Back to Algebra Page

On this page:

• Introduction
• Characteristic
• Field Extensions
• Algebraic Extensions
• Splitting Fields
• Algebraic Closure
• Separable Extensions
• Cyclotomic Extensions
• Worked Exercises

Introduction

On this page, we investigate the basic theory of field extensions. For motivation, field theory should (obviously?) be the study of the category of fields. Consider a morphism of fields $E \to F$. Such a morphism preserves the unit and is (obviously) a ring homomorphism. Thus, the kernel of this map is an ideal of $E$ which implies the kernel is trivial (fields have exactly two ideals: the zero ideal and itself, and I require that ring homomorphisms send $1_E \mapsto 1_F$). So the morphism $E \to F$ is actually an embedding of $E$ into $F$. This implies that the study of the category of fields is equivalent to studying a field and its subfields (or superfields — which are called field extensions). Hence, we come to the study of field extensions.

Characteristic

[definition] [deftitle]Definition %counter%[/deftitle]

The characteristic $\char(F)$ of the field $F$ is defined to be the smallest positive integer $p$ for which $p\cdot 1_F = 0$. If no such $p$ exists, we define $\char(F) = 0$. [/definition]

The next proposition is pretty obvious:

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

The characteristic of a field is always $0$ or a prime integer $p$. [/theorem]

[proof] Clearly there are fields of characteristic $0$. If not, then there is some $p$ for which $\char(F) = p$. If $p$ is not prime, then we can factor it into $p = ab$ where $a, b > 1$ and so

Since $a, b < p$, it follows that one of the factor above is $0$ which contradicts the minimality of $p$. [/proof]

Notice that there is a natural $\ZZ$-action on $F$ via the ring homomorphism

It is clear that $\ker \phi = \char(F)\ZZ$. Quotienting out $\ZZ$ by the kernel of $\phi$ gives an injection of $\ZZ$ or $\ZZ/p\ZZ$ into $F$. Thus, $F$ always contains a subfield isomorphic to $\QQ$ (the field of fractions of $\ZZ$) or $\ZZ/p\ZZ$ (which is its own field of fractions) by the minimality condition placed on the field of fractions. That is,

[definition] [deftitle]Definition %counter%[/deftitle]

The prime subfield of a field $F$ is the subfield of $F$ generated by $1_F$. By the above, it is isomorphic to $\QQ$ if $\char(F) = 0$ or $\ZZ/p\ZZ$ if $\char(F) = p$. [/definition]

Field Extensions

[definition] [deftitle]Definition %counter%[/deftitle]

Let $K$ be a field containing the field $F$. Then $K$ is an extension (field) of $F$, denoted $K/F$ (read “$K$ over $F$”). Sometimes we may always write $F \subseteq K$ or

[/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

The degree $[K:F]$ of the field extension $K/F$ is the dimension of $K$ as an $F$-vector space. If $[K:F] < \infty$, then we say that $K/F$ is finite. Otherwise, we say that $K/F$ is infinite. [/definition]

Admittedly, the above definition is a confusing one. It becomes clarified later on. An important class of field extensions are the ones that are obtained through solving polynomial equations over a given field $F$. If the polynomial fully splits over $F$, then there is obviously no need to talk about field extensions (given that the only goal is to solve the equation). On the other hand, if a polynomial fails to have a root, we may want to construct some larger field $K$ for which $K$ contains $F$ and the desired root.

This can be formalized as follows. Let $p(x) \in F[x]$. Then we will set

So long as $p(x)$ is irreducible, it follows that $K$ is a field (because $F[x]$ is a PID and ideals generated by an irreducible element are maximal in a PID as irreducibles are prime elements). It is easy to see that $F$ is a subfield of $K$. More importantly,

in $K$. As $\overline{p(x)} = p(\overline{x})$, it follows that $\overline{x}$ is a root of $p(x) \in K[x]$. Let us formally state this this result now.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $F$ be a field and let $p(x) \in F[x]$ be irreducible. Then there exists a field $K$ containing an isomorphic copy of $F$ in which $p(x)$ has a root. [/theorem]

We know in particular that $F[x]$ is a Euclidean domain and the resultant remainders due to Euclidean division are precisely the elements of $F[x]$ of degree $\le \deg p(x)$. Exploiting this fact allows us to talk about the field $K$ in terms of a basis, furthering our understanding of the structure of $K$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $F$ be a field and let $p(x) \in F[x]$ be an irreducible degree $n$ polynomial. Then

is a basis for $K \coloneqq F[x]/\generator{p(x)}$ as an $F$-vector space. In particular, this says that $[K:F] = n$. [/theorem]

[proof] (Spanning) By the division algorithm, we know that

where $\deg r(x) < n$ or $r = 0$ for any $f(x) \in F[x]$. Thus, we obtain spanning.

