, such that it’s domain will be $K$.

At first sight, this question may look kind of boring… I mean, why would we care about extending embeddings? How is that useful? Well… let’s just say that by the end of the day – that’s probably the only thing we would want to do!

We will denote the number of possible extensions as $n_{F\overset{\varphi}{\to}E}^K$.

For example, Consider the following diagram:

\begin{array}{ccc}
\mathbb{Q}[i] & \overset{\varphi_{1}}{\longrightarrow} & \mathbb{C}\\
\cup &  & \shortparallel\\
\mathbb{Q} & \overset{\varphi}{\longrightarrow} & \mathbb{C}
\end{array}

Here, $\varphi$ is just the inculsion embedding: $\varphi(q) = q$. Let’s try to find out in how many ways we can define $\varphi_1$. First, pick some $a+bi\in\mathbb{Q}[i]$. Now, using the properties of a ring homomorphism yields that:

\varphi_1(a+bi)=\varphi_1(a)+\varphi_1(bi)=\varphi_1(a)+\varphi_1(b)\varphi_1(i)
\overset{\varphi_1|_\mathbb{Q}=\varphi}{=}\varphi(a)+\varphi(b)\varphi_1(i)=a+b\varphi_1(i)

It turns out that $\varphi_1$ is defined entirely by the image of $i$.

As it turns out, there are exactly two options for the image of $i$: $\varphi_1(i) = \pm i$, therefore $n_{\mathbb{Q}\overset{\varphi}{\to}\mathbb{C}}^{\mathbb{Q}[i]}=2$. We shall see why it’s true in a moment, however, I need to present a short statement in ring theory first.

### Ring theory break

Let $\varphi:R\to S$ be a ring homomorphism and suppose that $I\vartriangleleft R$ is some ideal of $R$.

We can use $\varphi$ do define a map $\varphi^\prime:R/I\to S$, defined as $\varphi^\prime(r+I)=\varphi(r)$.

\begin{array}{ccc}
I\vartriangleleft R & \overset{\varphi}{\longrightarrow} & S\\
\downarrow &  & \shortparallel\\
R/I & \overset{\varphi^{\prime}}{\longrightarrow} & S
\end{array}

However, this map may not even be well-defined! If it is though, then this map is a ring homomorphism as well (it follows immediately from the fact that $\varphi$ is a homomorphism).

Let’s think for a second why it may not be well defined: suppose that $r_1+I=r_2+I$, then if $\varphi^\prime$ is well defined then $\varphi^\prime(r_1+I)=\varphi^\prime(r_1+I)$, thus $\varphi(r_1)=\varphi(r_2)$, and that’s not trivial at all! There is no guarantee that elements from the same coset in the qutient ring will be sent to the same element in $S$.

However, if $I\sub \ker\varphi$, then $\varphi(r_1)$ and $\varphi(r_2)$ are indeed the same. Here is why:

r_1+I=r_2+I\Rightarrow r_1-r_2\in I\overset{I\sub\ker\varphi}{\Rightarrow}\varphi(r_1-r_2)=0
\Rightarrow \varphi(r_1)-\varphi(r_2)=0\Rightarrow\varphi(r_1)=\varphi(r_2)\Rightarrow\varphi^\prime(r_1+I)=\varphi^\prime(r_2+I)

Great, so we can now conclude that if $I\sub \ker\varphi$ then $\varphi^\prime$ is well defined! (the opposite direction is also true, but I won’ t use it – you can try and prove it yourself if you want, it is not complicated at all).

#### Back to embedding extentions

We are now ready to prove our first result:

Suppose that $\varphi:F\to E$ is an embedding and $F_1=F[a]$. Let $p\in F[x]$ be the minimal polynomial of $a$ over $F$. Then $n_{F\to E}^{F_1}$ is exactly the number of roots of $\varphi(p)$ in the field $E$.

This theorem explains why there were only 2 extentions in the example. The minimal polynomial of $i$ over $\mathbb{Q}$ is $x^2+1$, and it’s root are $\pm i$.

#### The proof

The idea is to find a correspondence between the roots of $\varphi(p)$ and the extensions of $\varphi$.

First, let’s match a root to a given extension $\varphi^\prime$.

