Introduction
In the previous page, we stood at a crossroads. Having
established that we can always expand a field to find the roots of any nonconstant polynomial, we must now
ask: how do we measure the "size" or the "degrees of freedom" of this new, extended universe?
We now take the path toward continuous symmetry and geometry. Before we can understand the smooth, continuous dimensions of Lie groups and manifolds
used in modern ML, we must first understand dimension in its purest, discrete algebraic form. When we adjoin a new element to a field,
the enlarged field becomes a vector space over the original one, and its dimension measures how much the element adds.
Connection to Geometric Deep Learning (GDL)
In robotics and 3D computer vision, we constantly track degrees of freedom (for instance, a rigid
body in \(SE(3)\) has 6 degrees of freedom). In abstract algebra, the "degree of an extension" tracks
the exact dimensionality of a new field over its base. The rules governing how these discrete dimensions
stack and multiply, above all the Tower Rule, are the rigorous mathematical ancestors of
dimension counting in the smooth manifolds and Lie groups studied in the pages ahead.
Characterization of Extensions
When we expand a base field \(F\) by adding a new element \(a\) to create an extension \(E\), the structure
of this new universe depends entirely on the nature of \(a\). Specifically, we must ask if \(a\) is bound by
the algebraic rules of \(F\), or if it is entirely independent. Throughout, fields are assumed to satisfy
\(1 \neq 0\).
Definition: Algebraic & Transcendental Extensions
Let \(E\) be an extension field of a field \(F\). Then \(a \in E\) is called:
-
algebraic over \(F\) if \(a\) is the zero of some nonzero polynomial in \(F[x]\).
-
transcendental over \(F\) if \(a\) is not algebraic over \(F\).
An extension \(E\) of \(F\) is called an algebraic extension of \(F\) if every element
of \(E\) is algebraic over \(F\). If \(E\) is not an algebraic extension of \(F\), it is called a
transcendental extension of \(F\).
If an element is algebraic (like \(\sqrt{2}\) over \(\mathbb{Q}\)), it brings a finite, measurable amount
of new information to the field. If it is transcendental (like \(\pi\) or \(e\)), it brings infinitely many
linearly independent powers, and the dimension of the field becomes infinite. Because we are building toward
the study of finite-dimensional continuous groups, we care deeply about the algebraic case.
Theorem: Characterization of Extensions
Let \(E\) be an extension of the field \(F\) and let \(a \in E\). If \(a\) is transcendental over
\(F\), then \(F(a) \cong F(x)\), the field of rational functions in \(x\) over \(F\) (the field of
fractions of \(F[x]\)).
If \(a\) is algebraic over \(F\), then \(F(a) \cong F[x]/\langle p(x) \rangle\), where \(p(x)\) is the
unique monic irreducible polynomial in \(F[x]\) of minimal degree such that \(p(a) = 0\).
Proof:
We treat the two cases separately.
Algebraic case. Since \(a\) is algebraic, the set
\(I = \{f(x) \in F[x] \mid f(a) = 0\}\) is a nonzero ideal of \(F[x]\). As
\(F[x]\) is a principal ideal domain,
\(I = \langle p(x) \rangle\) for a unique monic polynomial \(p(x)\) of minimal degree. Uniqueness holds
because any two such generators differ by a unit, and the monic condition pins down the unit.
This \(p(x)\) is irreducible. If \(p = gh\) with both factors of lower degree, then \(g(a)h(a) = 0\)
in the field \(F(a)\) forces one factor to vanish at \(a\), contradicting the minimality of
\(\deg p\). The isomorphism \(F(a) \cong F[x] / \langle p(x) \rangle\) then follows from
the structure theorem for \(F(a)\)
proved on the previous page.