(Linear independence) Suppose that

where each $a_i \in F$. Then it follows that

Thus, $p(x) \mid a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}$. So either $a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1} = 0$ or $p(x)$ has degree less than $n$. The latter is impossible so we must have that $a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1} = 0$. [/proof]

[example] [extitle]Example %counter%[/extitle]

The classic example is the field extension $\RR[x]/\generator{x^2 + 1}$ of $\RR$. It is straightforward to verify that $\RR[x]/\generator{x^2 + 1} \cong \CC$. It follows that $\CC$ is a degree two extension of $\RR$. [/example]

[example] [extitle]Example %counter%[/extitle]

Another classic example is the field extension $\QQ[x]/\generator{x^2 + 1}$ of $\QQ$. It is straightforward to verify that $\QQ[x]/\generator{x^2 + 1} \cong \QQ(i)$. It follows that $\QQ(i)$ is a degree two extension of $\QQ$. [/example]

[example] [extitle]Example %counter%[/extitle]

Consider $\QQ[x]/\generator{x^2 - 2}$. By Eisenstein, $x^2 - 2$ is irreducible over $\QQ$ (by applying it to $p = 2$) so we have a degree two field extension. We claim that this field is isomorphic to $\QQ(\sqrt{2})$. Indeed, if we consider the morphism

the claim follows by the First Isomorphism Theorem. [/example]

[example] [extitle]Example %counter%[/extitle]

Consider $\QQ[x]/\generator{x^3 - 2}$. By Eisenstein, $x^3 - 2$ is irreducible over $\QQ$ (by applying it to $p = 2$) so we have a degree three field extension of $\QQ$. Notice that

Say we want to compute the inverse of $1 + \sqrt[3]{2}$ in this field. It is immediate that $1 + x$ and $x^3 - 2$ are relatively prime in $\QQ[x]$ (both are irreducible!) so Bezout’s identity implies that

for some $f(x), g(x) \in \QQ[x]$. So then

This implies that $f(\sqrt[3]{2})$ is the inverse of $1 + \sqrt[3]{2}$. Straightforward computations in the ring $\QQ[x]/\generator{x^2 - 3}$ show that $f(\sqrt{3}) = \frac{1}{3}(1 - \sqrt[3]{2} + \sqrt[3]{2^2})$. [/example]

Note that these examples show that the is much importance of the fields of the form $F(S)$ where $S \subseteq K$. We obtain this field by taking the intersection of all the fields containing $S$ and $F$ (which is, again, a field, that contains $F$ and $S$). This justifies the terminology in the following definition.

[definition] [deftitle]Definition %counter%[/deftitle]

Let $K$ be an extension of the field $F$ and let $\alpha, \beta, \ldots \in K$. Then the smallest subfield $F(\alpha, \beta, \ldots)$ of $K$ containing both $F$ and elements $\alpha, \beta, \ldots$ is called the field generated by $\alpha, \beta, \ldots$ over $F$.

If the field $K$ is generated by a single element $\alpha$ over $F$ (i.e. $K = F(\alpha)$ for some $\alpha\in K$), then we say that $K$ is a simple extension of $F$ and $\alpha$ is a primitive element for the extension $K/F$. [/definition]

As one can probably guess from the work we did above, we can construct the elements of $F(\alpha)$ explicitly. In particular, we know that there is a ring homomorphism

The kernel of this map is $\generator{p(x)}$ where $p(x) \in F[x]$ is irreducible and $\alpha$ is a root of $p(x)$. It is clear that $\generator{p(x)}$ is a subset of the kernel. Now, since $F[x]/\generator{p(x)}$ is a field, $\generator{p(x)}$ is maximal. Thus, the kernel is actually $\generator{p(x)}$ First Isomorphism Theorem takes care of the rest (it is easy to see our map is surjective.) So we obtain the following theorem:

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K$ an extension of the field $F$. If $p(x)\in F[x]$ is irreducible and $\alpha\in K$ is a root of $p(x)$ in $K$, then $F(\alpha) \cong F[x]/\generator{p(x)}$. [/theorem]

Notice that this says that $\alpha$ can be ANY root of $p(x)$. This means that complexity is roots is, in a certain sense, not distinguishable. For instance, $\QQ[x]/\generator{x^3 - 2}$ is isomorphic to $\QQ(\sqrt[3]{2})$, $\QQ(\omega\sqrt[3]{2})$, and $\QQ(\omega^2\sqrt[3]{2})$ where $\omega$ is the primitive cube root of unity

That being said, however, notice that we can still understand the structure of $F(\alpha)$ pretty easily (so easily, in fact, that we will omit the proof).