##### Extension to root

Just for comfort, let’s write $p$ as $x^n+b_{n-1}x^{n-1}+\cdots+b_0$. We know that $p(a)=0$, therefore, $0=\varphi^\prime(0)=\varphi^\prime(p(a))$. We can now conclude that:

0=\varphi^\prime(p(a))=\varphi^\prime(a^n+b_{n-1}\cdot a^{n-1}+\cdots+b_0)
=\varphi^\prime(a)^n+\varphi^\prime(b_{n-1})\varphi^\prime( a)^{n-1}+\cdots+\varphi^\prime(b_0)

However, $b_{i}\in F$ for every $i$, therefore $\varphi^\prime(b_i)=\varphi(b_i)$. Thus:

=\varphi^\prime(a)^n+\varphi(b_{n-1})\varphi^\prime( a)^{n-1}+\cdots+\varphi(b_0)=\varphi(p)(\varphi^\prime(a))

Would you look at that! we just found out that $\varphi^\prime(a)$, is a root of $\varphi(p)$. This means that every extension must send $a$ to a root of $\varphi(p)$.

Moreover, it’s not hard to see that $\varphi^\prime$ is defined entirely by the image of $\alpha$ (try to prove it! The process is really simiar to what we’ve just done, and to what we’ve seen in the example).

Those facts allow us to conclude that the number of the extensions is at most the number of roots of $\varphi(p)$, and since this root is unique to $\varphi(p)$ (again, the extension if defined entirely by it…) we can match each extension $\varphi^\prime$ to it’s corresponding root – $\varphi^\prime(a)$ (and this match is one-to-one).

On the other hand, we still need to show the opposite direction – find a way to match each root to an extension in a unique way.

##### Root to extension

Suppose that $a^\prime$ is a root of $\varphi(p)$. Now, consider the following ring homomorphism $\Phi:F[x]\to E$:

\Phi(a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0)=\varphi(a_n)\cdot(a^\prime)^n+\varphi(a_{n-1})\cdot(a^\prime)^{n-1}+\cdots +a_0

This is the homomorphism that maps $x$ to $\alpha^\prime$ and extends the homomorphism $\varphi$.

In addition, since $a^\prime$ is a root of $\varphi(p)$ we get:

\Phi(p)=\varphi(p)(a^\prime)=0

Thus, $\langle p\rangle \sub \ker\Phi$. This situation fits perefectly to what I’ve just proved about rings:

\begin{array}{ccc}
\langle p\rangle\vartriangleleft F[x] & \overset{\Phi}{\longrightarrow} & E\\
\downarrow &  & \shortparallel\\
F[x]/\langle p\rangle & \overset{\Phi^{\prime}}{\longrightarrow} & E
\end{array}

Therefore, the ring homomorphism $\Phi^\prime:F[x]/\langle p\rangle \to E$ defined as:

\Phi^\prime(f+\langle p\rangle)=\Phi(f)

is well defined. However, notice that $F[a]\cong F[x]/\langle p\rangle$, and since $F[a]$ is a field, this homomorphism is an embedding, moreover, by it’s definition, it extended $\varphi$. Now we can use the isomorphism:

i:F[a]\to F[x]/\langle p\rangle \\
a\mapsto x + \langle p\rangle 

And define the extension $\varphi^\prime$ as:

\varphi^\prime=\Phi^\prime\circ i: F[a]\to E

Indeed, $\varphi^\prime$ extends $\varphi$, and $\varphi^\prime(a)=a^\prime$.

And that’s it – we’ve found one-to-one correspondence from embedding extension to roots, and vice versa. Therfore, the correspondence are, in fact, bijections.

#### Conclustions

There are two immediate conclustions we can derive from this statement:

• $n_{F\to E}^{F_1}=(\text{\# roots in } E) \leq \deg p = [F_1:F]$. We have an upper bound on the number of extensions, which is the degree of $p$ (the minimal polynomial)
• There are extentions $\iff$ There are roots for $\varphi(p)$.

Even though this lemma is very useful, we still haven’t tackled the general case, when $F_1/F$ is not simple (a simple extension is a field extension $K/F$ such that $K=F[\alpha]$ for some $\alpha$).

### Last effort

I want to move on to the general case now, which will also prove the uniqueness of the splitting field, But I want to give a new definition first:

We say that a polynomial $g$ that splits in $E$ is separable if it’s roots are not the same. In other words, It has $\deg g$ different roots.