Transcendental case. Consider the evaluation homomorphism
\(\phi : F[x] \to F(a),\ f(x) \mapsto f(a)\). Since \(a\) is transcendental, \(\ker \phi = \{0\}\), so
\(\phi\) is injective and \(F[x]\) embeds into \(F(a)\). The image \(\phi(F[x])\) is an integral domain
inside the field \(F(a)\). The construction of the field of fractions of an integral domain, and the fact
that an injective homomorphism of an integral domain into a field extends to its field of fractions, are
used here without proof. Granting them, \(\phi\) extends to an embedding of \(F(x)\) into \(F(a)\). By
minimality of \(F(a)\) as the smallest field containing \(F\) and \(a\), we conclude \(F(a) \cong F(x)\).
In the algebraic case, as established on the previous page, the isomorphism
\(F(a) \cong F[x]/\langle p(x) \rangle\) gives us a concrete data structure. More importantly, it implies a
strict limit on the size of the new field. This brings us to the most crucial property of algebraic
extensions: their vector space dimension.
Finite Extensions & The Tower Rule
Here, abstract algebra meets linear algebra directly. Any extension field \(E\) can simply be viewed as a
vector space over its base field \(F\). For example, the complex numbers \(\mathbb{C}\) form a
2-dimensional vector space over the real numbers \(\mathbb{R}\), with the basis \(\{1, i\}\).
Definition: Degree of an Extension
Let \(E\) be an extension field of a field \(F\). If \(E\) has dimension \(n\) as a vector space over
\(F\), we say that \(E\) has degree \(n\) over \(F\) and write
\[
[E:F] = n.
\]
If \([E:F]\) is finite, \(E\) is called a finite extension of \(F\), and if not, \(E\) is
called an infinite extension of \(F\).
Restricting the vector space dimension forces every element of the field to be algebraic. There is simply not
enough "room" in a finite-dimensional space for an element to be transcendental.
Theorem: Finite Implies Algebraic
If \(E\) is a finite extension of the field \(F\), then \(E\) is an algebraic extension of \(F\).
Proof:
Suppose that \(E\) is a finite extension of \(F\) with dimension \([E:F] = n\), and
\(a \in E\). If two of the powers \(1, a, \ldots, a^n\) coincide, say \(a^i = a^j\) with
\(i \lt j\), then \(a\) is a zero of the nonzero polynomial \(x^j - x^i\) and is algebraic.
Otherwise \(\{1, a, \ldots, a^n\}\) is a set of \(n+1\) elements, and since \(E\) is
spanned over \(F\) by the \(n\) elements of a basis, it is linearly dependent by the
Spanning Set Bound.
Thus, there exist elements \(c_0, c_1, \ldots, c_n \in F\) (not all zero) such that
\[
c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0.
\]
Then \(a\) is a root of the nonzero polynomial
\[
f(x) = c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0.
\]
Therefore, by definition, \(a\) is algebraic over \(F\). Since this applies to any \(a \in E\), the entire
extension \(E\) is algebraic.
Remark: The Converse Fails
An algebraic extension need not be finite. The standard counterexample is the set
\(\overline{\mathbb{Q}}\) of all algebraic numbers, that is, of every \(\alpha \in \mathbb{C}\)
algebraic over \(\mathbb{Q}\). It is a field, as the last section shows, and it is algebraic
over \(\mathbb{Q}\) by construction, yet \([\overline{\mathbb{Q}} : \mathbb{Q}] = \infty\).
The elements \(\sqrt[n]{2}\) for \(n = 1, 2, 3, \ldots\) generate subfields of degrees
\(1, 2, 3, \ldots\), since each \(x^n - 2\) is irreducible over \(\mathbb{Q}\) by
Eisenstein's Criterion
with \(p = 2\). No finite \(\mathbb{Q}\)-basis can therefore exist.
When building complex architectures, we need to know how the total dimensionality behaves. Stacking layers in
a deep neural network and composing transformations in a robotic arm are two such cases. In field theory, we
stack extensions. The Tower Rule tells us exactly how these discrete dimensions multiply.
Theorem: Tower Rule
If \(K\) is a finite extension of the field \(E\), and \(E\) is a finite extension of the field \(F\), then
\(K\) is a finite extension of \(F\), and their degrees multiply:
\[
[K:F] = [K:E][E:F].
