Exploration 7: Fields

December 11, 2023.

Questions:

TODO

Recall that we can categorize each field $F$ as either an infinite field ( $\char F=0$ ) or a finite field ( $\char F=p$ ). Last time we discussed finite fields ( $\char F=p$ ). Now we extend the discussion to infinite fields ( $\char F=0$ ).

Theorem: Every infinite field (where

\char F=0

) contains the rationals

\QQ

In characteristic $0$ , the rationals $\QQ$ are generated by $1$ : first by generating the integers $\ZZ$ via repeated addition and subtraction, then treating each integer as a unit and taking closure under product and inverses of every integer to get $\QQ$ . Since every ring contains $1$ , which generates $\QQ$ , every field in characteristic $0$ contains $\QQ$ .

Now recall that constructing a finite field $\FF_q$ (for some prime power $q=p^n$ ) is done by quotienting $\ZZ_p$ (i.e. $\FF_p$ ) by an irreducible polynomial $\in\ZZ_p[x]$ of degree $n$ . We can do this with infinite fields as well. Let $F$ be a field, and let $f$ be an irreducible polynomial in $F[x]$ , so that $(f)$ is maximal and therefore $F[x]/(f)$ is a field. This exploration will be all about studying the characteristics of these quotient fields $F[x]/(f)$ .

In this section, we study quotient fields.

First of all, quotienting by $(f)$ sends $f\in F[x]$ to $0$ . Where does this quotient send $x\in F[x]$ ?

Therefore: for polynomials

f

irreducible in

F[x]

, the quotient

F[x]/(f)

maps

x

to a root of

f

that exists outside of the field

F

Let’s say we have $f=x^2+1$ , which is irreducible in $\RR[x]$ . Then the quotient $\RR[x]/(f)$ sends $f$ to the zero coset $[0]$ , meaning we have $[x^2+1]=[0]$ which implies $[x]^2+[1]=[0]$ .
This means that $x$ gets mapped to some element that is a root of the polynomial $x^2+[1]$ (where coefficients are cosets from the quotient $\RR[x]/(f)$ ).
In summary, $x^2+1$ has no roots $\RR$ but it does in $\RR[x]/(f)$ (the coset $[x]$ ). The idea is that the act of adjoining $x$ and then quotienting creates this root.

Since $\RR[x]/(f)$ contains $\RR$ as a subfield, $\RR[x]/(f)$ is an extension of $\RR$ . So just like with finite fields, our new field is “larger” in some sense, which we’ll learn how to measure later. The important point is that when we extend a field in this manner, we’re doing so by adding a new root of an irreducible polynomial $f\in F[x]$ to $F$ !

In this section, we show another perspective for extending fields.

Recall that one way to turn any ring into a field is to adjoin units to it. That is, for every nonzero element $a$ , adjoin an element $a^{-1}$ so that $a$ becomes a unit. This results in a field known as the field of fractions, the idea being that for every $a\in F$ , you adjoin $\frac{1}{a}$ to $F$ , so now everything in $F$ looks like a fraction.

The field of fractions gives us another way to extend fields. If we adjoin some new element $\alpha$ to an existing field $F$ , and then take the field of fractions of the result, you get a new field, which we denote $F(\alpha)$ . This operation is known as a simple extension of $F$ by $\alpha$ . In this case, $\alpha$ is the generator of the simple extension.

So we have two ways of extending a field $F$ : the first way is to quotient its polynomial ring by an irreducible polynomial, and the second way is to adjoin a new element and take the field of fractions. Surprisingly, these turn out ot be equivalent. We’ll first prove a very interesting lemma about simple extensions.

Lemma: If

\alpha

is the root of any polynomial

f\in F[x]

, then

F[\alpha]=F(\alpha)

WLOG we can assume $f$ is irreducible in $F[x]$ . This is because if $f$ is reducible, $\alpha$ must be a root of at least one of the irreducible factors of $f$ , and we can take that factor to be $f$ instead.

$F[\alpha]$ is exactly what you get when you evaluate every polynomial $F[x]$ at this new root $\alpha$ . To show that $F[\alpha]$ is a field, let $g(\alpha)$ be an arbitrary nonzero element of $F[\alpha]$ , where $g$ is some nonzero polynomial in $F[x]$ . WTS $g(\alpha)$ is a unit.