For example, the polynomial $f(x)=(x-1)(x-2)(x-3)$ is seperable, but on the other hand, the polynomial $g(x)=(x-1)^2(x-5)^3$ is clearly not seperable – It’s degree is 8 but he only has 2 different roots.

Ok, we are now ready to present the general case:

Suppose that $\varphi:F\to E$ is an embedding and $F\sub K$ is a finite extension. Then:

1. $n_{F\to E}^{K} \leq [K:F]$
2. If $K$ is generated over $F$ by roots of a polynomial $f$ such that $\varphi(f)$ splits in $E$, then
• There are extensions of $\varphi$ to $K$.
• If $\varphi(f)$ is seperable in $E$, then $n_{F\to E}^{K}=[K:F]$

What this theorem says exactly? well, the first part gives us an upper bound to the extension, which is great and very similar to what we’ve seen about simple extensions.

The second part assumes something beyond – It assumes that $K$ is not just an arbitrary extension, but it’s generated from roots of a polynomial that it’s image splits.

If that’s the case, we are guaranteed to find extensions to the embeding, at least one! But that’s where the story ends – if $f$ turns out to be separable, we know exactly how many extension are there – The number of the extensions is the degree of the field extension, $[K:F]$.

#### The proof

##### Part 1

In a similar way to other proofs I have brought today, this one is going to be done by induction as well – This time, the induction is on the degree of the extension.

First, we will represent $K$ as $K=F[a_1,\dots,a_n]$ where $a_i\in K$. Define $F_1=F[a_1]$, and now we have $F\sub F_1\sub K$.

\begin{array}{c}
K=F[a_1,\dots,a_k]\\
\vert\\
F_{1}=F[a_1]\\
\vert\\
F
\end{array}

Which yields the following diagram:

\begin{array}{ccc}
K & \overset{\varphi^\prime}{\longrightarrow} & E\\
\uparrow &  & \shortparallel\\
F_{1} & \longrightarrow & E\\
\uparrow &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E
\end{array}

We are interested in the number of ways to extend $\varphi$ to $\varphi^\prime$. We are going to use the fact that $F_1$ is a simple extension, thus $n_{F\to E}^{F_1}\leq [F_1:F]$.

Notice that for each extension of $\varphi$ to $\varphi^\prime$, we can restrict the domain of the it to be $F_1$. Then $\varphi\prime|_{F_1}$ is an extension of $\varphi$ to $F_1$!

So, in fact, any extension $\varphi^\prime:K\to E$ is in particular an extension of some extension $\varphi_1: F_1\to E$.

Therefore, the number of extensions of $\varphi$ to $\varphi^\prime$ is exactly:

n_{F\to E}^K=\sum_{\varphi_1:F_1\to E} n_{F_1\overset{\varphi_1}{\to} E}^K

Summing over all the numbers $n_{F_1\overset{\varphi_1}{\to} E}^K$ will give us exactly $n_{F\to E}^K$ (convince yourself!).

Now, by induction, we know that: $n_{F_1\overset{\varphi_1}{\to} E}^K\leq [K:F_1]$. Therefore:

n_{F\to E}^K=\sum_{\varphi_1:F_1\to E} n_{F_1\overset{\varphi_1}{\to} E}^K\leq \sum_{\varphi_1:F_1\to E} [K:F_1]

But there are exactly $n_{F\to E}^{F_1}$ such $\varphi_1$ extensions. Thus:

n_{F\to E}^K\leq \sum_{\varphi_1:F_1\to E} [K:F_1]=n_{F\to E}^{F_1}[K:F_1]

And we said before that $n_{F\to E}^{F_1}\leq [F_1:F]$, which yields:

n_{F\to E}^K\leq [F_1:F][K:F_1]=[K:F]

As desired (I used here the multiplicative property)

#### Part 2

Now, $K=F[a_1,a_2,\dots, a_n]$ where $a_i$ are the roots of the polynomial $f$. Denote $F_1=F[a_1]$.