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $K$ an extension of the field $F$. If $p(x)\in F[x]$ is irreducible and $\alpha\in K$ is a root of $p(x)$ in $K$, then

[/theorem]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $\phi:F \to F’$ be an isomorphism of fields. Let $p(x) \in F[x]$ be an irreducible polynomial. Then the polynomial $p’(x) \in F’[x]$, obtained by applying $\phi$ to the coefficients of $p(x)$, is irreducible. Furthermore, if $\alpha$ is a root of $p(x)$ and $\beta$ is a root of $p’(x)$, then there is an isomorphism

extending $\phi$. [/theorem]

[proof] Define $\widetilde{\phi}:F[x] \to F’[x]$ by sending

It is obvious that this is a ring isomorphism. Now write

Then

The isomorphism $\widetilde{\phi}$ maps the maximal ideal $\generator{p(x)}$ to the ideal $\generator{p’(x)}$ so the latter is maximal. Thus, $p’(x)$ is irreducible in $F’[x]$. So then passing to the quotient rings gives an isomorphism of fields

and the isomorphism of $F(\alpha)$ and $F’(\beta)$ is given by composing the isomorphisms. [/proof]

Algebraic Extensions

In this section, $F$ is a field and $K$ is an extension of $F$.

[definition] [deftitle]Definition %counter%[/deftitle]

The element $\alpha \in K$ is algebraic over $F$ if $\alpha$ is the root of some nonzero polynomial $f(x) \in F[x]$. If $\alpha$ is not algebraic over $F$, then $\alpha$ is transcendental over $F$. We say that $K/F$ is algebraic provided that every element of $K$ is algebraic over $F$; otherwise, $K$ is a transcendental extension. [/definition]

Obviously, if $\alpha$ is algebraic over $F$, then it is algebraic over any extension field $L \ni \alpha$ of $F$.

From a ring theoretic perspective, if there is at least one polynomial in $F[x]$ that annihilates $\alpha$, then we are interested in the generator of said annihilator ideal. Another way of stating this is the following: Suppose we define the map

Then $\ker\phi$ is an ideal of $F[x]$. As $F[x]$ is a Euclidean domain, it is also a PID so $\ker\phi = \generator{m_{\alpha,F}(x)}$ where $m_{\alpha,F}(x)$ is monic. This polynomial is so important that it has a name: the minimal polynomial for $\alpha$ over $F$. An equivalent way of obtaining this polynomial is the following:

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $\alpha$ be algebraic over $F$. Then there is a unique monic irreducible polynomial $m_{\alpha, F}(x) \in F[x]$ which has $\alpha$ as a root. A polynomial $f(x) \in F[x]$ has $\alpha$ as a root if and only if $m_{\alpha, F}(x)$ divides $f(x)$ in $F[x]$. [/theorem]

[proof] (First claim) The existence of $m_{\alpha, F}(x)$ is given by the fact that $F$ is a field and that $F[x]$ is a PID. If $m_{\alpha, F}(x)$ isn’t monic, we can always make it so by dividing out by the leading coefficient to obtain an associate. Uniqueness is given by monicity condition and the fact that two principal ideals are equal in an integral domain if and only if their generators are associates.

(Second claim) If $f(\alpha) = 0$, then $f(x) \in \generator{m_{\alpha,F}(x)}$ and the claim follows immediately. The converse is obvious. [/proof]

The minimal polynomial, in general, changes if we enlargen our field. For example, the minimal polynomial for $i$ over $\RR$ is just $x^2 + 1$ (this is easily seen by considering the usual ring homomorphism $\RR[x] \to \CC$); however, the minimal polynomial for $i$ over $\CC$ is $x - i$. This holds in much greater generality.

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

If $L/F$ is an extension of fields and $\alpha$ is algebraic over both $F$ and $L$, then $m_{\alpha,L}(x) \mid m_{\alpha, F}(x)$ in $L[x]$. [/theorem]

[proof] Since $F[x] \subseteq L[x]$, it follows that $m_{\alpha, F}(x) \in L[x]$. Thus, we apply the second claim of the previous theorem to give the desired claim. [/proof]

Let us finally give some formal definitions to everything.

[definition] [deftitle]Definition %counter%[/deftitle]

The polynomial $m_{\alpha,F}(x)$ is called the minimal polynomial for $\alpha$ over $F$. If the field is understood, we will instead write $m_\alpha(x)$. The degree of $\alpha$ is the degree of $m_\alpha(x)$. [/definition]

It should be clear that a monic polynomial over $F$ with $\alpha$ as a root is the minimal polynomial for $\alpha$ over $F$ if and only if it is irreducible over $F$. This observation gives us the following result which justifies the usage of our terminology above.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $\alpha$ be algebraic over $F$. Then

Thus,

[/theorem]

One thing that should be clear from the way we have defined degree is that $\alpha$ being algebraic means there is a minimal polynomial for it over $F$ and, so, the degree is finite. The converse requires a little bit more care which gives rise to the surprising (or at least it was to me) result.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The element $\alpha$ is algebraic over $F$ if and only if the simple extension $F(\alpha)/F$ is finite. [/theorem]

[proof] ($\implies$) This is obvious from the way the minimal polynomial is defined! We know that

which implies $F(\alpha)$ is a finite extension over $F$.

($\impliedby$) Suppose that $F(\alpha)$ is an $n$-dimensional $F$-vector space. Then the elements

are linearly dependent. Thus, there is some polynomial in $F[x]$ of degree $n$ for which $\alpha$ is a root. [/proof]

Since there is a natural embedding sequence

we obtain the nice following corollary!

