Notes - Rings and Modules HT24, Fields

$E$ is a field extension of $F$, i.e. $F \subseteq E$.

\[[E : F] := \dim_F(E)\]

i.e. viewing $E$ as a vector space over $F$, and then taking dimension.

Suppose:

• $E/F$ is a field extension
• $d = [E : F] = \dim _ F(E) < \infty$
• $V$ is a vector space over $E$

Then:

• $V$ can be viewed as a vector space over $F$
• $V$ is finite dimensional over $E$ iff $V$ is finite dimensional over $F$, and the dimensions are related by $\dim _ F(V) = [E : F] \dim _ E (V)$.

\[\dim_F(V) = [E : F] \dim_E (V)\]

Overall idea: Follow the definitions. Write out a basis for $E$ over $F$, and a basis for $V$ over $E$, and then multiply them together to get a basis for $V$ over $F$.

For $(1)$, this is immediate from restricting the multiplication map from $E$ to $F$, this will still be a vector space.

For $(2)$, the backward direction ($V$ finite dimensional over $F$ implies $V$ is finite dimensional over $E$) follows from the fact that a finite $F$-spanning set is in particular an $E$-spanning set.

The forward direction follows from showing that if $\{x _ 1, \cdots, x _ d\}$ is an $F$-basis of $E$ and $\{e _ 1, \cdots, e _ n\}$ is an $E$-basis of $V$, then

\[\\{x_i e_j \mid 1 \le i \le d, 1 \le j \le n\\}\]

is an $F$-basis of $V$.

Spanning: Consider $v \in V$. Write as

\[v = \sum^n_{i =1} \lambda_i e_i\]

for $\lambda _ i \in E$. Then, each $\lambda _ i$ can be written in terms of the $F$-basis of $E$, to get

\[\begin{aligned} v &= \sum^n_{i = 1} \left( \sum^d_{j = 1} \mu_{ij} x_j \right) e_i \\\\ &= \sum_{1 \le i \le n, 1 \le j \le d} \mu_{ij} (x_j e_i) \end{aligned}\]

So it spans.

Linearly independent: Consider the above in the case where $v = 0$. Then by the linear independence of the $E$-basis of $V$, we must have all $\lambda _ i = 0$. But then by the linear independence of the $F$-basis of $E$, we must have all $\mu _ {ij} = 0$. Hence the set is linearly independent.

Hence overall,

\[\\{x_i e_j \mid 1 \le i \le d, 1 \le j \le n\\}\]

is indeed an $F$-basis of $V$, and thus

\[\dim_F(V) = [E : F] \dim_E (V)\]

$[K : F]$ is finite iff $[E : F]$, $[K : E]$ are and

\[[K : F] = [K : E] [E : F]\]

We use the following result:

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Suppose:

• $E/F$ is a field extension
• $d = [E : F] = \dim _ F(E) < \infty$
• $V$ is a vector space over $E$

Then:

• $V$ can be viewed as a vector space over $F$
• $V$ is finite dimensional over $E$ iff $V$ is finite dimensional over $F$, and the dimensions are related by $\dim _ F(V) = [E : F] \dim _ E (V)$.

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Viewing $F$ as a vector space over $E$ (since $E/F$), $\dim _ E F = [E:F]$.

But by the result above, since $K$ is a field extension of $E$, $F$ can be viewed as a vector space over $K$. The the degree of the field extension $\dim _ K F = [K : F]$ must equal $[K : E] \dim _ E F = [K : E] [E : F]$. So

\[[K : F] = [K:E][E:F]\]

Furthermore If any of the intermediate extensions are infinite, then the overall one is.

The intersection of subfields of $\mathbb C$ containing elements in $T$, i.e. the smallest subfield of $\mathbb C$ containing all elements of $T$.

• $\mathbb Q$ has a finite extension containing $a$.
• …equivalently, $\mathbb Q(a)$ is finite

(and here being finite means having finite degree as an extension).

Suppose:

• $E/F$ is a finite extension of fields
• $\alpha \in E$

Then:

\[F(\alpha) \cong \frac{F[t]}{\langle f\rangle}\]

via some $f \in F[t]$ where $f$ is monic and irreducible such that the evaluation map $\phi : f \to f(\alpha)$ induces the isomorphism.

The long part of this proof is showing that $\text{Im } \phi$ is a field by using the properties of PIDs.

Since $\alpha \in E$, $F(\alpha)$ is finite.

Let $d = [F(\alpha): F] = \dim _ F(F(\alpha))$.

Then $\{\alpha^i \mid 0 \le i \le d \}$ is linearly dependent (it is a collection of $d+1$ elements), so $\exists \lambda _ i$ s.t. $\sum^d _ {i = 0} \lambda _ i \alpha^i = 0$

Taking $g(x) := \sum^d _ {i = 0} \lambda _ i x^i$ shows that the $\ker \phi \ne \{0\}$

Since $\ker \phi$ is a nonzero ideal in $F[t]$, it is generated by a unique monic polynomial, so $\ker \phi = \langle f \rangle$

Then $F[t] / \langle f \rangle \cong \text{Im } \phi$, so we need to show $\text{Im } \phi = F(\alpha)$

Since $\text{Im } \phi$ is a subfield of $E$, it is an integral domain. This implies that $\langle f \rangle$ is prime.

But if $\langle f \rangle$ is prime, it must be maximal since a polynomial field is a PID, and prime and maximal ideals are the same in a PID. So $\text{Im } \phi$ is actually a field.