By the extended Euclidean algorithm, we have $\gcd(f,g)=af+bg$ for some $a,b\in F[x]$ .
Since $f$ is irreducible in $F[x]$ , and therefore prime, $\gcd(f,g)$ must be $1$ or $f$ . Because $g(\alpha)$ is nonzero, $\alpha$ is not a root of $g$ , implying $g$ doesn’t have a factor $f$ (which does have $\alpha$ as a root.) Therefore $\gcd(f,g)$ must be $1$ .
Using the assumption that $\alpha$ is a root of $f$ , we get $f(\alpha)=0$ . Then apply the evaluation map at $\alpha$ to both sides: $\begin{aligned} \gcd(f,g)&=a\cdot f+b\cdot g\\ \gcd(f,g)(\alpha)&=a(\alpha)\cdot f(\alpha)+b(\alpha)\cdot g(\alpha)\\ 1&=a(\alpha)\cdot 0+b(\alpha)\cdot g(\alpha)\\ 1&=b(\alpha)\cdot g(\alpha)\\ \end{aligned}$ which shows that our arbitrary nonzero $g(\alpha)$ is a unit.

Therefore every nonzero element of $F[\alpha]$ is a unit, and thus $F[\alpha]$ is its own field of fractions equal to $F(\alpha)$ .

Now we show that the two methods of extending a field — quotienting by an irreducible and doing a simple extension — are equivalent.

Theorem: If

f

is irreducible over

F

, the quotient

F[x]/(f)

is isomorphic to a simple extension

F(\alpha)

where

\alpha

is some root of

f

Define a kind of evaluation map $g\mapsto g(\alpha)$ and call it $\varphi_\alpha:F[x]\to F(\alpha)$ .
The kernel of $\varphi_\alpha$ is every polynomial that evaluates to $0$ at $\alpha$ . $f$ is one such polynomial.
Since $f$ is irreducible, it generates a maximal ideal $(f)$ of $F[x]$ , which contains every polynomial that also has $\alpha$ as a root. Therefore the kernel of $\varphi_\alpha$ is exactly $(f)$ , since they all evaluate to $0$ at $\alpha$ .
As for the image of $\varphi_\alpha$ , note that evaluating all polynomials in $F[x]$ at $\alpha$ results in exactly $F[\alpha]$ , which by the above lemma is exactly $F(\alpha)$ .
Then by the first isomorphism theorem, which states $F[x]/\ker\varphi_\alpha\iso\im\varphi_\alpha$ , we have $F[x]/(f)\iso F(\alpha)$ .

Since there are two isomorphic ways to extend fields, in general, should we write the quotient field $F[x]/(f)$ or should we write simple extension $F(\alpha)$ ? A quotient field makes explicit that we add roots of $f$ to the field, so implicitly the result has the root $\alpha$ of $f$ . A simple extension makes explicit that we add $\alpha$ to the field, so implicitly we’ve added the roots of its polynomial $f$ .

In fact we can ignore this detail, and simply write $K/F$ or “ $K$ is a field extension of $F$ ” (not to be confused with the identical notation used for quotients). Unlike the above, this notation can express any size of field extension. Specifically, $K/F$ could denote that $K$ is a (perhaps infinite) composition of simple extensions of $F$ , for example, $K=F(\alpha)(\beta)(\gamma)(\delta)$ .

The general definition of a field extension of $F$ , written $K/F$ , is any field $K$ that contains $F$ as a subfield. Since you can theoretically build $K$ from $F$ by continually taking simple extensions of elements in $K$ not in $F$ , every field extension can be expressed as a (perhaps infinite) composition of simple extensions.

Another notation is useful when we start getting into field extensions of field extensions (called towers): $K\supseteq L\supseteq F$ . This notation pretty much only exists since $K/L/F$ is ambiguous (is the slash quotienting or a field extension?). Like $K/F$ , it doesn’t tell you how the extensions were obtained, but it does tell you about the existence of an intermediate extension $L$ .

But for the purposes of this exploration where we are explicitly focusing on the structure of quotient fields, we’ll use the $F[x]/(f)$ form throughout.

In this section, we explore the properties of the roots within the extended fields.