Since $\varphi(f)$ has roots in $E$, from this fact, we conclude that there are extensions of $\varphi$ to $F_1$, and by induction, the extensions from $F_1$ have extensions to $K$and we have found extensions! In a more detailed way – after we’ve found an extension to $F_1$, we can extend is to $F_2=F[a_1,a_2]=(F[a_1])[a_2]$ which is a simple extension of $F_1$. In a similar way we define $F_3,F_4,F_5,...,F_n$ but $F_n$ is just the field $K$!

\begin{array}{ccc}
K & \overset{\varphi^\prime}{\longrightarrow} & E\\
\uparrow &  & \shortparallel\\
F_{n-1} & \longrightarrow & E\\

\vdots &  & \vdots\\
F_{2} & \longrightarrow & E\\
\uparrow &  & \shortparallel\\
F_{1} & \longrightarrow & E\\
\uparrow &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E
\end{array}

It’s just like climbing a ladder – step-by-step. The reason we can proceed at each step is because every $F_k$ ‘yields’ a new root (or changes nothing – this happens when there are mutiple roots, pause and ponder about it for a moment – this is exactly the case where we can’t get the maximal number of extensions!), and we always have a root in $E$ that we can send it to – all the roots of $\varphi(f)$ are in $E$ and the number of the different roots in $E$ is the same number of different roots of $f$:

Both polynomials have degree $n$, and if one has $k$ different roots, than so as the other.

For example, think about the case where $\varphi(f)=(x-a^\prime)^n$. Here, we have to map $a_1$ to $a^\prime$. So we get that

\varphi(f)=\varphi^\prime(f)=\varphi^\prime((x-a_1)^n)=(x-a^\prime)^n

And since $\varphi^\prime$ is one-to-one – $f=(x-a_1)^n$ then $K=F_1$ and $a_1=\cdots=a_n$. There is only one embedding extension, but the degree of the extension is $n$ – the degree of the polynomial $f$.

And that’s exactly the reason that we need $\varphi(f)$ to be separable in order to get as much extensions as possible:

Indeed, if $\varphi(f)$ is separable in $E$, we conclude that $[F_1:F]=n_{F\to E}^{F_1}$. And we can now use the induction and repeat the same process from the first part of the proof to get the equality.

That’s it! However I want to remark one important insight, and this is a really, really important insight:

We know exactly how extension $\varphi_1:F_1\to E$ behaves – it maps a root of $f$ – let’s call it $a$ to a root of $\varphi(f)$, which we will call $a^\prime$.

So, when we extend $\varphi$ to $K$ while ‘going through’ $\varphi_1$ we get an embedding $\varphi^\prime:K\to E$ such that $\varphi^\prime(a)=a^\prime$.

Once again, it may not be so clear why this insight is so important, but trust me when I am telling you – This insight will help us a lot, and in my opinion – it’s the most important thing that you have to understand in galois theory (which we haven’t even discussed yet – when will start discussing it, you’ll see the name galois almost everywhere…)

## Finally – Uniqueness of the splitting field

From what we just proved, the uniqueness follows immediately: First, notice that every splitting field $K$ of $f$ is embedded in any field $E$ over $F$ where $f$ splits in.

Why? This situation fits perfectly the second part of the theorem, we can apply it to the diagram:

\begin{array}{ccc}
K & \overset{\varphi_{1}}{\longrightarrow} & E\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E
\end{array}

As I mentioned, the splitting field is generated by the roots of $f$, and by our assumption, $\varphi(f)=f$ (since $F\sub E$, we will pick the inclusion embedding). The second part of the theorem assures us that there are extensions. Thus, $K$ is embedded in $F$.

Now suppose that there exists another splitting field $K^\prime$. we can substitute $K^\prime = E$ to conclude that there is an embedding $\Phi$ from $K$ to $K^\prime$.

\begin{array}{ccc}
K & \overset{\Phi}{\longrightarrow} & K^\prime\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} &K^\prime
\end{array}

The image of $K$ in $K^\prime$ splits $f$ as well (why?), by the minimality of $K^\prime$, we get $\Phi(K)=K^\prime$, therefore, the embedding is actually an isomorphism, and the uniqueness is finally proved!

## Summary

So that was a long post, we’ve seen a lot today, learned about degrees of extensions, the spltting field and it’s uniquness. And it might not seem like it’s the most important thing – but we also successfully tackled the problem of extending embeddings, and I’ll say it one more time: This is the most important thing in this theory!