$\text{Im } \phi$ is a field that contains $F$ and $\alpha$, so then $F(\alpha) \subseteq \text{Im } \phi$. But then also $F(\alpha) \supseteq \text{Im }\phi$ since the elements of $\text{Im } \phi$ are $F$-linear combinations. Hence

\[F(\alpha) \cong \frac{F[t]}{\langle f\rangle}\]

as required.

$m _ \alpha(x)$ is irreducible, so $\langle m _ \alpha(x) \rangle$ is a maximal ideal, hence the quotient is a field.

We have that

\[\mathbb Q[A] \cong \frac{\mathbb Q[x]}{\langle m(x)\rangle}\]

by the evaluation homomorphism $f(x) \mapsto f(A)$, and likewise

\[\mathbb Q[\alpha] \cong \frac{\mathbb Q[x]}{\langle m(x)\rangle}\]

by the evaluation homomorphism $f(x) \mapsto f(\alpha)$. Chaining isomorphisms gives the result.

• This is equivalent to finding a field extension of $\mathbb F _ 4$ that has 16 elements, say $\mathbb F _ {16}$.
• This can be done by finding an irreducible polynomial of degree 2 in $\mathbb F _ 4$, since we then have the field

\[\frac{\mathbb F[x]}{\langle m(x)\rangle}\]

• which has a basis of $2$ elements, $\{1, x\}$.
• Hence there are $2$ independent choices for $k$ coefficients of $x$ in this new field, so $4^2$ elements.
• Then can find a matrix with minimal polynomial $m(x)$.

Pick a matrix with $f(x)$ as its minimal polynomial, e.g. the companion matrix

\[C(f) = \begin{pmatrix} 0 & \cdots & \cdots & 0 & -a_0 \\\\ 1 & 0 & \ddots & \vdots & - a_1 \\\\ 0 & 1 & \ddots & \vdots & \vdots \\\\ \vdots & \ddots & \ddots & 0 & -a_{n-2} \\\\ 0 & \cdots & 0 & 1 & -a_{n - 1} \end{pmatrix}\]

Fix some $a \in R \setminus \{0\}$ for which we want to find a multiplicative inverse. Then consider the map $T : R \to R$ given by $x \mapsto ax$. Since $R$ is an integral domain, this map is injective. But also since we can view $R$ as a finite dimensional vector space, any injective linear map must also be surjective. Hence $\exists x$ s.t. $ax = 1$.

Consider

\[\mathbb C(t) = \\{p(t) / q(t) \mid p \in \mathbb C[x], q \in \mathbb C[x] \setminus \\{0\\}\\}\]

with the standard definition of addition and multiplication of fractions. Then this is a field containing $\mathbb C$.

Consider the unique ring homomorphism $\phi : \mathbb Q[t] \to \mathbb C$ where $\phi(t) := \sqrt n$. $\text{Im}(\phi)$ is a subring of $\mathbb C$ and is in fact to equal to $\mathbb Q[\sqrt n]$ (consider powers of $\sqrt n$).

Since $\mathbb C$ is a field, it is in particular an integral domain. Then $\text{Im}(\phi)$ must also be an integral domain, and so $\ker \phi$ is a prime ideal of $\mathbb Q[t]$. Since $\mathbb Q[t]$ is a PID, all non-zero prime ideals are maximal ideals (to prove this, you first show every prime ideal corresponds to an irreducible element, and then show that irreducible elements generate maximal ideals). Since $t^2 - n \in \ker \phi$, the kernel is nonzero.

So

\[\text{Im}(\phi) \cong \frac{\mathbb Q[t]}{\ker \phi}\]

is actually a field.

Therefore, it must be the subfield $F _ n$ of $\mathbb C$ generated by $\sqrt n$. Since $\mathbb Q[t]$ is a PID, $\ker \phi$ is a principal ideal, so $\ker \phi = \langle f \rangle$ for a unique monic irreducible polynomial $f$ (since irreducibles are prime in a PID). We must have $f \mid t^2 - n$.

Since $\text{content}(t^2 - n) = 1$, $t^2 - n$ is irreducible over $\mathbb Q[t]$ iff it is irreducible in $\mathbb Z[t]$. But since it is a quadratic, this is true iff it has an integer root, which happens iff $n$ is square.

If $n$ is square, then $R _ n = \mathbb Z$ and $F _ n = \mathbb Q$.

Otherwise, $f = t^2 - n$, and so $\ker \phi = \langle t^2 - n \rangle$. Then $F _ n \cong \frac{\mathbb Q[t]}{\langle t^2 - n\rangle}$ is degree $2$ over $\mathbb Q$.

Since $R$ is a Euclidean domain, we can write $x = a/b$ where $a, b \in R$ and $a, b$ coprime.

Then the equation implies

\[\begin{aligned} &c_0 + c_1 \frac a b + \cdots + c_{n-1} \frac{a^{n-1} }{b^{n-1} } + c_n \frac{a^n}{b^n} = 0 \\\\ \implies& c_0b^n + c_1 b^{n-1}a + \cdots +c_{n-1}ba^{n-1} + a^n = 0 \\\\ \implies&a^n = -b(c_0 b^{n-1} + c_1 b^{n-2} + \cdots + c_{n-1}a^{n-1}) \end{aligned}\]

so $b \mid a^n$, which since $a$ and $b$ are coprime, implies that $b$ is a unit. Then $x = a/b$ is in $R _ n$.

Since the highest common factor always exists in a Euclidean domain, you can write

\[x = \frac a b\]

where $a, b$ are coprime.