The lemma we proved earlier about $F(\alpha)=F[\alpha]$ is actually quite foundational. In fact, we can prove a kind of converse:

Theorem:

F(\alpha)\iso F[\alpha]

iff

\alpha

is the root of some polynomial

f\in F[x]

To show the forward direction, note that if $F(\alpha)\iso F[\alpha]$ , then $F[\alpha]$ is a field, so in particular $\alpha^{-1}$ is an element in $F[\alpha]$ corresponding to some polynomial $g\in F[x]$ .
Since $\alpha^{-1}\alpha=1$ , we have $\alpha^{-1}\alpha-1=0$ .
Then the corresponding polynomial $f=gx-1\in F[x]$ has $\alpha$ as a root, by construction.
The aforementioned lemma proves the backward direction.

This is a pretty important equivalence to have, so important that when $\alpha$ is the root of a polynomial in $F[x]$ , where $F$ is a field, we say that $\alpha$ is algebraic over $F$ .

Corollary: $F(\alpha)=F[\alpha]$ iff $\alpha$ is algebraic over $F$ ..

Whenever we have an element $\alpha$ that is algebraic over $F$ , we know that the simple extension $F(\alpha)$ , by the theorem we proved, results in a field isomorphic to $F[x]/(f)$ for some irreducible polynomial $f$ with root $\alpha$ .

Consider the relationship between the root $\alpha$ and its irreducible polynomial $f$ . It’s easy to identify the root $\alpha$ given $f$ — $\alpha$ is exactly the coset $[x]$ in $F[x]/(f)$ . But given $\alpha$ , can we identify a irreducible polynomial $f$ where $\alpha$ is a root?

For this, we need look no further than the kernel of the evaluation map $\varphi_\alpha$ , which is defined to be all the polynomials $\in F[x]$ for which $f(\alpha)=0$ . In fact, since $F[x]$ is an Euclidean domain and therefore a PID, the ideal $\ker\varphi_\alpha$ must be principal i.e. generated by some polynomial $f$ . If we can identify $f$ , then we have identified a polynomial where $\alpha$ is a root. We can deduce a couple things about $f$ . First of all, $f$ is irreducible:

Theorem: Over a field

F

, the polynomial

f\in F[x]

generating

\ker\varphi_\alpha

is irreducible.

Since $F$ (being a field) has no zero divisors, the equation $\varphi_\alpha(f)=0$ means any nonzero factor of $\varphi_\alpha(f)$ would constitute a zero divisor, therefore $\varphi_\alpha(f)$ is irreducible and so $f$ is irreducible.

Second, $f$ is unique up to associates:

Theorem: Over a field

F

, the polynomial

f\in F[x]

generating

\ker\varphi_\alpha

is unique (up to associates).

If $g$ also generates $\ker\varphi_\alpha$ , then $f$ and $g$ divide each other and therefore differ by a unit. Then $f$ is unique up to associates. (If we make $f$ monic, then it is unique, period.)

Corollary: Over a field $F$ , the monic polynomial $f\in F[x]$ generating $\ker\varphi_\alpha$ is unique.

Thus, is some unique monic irreducible polynomial $f$ that generates the kernel of $\varphi_\alpha$ , and it is known as the minimal polynomial of $\alpha$ over $F$ . So indeed, every root $\alpha$ has a minimal polynomial $f\in F[x]$ that is unique, irreducible over the base field $F$ , and has $\alpha$ as a root.

Thus every time we quotient by some irreducible polynomial $f$ (i.e. $F[x]/(f)$ ) we’re actually adjoining a new root $\alpha=[x]$ , and any time we adjoin a root $\alpha$ (i.e. $F(\alpha)$ ), we’re actually quotienting $F$ by its minimal polynomial $f$ .

Corollary: $F(\alpha)\iso F[x]/(f)$ whenever $f$ is the minimal polynomial of $\alpha$ .

This lets us prove an interesting fact about simple extensions. If a minimal polynomial $f$ has two roots $\alpha,\beta$ , then adjoining $\alpha$ has the same effect as adjoining $\beta$ . The converse is also true.

Theorem: Any two simple extensions by algebraic elements are isomorphic,

F(\alpha)\iso F(\beta)

, iff

\alpha

and

\beta

have the same minimal polynomial.

If $\alpha$ and $\beta$ have the same minimal polynomial $f$ , then both $F(\alpha)$ and $F(\beta)$ are both isomorphic to $F[x]/(f)$ , and therefore to each other.