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

If $K/F$ is a finite extension, then $K/F$ is algebraic. [/theorem]

[proof] If $\alpha$ is any element of $K$, then we know that $F(\alpha)$ is an $F$-vector subspace of $K$. So certainly the degree of $F(\alpha)/F$ is finite which means that $\alpha$ is algebraic. It follows that every element of $K$ is algebraic over $F$ and we are done. [/proof]

[example] [extitle]Remark %counter%[/extitle]

The converse to this statement is FALSE. For instance, the algebraic closure $\overline{\QQ}$ of $\QQ$ has infinite degree despite being algebraic over $\QQ$. [/example]

Let us now pivot a bit and talk about a theorem in field theory that is immediate from linear algebra.

[theorem] [thmtitle]Theorem %counter% (Tower Theorem)[/thmtitle]

Let $F \subseteq K \subseteq L$ be fields. Then

[/theorem]

[proof] Let $\alpha_1, \ldots, \alpha_n$ be a basis for $K$ over $F$ and $\beta_1, \ldots, \beta_m$ be a basis for $L$ over $K$. We claim that the collection $\alpha_i\beta_j$ forms a basis for $K$ over $F$.

(Spanning) Let $\gamma \in L$. Then, as $L$ is a $K$-vector space, we can write

Furthermore, since $K$ is an $F$-vector space, we can write each $b_j$ as a linear combination of $\alpha_i$’s:

Thus, we obtain spanning as desired.

(Linear independence) If

Then noting the $\beta_j$’s being a basis for $L$ over $K$ implies that the $\sum_{i=1}^n a_{ij}\alpha_i$ are all $0$. But $\alpha_i$’s being a basis of $K$ over $F$ implies that all the $a_{ij}$’s are $0$. [/proof]

An obvious corollary that follows is the following:

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $F \subseteq K \subseteq L$ be fields. Then $[K:F]$ divides $[L:F]$. [/theorem]

We can also characterize the finite extensions of a field $F$:

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The extension $K/F$ is finite if and only if $K = F(\alpha_1, \ldots, \alpha_n)$ where each $\alpha_i$ is algebraic over $K$. Additionally,

[/theorem]

[proof] Granted the first claim, we claim that

Indeed, since $F(\alpha_1, \ldots, \alpha_{i}) = F(\alpha_1, \ldots, \alpha_{i-1})(\alpha_i)$, $\alpha_i$ is algebraic over $F(\alpha_1, \ldots, \alpha_{i-1})$. Its minimal polynomial over $F(\alpha_1, \ldots, \alpha_{i-1})$ must divide its minimal polynomial over $F$ so this implies the above. The rest follows.

($\implies$) Pick out any basis for $K$. Obviously $K$ is generated by that basis and each basis element $\alpha_i$ is algebraic over $F$.

($\impliedby$) Immediate by the tower theorem. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

If $\alpha$ and $\beta$ are algebraic over $F$, then so are $\alpha \pm \beta$, $\alpha\beta$, and $\alpha/\beta$. [/theorem]

[proof] All of these are in $F(\alpha, \beta)$ and the theorem shows this extension is finite. Finite extensions are algebraic so the claim follows. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $L/F$ be an extension. Then the collection of elements of $L$ that are algebraic over $F$ form a subfield $K$ of $L$ containing $F$. [/theorem]

[proof] Obviously every element of $F$ is algebraic over $F$ so $K$ definitely contains $F$. The previous theorem shows that $K$ is closed under the normal operations of a field so the claim follows. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

If $K$ is algebraic over $F$ and $L$ is algebraic over $K$, then $L$ is algebraic over $F$. [/theorem]

[proof] Let $\alpha \in L$. Since $L$ is algebraic over $K$, it follows that there is some polynomial

in $K[x]$ that has $\alpha$ as a root. Since $K$ is algebraic over $F$, each $k_i$ is algebraic over $F$. This tells us that the extension

is finite and we know that finite extensions are algebraic. So then

is a finite extension as $\alpha$ is algebraic over $F(k_0, \ldots, k_n)$. Thus, $\alpha$ itself is algebraic over $F$ and the claim follows. [/proof]

Any now we come to our final topic in this section: composite fields.

[definition] [deftitle]Definition %counter%[/deftitle]

Let $K_1$ and $K_2$ be two subfields of a field $K$. Then the smallest subfield of $K$ containing both $K_1$ and $K_2$ is called the composite field of $K_1$ and $K_2$, denoted $K_1K_2$. [/definition]

Notice that this gives rise to the latice

Intuitively, if $K_1$ and $K_2$ are finite extensions of $F$, then there are bases $\alpha_1, \ldots, \alpha_n$ for $K_1$ and $\beta_1, \ldots, \beta_m$ for $K_2$ over $F$. Then, by definition,

So $\alpha_1,\ldots, \alpha_n, \beta_1, \ldots, \beta_m$ span all of $K_1K_2$ as an $F$-vector space. This gives the following proposition.

