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:


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:


We now have:


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:


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:


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:


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:


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:


where \deg(r(x))<deg((x-\alpha))=1. 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:


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.
E & _{(f_{2})}\\
K & _{(f_{1})}\\
  • 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


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.

F & \overset{\varphi}{\longrightarrow} & E

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

K & \overset{\varphi_{1}}{\longrightarrow} & E\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E

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:

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

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:


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).

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

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:


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

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

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.


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.


Which yields the following diagram:

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

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 Kand 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!

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

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


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:

K & \overset{\varphi_{1}}{\longrightarrow} & E\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} & E

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.

K & \overset{\Phi}{\longrightarrow} & K^\prime\\
\cup &  & \shortparallel\\
F & \overset{\varphi}{\longrightarrow} &K^\prime

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!


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!

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