Corollary: Equivalently, if $\alpha$ and $\beta$ are roots of the same irreducible polynomial $f$ over $F$ , then $F(\alpha)\iso F(\beta)$ .

There is also a deeper theorem about automorphisms of field extensions that we’ll use later. We prove it here now:

Theorem: Every automorphism

\sigma:F(\alpha)\to F(\alpha)

that fixes

F

will permute the roots of the minimal polynomial

f

\alpha

Suppose $f(\alpha)=0$ for $f$ a minimal polynomial over $F$ . Every automorphism $\sigma:K\to K$ that fixes $F$ must satisfy $\begin{aligned} f(\alpha)&=0\\ \sigma(f(\alpha))&=\sigma(0)\\ f(\sigma(\alpha))&=0&\text{ since coefficients}\in F\text{ are fixed by }\sigma \end{aligned}$ implying that $\sigma(\alpha)$ is a root of $f$ as well.

In summary, what we’ve shown here is that given any algebraic element $\alpha$ , we can refer to its minimal polynomial, which is unique and always exists. Different simple extensions, then, can be identified with the minimal polynomial of its generator.

What about larger extensions? Are we able to identify extensions like $F(\alpha)(\beta)=F(\alpha,\beta)$ with some kind of minimal polynomial?

In this section, we demonstrate how non-simple extensions can be reduced to a simple extension.

If we find that $F(\alpha,\beta)$ is equal to some simple extension $F(\gamma)$ , then we can classify the extension with the minimal polynomial of $\gamma$ . But when does such a $\gamma$ exist?

Lemma: Every extension

F(\alpha,\beta)

by two elements

\alpha,\beta

algebraic over

F

is equal to a simple extension, provided that the minimal polynomials of

\alpha

and

\beta

have distinct roots.

First, if $F$ is a finite field, then $F(\alpha,\beta)$ is the same as $F(\alpha)$ if $\alpha$ has a larger order, and $F(\beta)$ otherwise. (proof) This trivially makes $F(\alpha,\beta)$ a simple extension.

Now we prove the case where $F$ is infinite. Consider an arbitrary two-element extension $F(\alpha,\beta)$ by algebraic elements $\alpha,\beta$ . We will argue that this extension is equal to the simple extension $F(\alpha+c\beta)$ for some nonzero $c\in F$ .

The goal is to choose $c$ such that $\beta\in F(\alpha+c\beta)$ . This is because then $(\alpha+c\beta)-c\beta=\alpha\in F(\alpha+c\beta)$ as well, making $F(\alpha+c\beta)$ equal to $F(\alpha,\beta)$ .

To do this, we will prove that the number of nonzero $c\in F$ for which $\beta\notin F(\alpha+c\beta)$ is finite. Since the base field $F$ is infinite, that leaves infinitely many $c$ for which $\beta\in F(\alpha+c\beta)$ .

Since $\beta\notin F(\alpha+c\beta)$ , $F(\alpha+c\beta)$ is a subfield of $F(\alpha,\beta)$ .
Adjoin every root of the the minimal polynomials $f,g$ of $\alpha,\beta$ respectively to $F(\alpha+c\beta)$ . This results in a field $K$ , which has $F(\alpha+c\beta)$ as a subfield. Then every automorphism $\sigma$ that fixes $F(\alpha+c\beta)$ must permute the roots of $f$ and $g$ .
Then we have: $\begin{aligned} \alpha+c\beta&=\sigma(\alpha+c\beta)\\ \alpha+c\beta&=\sigma(\alpha)+c\sigma(\beta)\\ \alpha-\sigma(\alpha)&=c\sigma(\beta)-c\beta\\ \alpha-\sigma(\alpha)&=c(\sigma(\beta)-\beta)\\ \frac{\alpha-\sigma(\alpha)}{\sigma(\beta)-\beta}&=c\\ \end{aligned}$ For such a nonzero element $c$ to exist, we must ensure that both the numerator and denominator are nonzero, i.e. $\sigma(\alpha)$ must not map to $\alpha$ , and $\sigma(\beta)$ must not map to $\beta$ . Here we must use the assumption that $f$ and $g$ have distinct roots, so that there exists some $\sigma(\alpha)\ne\alpha$ and $\sigma(\beta)\ne\beta$ .
This implies there are finitely many $c$ that exist, since both $f,g$ have finitely many other roots ( $\deg f-1,\deg g-1$ that $\alpha,\beta$ can be permuted to.