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Let $K_1$ and $K_2$ be finite extensions of $F$. Then

[/theorem]

[proof] By the tower theorem, we know that

It remains to show that $[K_1K_2:K_1] \le [K_2:F]$. Since $K_1/F$ and $K_2/F$ are finite extensions, it follows that there are bases $\alpha_1, \ldots, \alpha_n$ of $K_1$ over $F$ and $\beta_1, \ldots, \beta_m$ of $K_2$ over $F$. Since

it follows that $K_1K_2$ is spanned by $\beta_1, \ldots, \beta_m$ over $K_1$. Since this is a spanning set and not necessarily a basis, it follows that $[K_1K_2:K_1] \le m = [K_2:F]$ and the claim follows. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $K_1/F$ and $K_2/F$ be finite extensions. If $[K_1:F]$ and $[K_2:F]$ are relatively prime, then

[/theorem]

[proof] By the tower law, we know that

Thus, $[K_1 : F]$ is a factor of $[K_1K_2 : F]$. Similarly, $[K_2 : F]$ is a factor of $[K_1K_2: F]$. Thus, the least common multiple of these two is a factor of $[K_1K_2:F]$. As $[K_1:F]$ and $[K_2:F]$ are relatively prime, their least common multiple is $[K_1:F][K_2:F]$. Thus,

where the second inequality follows from the above proposition. [/proof]

Splitting Fields

[definition] [deftitle]Definition %counter%[/deftitle]

A minimal extension field $K$ of $F$ is called a splitting field for $f(x) \in F[x]$ if $f(x)$ factors completely into linear factors. [/definition]

It should be fairly clear that the splitting field of a polynomial always exists — we just keep adjoining roots inductively until none are left. As such, we will omit the proof of the following theorem.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $F$ be a field and $f(x) \in F[x]$. Then there is a splitting field for $f(x)$. Furthermore, all such splitting fields are isomorphic. [/theorem]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

The splitting field $K$ of a polynomial of degree $n$ over $F$ is of degree at most $n!$ over $F$. [/theorem]

[proof] The polynomial $f(x)$ has at most $n$ distinct roots so at most $n$ adjoins are required to obtain all roots. Thus, the claim follows. [/proof]

One theorem that I will omit here (but is intuitively obvious) is that the splitting field is unique.

Algebraic Closure

I will not say much about the construction of the algebraic closure $\overline{F}$ of $F$.

[definition] [deftitle]Definition %counter%[/deftitle]

The field $\overline{F}$ is called the algebraic closure of $F$ if $\overline{F}/F$ is algebraic and every polynomial $f(x) \in F[x]$ splits completely over $\overline{F}$. We say that $F$ is algebraically closed if $\overline{F} = F$. [/definition]

The usage of *the* algebraic closure is justified — all algebraic closures are isomorphic (which we will omit a proof of).

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

$\overline{F}$ is algebraically closed over itself. [/theorem]

[proof] Let $f(x) \in \overline{F}[x]$ and $\alpha$ be a root of $f(x)$. Since $\overline{F}(\alpha)$ is algebraic over $\overline{F}$ and $\overline{F}$ is algebraic over $F$, it follows that $\overline{F}(\alpha)$ is algebraic over $F$. So $\alpha$ is algebraic over $F$ which implies that $\alpha\in\overline{F}$. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

$\overline{F}$ always exists. [/theorem]

Separable Extensions

Also of much interest in field theory is if all the roots of $f(x)$ are simple (i.e. distinct).

[definition] [deftitle]Definition %counter%[/deftitle]

Let $F$ be a field. We say that $f(x) \in F[x]$ is separable over $F$ if all its roots (in its splitting field) are distinct. A polynomial which is not separable is called inseparable. [/definition]

This leads to the following definition.

[definition] [deftitle]Definition %counter%[/deftitle]

We say that $K/F$ is a separable extension of $F$ if every element of $K$ is the root of a separable polynomial over $F$. Otherwise, we say that $K$ is inseparable. [/definition]

In analysis, we know that if a holomorphic function $f$ has a zero $c$ of order $r$, we can rewrite it as

where $g$ is holomorphic and nonzero at $z = c$. This arises by Laurent-expanding $f$ at $z = c$ and then factoring out $(z - c)^r$. Furthermore, we know that $f^{(k)}$ has a zero at $z = c$ for $0 \le k < r$ which is trivial by the above representation of $f$. This fact generalizes to polynomials, in general, by considering the formal derivative $D:F[x] \to F[x]$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

A polynomial $f(x)$ has a multiple root $\alpha$ if and only if $\alpha$ is also the root of $Df(x)$. Equivalently, $f(x)$ is separable if and only if it is relatively prime to $Df(x)$. [/theorem]

[proof] ($\implies$) If $\alpha$ is a multiple root, then we can write

where $g(x) \in F[x]$. It is trivial to prove the usual product rule and chain rules which shows that

Clearly $\alpha$ is a root of $Df(x)$ and the claim follows.