Since $\beta\notin F(\alpha+c\beta)$ implies that there are only finitely many choices for $c$ , the contrapositive says that all other choices of $c\in F$ imply $\beta\in F(\alpha+c\beta)$ . Since $F$ is an infinite field, there are infinitely many choices, so such a $c$ exists such that $\beta\in F(\alpha+c\beta)$ , implying $F(\alpha+c\beta)=F(\alpha,\beta)$ .

Given a couple of definitions, this lemma implies our main result:

Primitive Element Theorem: Every finite separable extension of

F

is simple.

A finite extension is one obtained by adjoining finitely many elements TODO they have to be algebraic. $F(\alpha,\beta,\ldots)$ . We can iteratively apply the lemma to show that all finite extensions of $F$ are simple.

A algebraic extension is simply a field extension by algebraic elements. All finite extensions are algebraic.

When a polynomial has distinct roots (i.e. no multiple roots) in some extension field, we call it a separable polynomial. Similarly, when every minimal polynomial is separable for elements in a field extension $K/F$ , we say it is a separable extension.

Then we can reword the above lemma concisely as a key theorem:

This is called the Primitive Element Theorem because when a field extension $K/F$ can be expressed as a simple extension $F(\alpha)$ , we say that $\alpha$ is a primitive element for the extension $K/F$ , since $\alpha$ generates the extension $K/F$ . So another way to word it is:

Primitive Element Theorem: Every finite separable extension has a primitive element.

In this section, we discuss separability.

So how do we prove separability?

In characteristic $0$ , recall that one can identify multiple roots using the derivative of a polynomial. Recall that the derivative of a polynomial $f=a_nx^n+\ldots+a_2x^2+a_1x+a_0$ is defined $f'=na_nx^{n-1}+\ldots+2a_2x+a_1$ The product rule of derivatives states that the derivative of a product is always $(fg)'=f'g + fg'$

The implication for multiple roots is this. If $f$ has a multiple root $\alpha$ , then it factors into $(x-\alpha)^2g$ . The product rule guarantees that the derivative of $f$ is $\begin{aligned} &~((x-\alpha)^2g)'\\ =&~((x-\alpha)(x-\alpha))'g+(x-\alpha)^2g'\\ =&~2(x-\alpha)g+(x-\alpha)^2g'\\ =&~(x-\alpha)(2g+(x-\alpha)g')\\ \end{aligned}$

This implies that if $f$ has multiple root $\alpha$ , then $f'$ also has $\alpha$ as a root. Since the converse is also true, this means having multiple roots is the same as sharing roots with the derivative.

In other words, to check if $f$ is separable, all we need to do is check if $f$ shares any roots with its derivative $f'$ in its splitting field. As a matter of fact, this is true for any irreducible polynomial in characteristic $0$ .

Theorem: In characteristic

0

, every irreducible polynomial is separable.

It is enough to prove that $\gcd(f,f')$ is constant in the splitting field, because that means they cannot share any factor $(x-\alpha)$ , and therefore $f$ has no multiple root.
Since $f$ is irreducible, $\gcd(f,f')$ is equal to either $f$ or a constant polynomial. If it’s constant, we are done. Otherwise, $\gcd(f,f')=f$ is only possible if $f'=0$ , since $f'$ has smaller degree than $f$ .
But in characteristic $0$ , the only polynomials that have a zero derivative are the constant polynomials. This is because any nonconstant $a_kx_k^k$ would contribute the term $ka_kx_k^{k-1}$ , which is always nonzero in characteristic $0$ .
Since $f$ is irreducible, it is not a constant polynomial, and therefore $f'$ is nonzero, and thus $\gcd(f,f')$ cannot be $f$ and must be a constant.

Corollary: Every field extension in characteristic

0

is separable.

Since minimal polynomials are irreducible, every minimal polynomial is separable by the above theorem, and therefore every extension in characteristic $0$ is separable.

Since characteristic $0$ fields are precisely the infinite fields, this gives us a corollary of the Primitive Element Theorem: Every finite extension of an infinite field has a primitive element.