($\impliedby$) Since $\alpha$ is a root of $f(x)$, it follows that we may write

for some $g(x) \in F[x]$. So then we have

Since $\alpha$ is a root of $Df(x)$, it follows that $\alpha$ is a root of $g(x)$ so we can reexpress $g(x)$ as $(x - \alpha)h(x)$ for some $h(x) \in F[x]$. So it follows that

and the claim follows. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Every irreducible polynomial over a field of characteristic $0$ is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials. [/theorem]

[proof] Let $F$ be a field of characteristic $0$ and let $f(x)\in F[x]$ be irreducible. For the first claim, consider the fact that $Df(x)$ is a polynomial of degree $n-1$ (because we are working in characteristic $0$). Since $f(x)$ is irreducible, the only factors (up to constant multiples) of $f(x)$ are $1$ and $f(x)$. As $Df(x)$ is neither of those, the claim follows. Now let us show the second claim.

($\implies$) If $f(x)$ is separable, then it splits into linear factors over the splitting field of $f(x)$ over $F$. Since linear factors are trivially irreducible, and we cannot have any multiple roots, the claim follows immediately.

($\impliedby$) Since each irreducible polynomial itself is separable, each distinct irreducible polynomial contributes distinct linear factors over the splitting field over $F$ so the claim follows. [/proof]

Cyclotomic Extensions

Before we jump into cyclotomic extensions, we will refresh a bit on elementary number theory.

[definition] [deftitle]Definition %counter%[/deftitle]

Euler’s totient function $\phi:\ZZ_{>0} \to \ZZ_{>0}$ is the function defined by setting $\phi(n)$ to be the number of positive integers less than $n$ relatively prime to $n$. [/definition]

Now, as it turns out, $\phi(n)$ is actually the order of the multiplicative group $(\ZZ/n\ZZ)^\times$ consisting of the units of $\ZZ/n\ZZ$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The order of the group $(\ZZ/n\ZZ)^\times$ is $\phi(n)$. [/theorem]

[proof] Recall that $(\ZZ/n\ZZ)^\times$ are the units of the ring $\ZZ/n\ZZ$. Now, if $a$ is not relatively prime to $n$, then $a$ and $n$ have some nontrivial common divisor $d$ and it follows that $ad = 0$ in $\ZZ/n\ZZ$. Since zerodivisors cannot be units, it follows that $a$ is not a unit. On the other hand, if $a$ is relatively prime to $n$, then Bezout’s identity says there exist integers $x, y \in \ZZ$ for which

Modding out by $n$ gives $ax \equiv 1\pmod{n}$ so it follows that the congruence class of $x$ modulo $n$ is the multiplicative inverse of $a$ in $\ZZ/n\ZZ$ and we are done. [/proof]

Let $\mu_n$ denote the group of $n$th roots of unity. Obviously, $\mu_n \cong \ZZ/n\ZZ$. Suppose that $\omega$ is the primitive $n$th root of unity

We call the minimal polynomial $\Phi_n(x)$ of $\omega$ over $\QQ$ the $n$th cyclotomic polynomial. Since it doesn’t matter which root we choose (so long as we get a primitive $n$th root of unity), we obtain the following theorem.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $n$ be a positive integer. Then

[/theorem]

[proof] Note that any solution to $x^n - 1$ is a root of unity and, by Lagrange’s theorem, any primitive $k$th root of unity that solves $x^n - 1 = 0$ meets the requirement that $k\mid n$. So then, collecting all the distinct primitive roots of unity of their respective orders gives the desired claim. [/proof]

This theorem bears importance because we can compute the cyclotomic polynomials recursively. Obviously

Then

Notice that in particular, for a prime $p$:

Now we come to some pretty important statements:

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The cyclotomic polynomial $\Phi_n(x)$ is a monic polynomial in $\ZZ[x]$ of degree $\phi(n)$. [/theorem]

[proof] It is immediate that $\Phi_n(x)$ is monic since $x^n - 1$ itself is monic in $\QQ[x]$. It is also immediate that the degree of $\Phi_n(x)$ is $\phi(n)$. To show that $\Phi_n(x)\in \ZZ[x]$, we apply induction on $n$. Obviously the statement is true for $n = 1$. Now suppose that $\Phi_d(x) \in \ZZ[x]$ for all $d \in [n-1]$. Consider

Clearly, $f(x)$ divides $x^n - 1$ in $\QQ(\omega_n)$ of $n$th roots of unity. Since both $f(x)$ and $x^n - 1$ have coefficients in $\QQ$, it follows that $f(x)$ divides $x^n - 1$ in $\QQ[x]$. Gauss’ Lemma implies that $f(x)$ divides $x^n - 1$ in $\ZZ[x]$ and the claim follows. [/proof]

Something that is pretty important that we will skip the proof of is the following.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The cyclotomic polynomial $\Phi_n(x)$ is an irreducible monic polynomial in $\ZZ[x]$ of degree $\phi(n)$. [/theorem]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $\QQ(\omega_n)/\QQ$ be the extension over $\QQ$ containing the $n$th roots of unity. Then

[/theorem]

[proof] $\Phi_n(x)$ is the minimal polynomial of $\omega_n$ (where $\omega_n$ is a primitive $n$th root of unity). Since this has degree $\phi(n)$, the claim follows. [/proof]

Worked Exercises

[example] [extitle]TODO: Cite[/extitle]

Let $p(x) = x^4 + x + 1 \in (\ZZ/2\ZZ)[x]$.

1. Verify that $p(x)$ is irreducible.
2. Let $\alpha$ be a root of $p(x)$ in the field extension $K = (\ZZ/2\ZZ)[x]/\generator{p(x)}$. Show that $\alpha$ generates the multiplicative group $K^\times$.
3. Show that $p(x)$ splits into linear factors in $K$, and find its roots in terms of $\alpha$. [/example]

[solution] (1) It is clear that $p(x)$ has no roots in $\ZZ/2\ZZ$. So it follows that if $p(x)$ does factor, it does so as

This implies that $bd = 1$ so $b = d = 1$. Thus, $a + c = 1$ and $a + c = 0$, contradiction.

(2) We know that

The rest is pretty straightforward. There are only 15 elements to check and computing the powers of $\alpha$ gives them all (keeping in mind we are working in characteristic $2$).

(3) Not difficult by utilizing part (2) but tedious. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Let $F$ be a field of characteristic $\neq 2$. Let $a$ and $b$ be elements of $F$, neither of which is a square in in $F$.

1. Prove that $K = F(\sqrt{a}, \sqrt{b})$ is of degree $4$ over $F$ if $ab$ is not a square in $F$, and is of degree $2$ over $F$ otherwise.
2. Assume for the rest of the problem that $[K:F] = 4$. Prove that $\gamma = \sqrt{a} + \sqrt{b}$ is a primitive element for $K$ over $F$.
3. Prove that the minimal polynomial of $\gamma$ over $F$ is $g(x) = x^4 - 2(a+b)x^2 + (a-b)^2$.
4. Deduce that $x^4 - 10x^2 + 1$ is irreducible over $\QQ$. [/example]

[proof] (1) First we show that $[F(\sqrt{a}): F] = 2$. It is clear that $\sqrt{a} \notin F$ so there is no polynomial of degree one over $F$ for which $\sqrt{a}$ is a root. On the other hand, $x^2 - \sqrt{a}$ does the trick (this is the minimal polynomial). The claim that $[F(\sqrt{a}, \sqrt{b}):F(\sqrt{a})]$ being equal to one if $ab$ is a square and two if not follows from an identical argument.

(2) This is not difficult, but it is tedious. Notice that

is a linear system. As $F$ does not have characteristic $2$, this system can be solved for $\sqrt{a}$ and $\sqrt{b}$ which we will omit the calculations for. Doing so gives $\sqrt{a}, \sqrt{b} \in F(\gamma)$.

(3) It is straightforward to show that $\gamma$ satisfies $g(x)$. Since we know the degree of $F(\sqrt{a}, \sqrt{b})$ over $F$ is four, the claim follows.

(4) This is the minimal polynomial when we set $a = 3$ and $b = 2$ and the criterion for this being the minimal polynomial is that $\gamma$ is root if and only if the minimal polynomial is irreducible over $\QQ$. So the claim follows. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Prove that if $\alpha\beta$ and $\alpha + \beta$ are algebraic over $F$, then so are $\alpha$ and $\beta$. [/example]

[proof] Notice that $\alpha$ and $\beta$ are algebraic over $K\coloneqq F(\alpha\beta, \alpha + \beta)$ as both are roots to

Since $\alpha$ is algebraic over $K$, it follows that

is finite. Futhermore, we also see that $[K:F]$ is finite because $\alpha + \beta$ and $\alpha\beta$ are algebraic over $F$. So the tower law implies that

But since the tower law also implies that

it follows that $[F(\alpha):F]$ is finite so $\alpha$ is algebraic over $F$. A very similar claim follows for $\beta$. [/proof]

[example] [extitle]TODO: Cite[/extitle]

1. Determine the splitting fields of $x^4 - 4$ and $x^4 - 16$ over $\QQ$, including their degrees.
2. Let $a$ be a positive integer. Determine the degree of the splitting field of $s(x) = x^4 - a$ over $\QQ$, in terms of $a$. [/example]

[solution] (1) Noting that

Thus, the splitting field of this polynomial is $\QQ(\sqrt{2}, i\sqrt{2})$. The minimal polynomial of $\QQ(\sqrt{2}, i\sqrt{2})$ over $\QQ(\sqrt{2})$ is just $x^2 + 2$ so Tower Law says the degree of the splitting field of $x^4 - 4$ over $\QQ$ is 4.

For $x^4 - 16$, similar reasoning shows that $\QQ(2i)$ (the splitting field) is degree two over $\QQ$.