In summary, since this exploration deals only with finite extensions of infinite fields, we can pretty much always assume that field extensions are simple, unless otherwise stated.

In this section, we go wild with adding roots to a field.

When we add a root $\alpha$ to a field via quotienting by some polynomial $f$ irreducible over $F$ , we’re also making it so that $f$ is actually reducible over the resulting field, in the sense that $f$ now factors into $(x-\alpha)g$ for some $g$ . You can imagine factoring $g$ into irreducibles, and repeating the process of adjoining their roots to make more linear factors $(x-\beta)$ , until you’ve decomposed the original $f$ into linear factors over an extension field $K/F$ .

The resulting field is known as a splitting field of $f$ over $F$ . If $K$ is a splitting field of $f$ , we say that $f$ splits over $K$ , or equivalently $f$ splits in $K[x]$ . For example: $x^2+1$ splits over $\ZZ(i)$ , because $x^2+1=(x+i)(x-i)$ .

You could go even further than splitting fields. Imagine constructing an extension field $K$ by splitting every polynomial over $F$ , and then every polynomial over the resulting splitting field, and repeating until you’ve created a field extension $K/F$ so large such that every polynomial in $K[x]$ splits.

Then we’ve taken the algebraic closure of $F$ . We say $E$ is an algebraically closed field, i.e. one in which every nonconstant polynomial in $E[x]$ has a root in $E$ . The canonical example of an algebraically closed field is $\CC$ . The proof that $\CC$ is algebraically closed is called the Fundamental Theorem of Algebra, which we’ve previously proved.

In this section, we examine how to compute the splitting field of a given polynomial over $F$ .

How can we identify the splitting field of $f$ over $F$ ? It turns out that splitting fields are unique (up to isomorphism):

Lemma: For any symbols

\alpha,\beta

F(\alpha)(\beta)=F(\beta)(\alpha)

When we do a simple extension by an element $F(\alpha)$ , we adjoin the element $\alpha$ to $F$ and take the field of fractions. By definition, this is the smallest field containing $F$ and $\alpha$ .
Then $F(\alpha)(\beta)$ is the smallest field containing $F(\alpha)$ and $\beta$ , i.e. it is the smallest field containing $F$ , $\alpha$ , and $\beta$ .
Likewise, $F(\beta)(\alpha)$ is the smallest field containing $F(\beta)$ and $\alpha$ , i.e. it is the smallest field containing $F$ , $\beta$ , and $\alpha$ .
Therefore these are the exact same field.

Theorem: The splitting field of

f

over

F

is unique up to isomorphism.

The splitting field always exists, since we can always quotient by the non-linear irreducible part of $f$ to adjoin a new root $\alpha$ .
So the process takes $f$ and adjoins a root to $F$ to get $f=(x-\alpha)g$ for some $g\in F(\alpha)$ .
Repeating this, we get a splitting field isomorphic to $F(\alpha_1)(\alpha_2)\ldots(\alpha_n)$ .
By the lemma, all reorderings of simple extensions are equal to each other. Since that means constructing a splitting field gives you a field isomorphic to $F(\alpha_1)(\alpha_2)\ldots(\alpha_n)$ regardless of the order you adjoin roots, all splitting fields of $f$ are isomorphic.

Just like how we can refer to the minimal polynomial of a given root $\alpha$ , this theorem lets us refer to the splitting field of a given polynomial $f$ .

As an example, take the splitting field of $x^3-2$ over $F$ , which is irreducible over $F$ . The roots of this polynomial (in $\CC$ ) are $\sqrt[3]{2}$ , $\zeta_3\sqrt[3]{2}$ , and $-\zeta_3\sqrt[3]{2}$ , where $\zeta_3$ is the third root of unity. None of these roots exist in $F$ , since $x^3-2$ is irreducible over $F$ .

In general, taking the splitting field of a polynomial $f$ over $F$ , means finding the minimal extension field $K/F$ that contains all $\deg f$ roots of $f$ , so that $f$ splits into linear factors in $K$ . When $f$ is irreducible, this means adding all $\deg f$ roots to $F$ .

Since splitting fields are unique up to isomorphism, we can add the roots in any order. First, let’s adjoin $\sqrt[3]{2}$ to obtain $F(\sqrt[3]{2})$ . If the base field contains $\zeta_3$ , this would add all the other roots as well, and we are done. Otherwise, the other roots are not added, since it is impossible to express $\zeta_3\sqrt[3]{2}$ in $F(\sqrt[3]{2})$ without $\zeta_3$ .