(2) Note $x^4 - a$ factors as

So the splitting field of $x^4 - a$ is just $\QQ(\sqrt[4]{a}, i\sqrt[4]{a})$. The degree is either $2$, $4$, or $8$ depending on if $a$ is a fourth power, square, or neither in $\QQ$. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Determine the splitting field of $x^4 + 2$ over $\QQ$, including its degree. [/example]

[solution] A root of $x^4 + 2$ satisfies the equation

So then we want to compute. That is, the roots $\xi_j$ are of the form

That is, the roots are of the form

With some effort, one can show this implies that the splitting field contains $\sqrt[4]{2}$ and $i$ so the splitting field is just $\QQ(\sqrt[4]{2}, i)$. So then the Tower Law implies

The later factor is just 4 (because $x^4 - 2$ is irreducible over $\QQ$ by Eisenstein) and the former is $2$ (because $x^2 + 1$ is irreducible over $\QQ(\sqrt[4]{2})$). Thus, $[\QQ(\sqrt[4]{2}, i) : \QQ] = 8$. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Determine the splitting field of $x^4 + x^2 + 1$ over $\QQ$, including its degree. [/example]

[solution] With a little bit of algebra (or complex analysis), it is straightforward to show that the roots of this polynomial are

So it follows that the splitting field is just $\QQ(i\sqrt{3})$. It’s also easy to see that the minimal polynomial of $i\sqrt{3}$ over $\QQ$ is $x^2 + 3$ so the splitting field has degree $2$ over $\QQ$. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Determine the splitting field of $x^6 - 4$ over $\QQ$, including its degree. [/example]

[solution] It is straightforward to see that the roots of this polynomial are of the form $\sqrt[6]{4}\omega^k$ where $k = 0, \ldots, 5$ and $\omega$ is the principal 6th root of unity. Thus, we claim that the splitting field is $\QQ(\sqrt[6]{4}, \omega)$. This is easy to see since clearly

For the reverse direction, we just take $\sqrt[6]{4}\omega$ and multiply by $(\sqrt[6]{4})^5$ to obtain $\omega$. By the tower law, we have that

Now, since $\sqrt[6]{4} = \sqrt[3]{2}$, it’s apparent that the second factor is just $3$ because $\sqrt[3]{2}$ has minimal polynomial $x^3 - 2$ over $\QQ$ (which is irreducible over $\QQ$ by Eisenstein). For the former, the minimal polynomial is

Thus, the splitting field has degree 6 over $\QQ$. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Determine the splitting field of $x^6 - 2$ over $\QQ$, including its degree. [/example]

[solution] Let $\omega$ be the principal 6th root of unity. Then we claim that the splitting field of $x^6 - 2$ over $\QQ$ is $\QQ(\sqrt[6]{2}, \omega)$. Showing that this field is minimal is trivial so we omit the proof of doing so. Now, $x^6 - 2$ is irreducible over $\QQ$ by Eisenstein and $\sqrt[6]{2}$ is certainly a root of this polynomial so this polynomial is the minimal polynomial of $x^6 - 2$. It follows that $[\QQ(\sqrt[6]{2}):\QQ] = 6$.

Now, on the other hand, we know that $x^2 - x + 1$ is an irreducible polynomial over $\QQ(\sqrt[6]{2})$ because if it did factor, $\QQ(\sqrt[6]{2})$ would contain $\omega$. Since the elements of $\QQ(\sqrt[6]{2})$ are real, we would arrive at a contradiction. So this polynomial is the minimal polynomial of $\omega$ over $\QQ(\sqrt[6]{2})$ and so it follows that $[\QQ(\sqrt[6]{2}, \omega):\QQ(\sqrt[6]{2})] = 2$.

By the tower law, we conclude that

[/solution]

[example] [extitle]TODO: Cite[/extitle]

Prove that $d$ divides $n$ if and only if $x^d - 1$ divides $x^n - 1$. [/example]

[proof] Recall that

So in the $\implies$ direction, any factor of $d$ is also a factor of $n$ and the claim follows immediately. In the $\impliedby$ direction, every factor of $d$ must be a factor of $n$. So $d\mid n$. [/proof]

[example] [extitle]TODO: Cite[/extitle]

For any prime $p$ and any nonzero $a \in F\coloneqq \ZZ/p\ZZ$, prove that $g(x) = x^p - x + a$ is irreducible and separable over $\ZZ/p\ZZ$. [/example]

[proof] (Separability) Computing the derivative of $g(x)$ gives $g’(x) = 1$. This polynomial has no roots so $g(x)$ is separable.

(Irreducibility) Suppose that $g(x)$ has a root $\alpha$ (in the splitting field). Then

Thus, by induction,

Since $g(x)$ is separable, we know that each of the linear factors above is distinct. Now if $\alpha \in F$, then so is each $\alpha + k$ which means that $0$ is a root of $g(x)$. This implies that $g(0) = a = 0$ which is a contradiction. Assume that none of the roots are in $F$, then the minimal polynomial of each $\alpha + j$ over $F$ is $g(x)$. [/proof]