The next step is to adjoin $\zeta_3\sqrt[3]{2}$ . But then taking the closure means this operation adds the remaining root is $-\zeta_3\sqrt[3]{2}$ , and since we’ve added all roots, the result is the splitting field of $x^3-2$ .

Therefore, the process of constructing a splitting field for a polynomial $f\in F[x]$ is inherently tied to knowing the roots of the polynomial, and involves the following steps:

Find new roots $\alpha_i$ of $f$ that generate extensions $F(\alpha_i)$ which are linearly disjoint over $F$ : extensions where each element in $\alpha_1,\alpha_2,\ldots$ cannot be expressed in terms of the others.
Adjoin each $\alpha_i$ to the base field $F$ , obtaining $F(\alpha_1,\alpha_2,\ldots)$ .
Done!

Now the problem becomes: how do we find roots that generate linearly disjoint extensions? What does it take for two roots to be linearly disjoint?

In this section, we discover the conditions for when two elements generate linearly disjoint extensions.

Recall that given a base field $F$ , all simple extensions of $F$ can be distinguished by its minimal polynomial, in the sense that different simple extensions have different minimal polynomials.

The degree of an extension is the degree of its minimal polynomial, or infinity if the extension is infinite. If the extension is written $K/F$ , then its degree is written $[K:F]$ .

Theorem: Finite field extensions

K/F

have finite degree

[K:F]

Since all algebraic elements have a minimal polynomial (necessarily of finite degree), every extension by algebraic element(s) are finite. This means TODO

Theorem: Two simple extensions

F(\alpha)

and

F(\beta)

are linearly disjoint iff the degree

[F(\alpha,\beta):F]

is equal to the product of

[F(\alpha):F]\cdot[F(\beta):F]

To be linearly disjoint, the two extensions must have the property that $F(\alpha)$ doesn’t contain $\beta$ and $F(\beta)$ does not contain $\alpha$ .

WLOG assume the first case where $F(\alpha)$ doesn’t contain $\beta$ . That would mean the minimal polynomial of $\beta$ over $F(\alpha)$ is the same as the minimal polynomial of $\beta$ over $F$ , i.e. $[F(\alpha)(\beta):F(\alpha)]=[F(\beta):F]$ . By a symmetric argument, we have $[F(\beta)(\alpha):F(\beta)]=[F(\alpha):F]$ . This implies that $[F(\alpha,\beta):F]=[F(\alpha)(\beta):F(\beta)]\cdot[F(\beta):F]=[F(\alpha):F]\cdot[F(\beta):F]=$ .

TODO verify this

TODO left off here

we’ve covered:

primitive element theorem: every finite extension is a simple extension, which is generated by a primitive element

field has degree equal to the minimal polynomial of its primitive element

two extensions of $F$ are linearly disjoint iff its degree is equal to the product of degrees of its components

splitting field definition based on linear disjointness

theorem: finite field extensions have finite degree

An element that is not algebraic over $F$ is called transcendental over $F$ . A key property of transcendental elements $\alpha$ is that the extensions $F(\alpha)$ they generate are linearly disjoint from every extension of $F$ (except the ones that include $\alpha$ ).

Theorem: If

\alpha

is transcedental over

F

, then

F(\alpha)

is linearly disjoint from any extension of

F

that doesn’t include

\alpha

TODO

But in general if you have two algebraic elements $\alpha$ and $\beta$ , then whether they are linearly disjoint depends on their minimal polynomials.

Theorem: Two algebraic elements

\alpha

and

\beta

generate linearly disjoint extensions if their minimal polynomials are coprime. TODO this only works one direction, non-coprime polynomials can still generate linearly disjoint extension

Let $f$ and $g$ be the minimal polynomials of $\alpha$ and $\beta$ respectively.

To be honest, this rigorous construction of the splitting field of $f$ is just a minor detail. Just like with minimal polynomials, we can ignore this actual construction, and just refer to the fact that given any polynomial $f$ , we can refer to its splitting field, which is unique (up to isomorphism) and always exists.

< Back to category Exploration 7: Fields (permalink)
Exploration 6: Finite fields Exploration 8: Galois theory