\]
Proof:
Let \(X = \{x_1, x_2, \ldots, x_n\}\) be a basis for \(K\) over \(E\), and
let \(Y = \{y_1, y_2, \ldots, y_m\}\) be a basis for \(E\) over \(F\).
We claim that the set
\[
YX = \{y_j x_i \mid 1 \leq j \leq m, \, 1 \leq i \leq n\}
\]
forms a basis for \(K\) over \(F\). To prove this, we must show that \(YX\) spans \(K\) over \(F\), and that
\(YX\) is linearly independent over \(F\).
First, let \(a \in K\). Then there are elements \(b_1, b_2, \ldots, b_n \in E\) such that
\[
a = b_1 x_1 + b_2 x_2 + \cdots + b_n x_n
\]
and, for each \(i = 1, 2, \ldots, n\), there are elements
\(c_{i1}, c_{i2}, \ldots, c_{im} \in F\) such that
\[
b_i = c_{i1} y_{1} + c_{i2} y_2 + \cdots + c_{im} y_{m}.
\]
Thus,
\[
\begin{align*}
a &= \sum_{i=1}^n b_i x_i \\\\
&= \sum_{i=1}^n \left(\sum_{j=1}^m c_{ij} y_j \right) x_i \\\\
&= \sum_{i, j} c_{ij} (y_j x_i).
\end{align*}
\]
This shows that \(YX\) spans \(K\) over \(F\).
Next, suppose there are elements \(c_{ij} \in F\) such that
\[
\begin{align*}
0 &= \sum_{i, j} c_{ij} (y_j x_i) \\\\
&= \sum_i \left(\sum_j (c_{ij} y_j)\right) x_i.
\end{align*}
\]
Then, since each \(\sum_j c_{ij} y_j \in E\) and \(X\) is a basis for \(K\) over \(E\), for each
\(i\), we have
\[
\sum_j c_{ij} y_j = 0.
\]
As each \(c_{ij}\) lies in \(F\) and \(Y\) is a basis for \(E\) over \(F\), each \(c_{ij} = 0\). Therefore,
the set \(YX\) is linearly independent over \(F\).
The \(mn\) products \(y_j x_i\) are pairwise distinct. If \(y_{j_1} x_{i_1} = y_{j_2} x_{i_2}\) with
\((j_1, i_1) \neq (j_2, i_2)\), the difference would give a nontrivial linear dependence among the elements
of \(YX\), contradicting the independence just proved. Therefore \(YX\) is a basis for \(K\) over \(F\) of
cardinality \(mn\), and we conclude
\[
[K:F] = mn = [K:E][E:F].
\]
If we extend a field with multiple algebraic elements, say \(F(a, b)\), do we really need to track both
generators? In characteristic \(0\), no. The Primitive Element Theorem acts as a form of
generator compression. The degree of the extension is unchanged, but a single element suffices to generate it.
Theorem: Primitive Element Theorem
If \(F\) is a field of characteristic \(0\), and \(a\) and \(b\) are algebraic over \(F\), then
there exists a single element \(c \in F(a, b)\) such that \(F(a, b) = F(c)\).
Proof Sketch:
Let \(p(x)\) and \(q(x)\) be the minimal polynomials of \(a\) and \(b\) over \(F\), with zeros
\(a_1 = a, a_2, \ldots, a_m\) and \(b_1 = b, b_2, \ldots, b_n\) in some extension \(K\) of \(F\).
Minimal polynomials are irreducible, so by
No Repeated Zeros in Characteristic \(0\)
neither \(p(x)\) nor \(q(x)\) has a repeated zero in \(K\). Hence the \(a_i\) are distinct and so are the
\(b_j\). Since \(F\) has characteristic \(0\), \(F\) is
infinite, so we can choose \(d \in F\) avoiding the finitely many values \((a_i - a)/(b - b_j)\) for
\(i \geq 1, j \gt 1\). Set \(c = a + db\). The choice of \(d\) ensures that \(a_i \neq a + d(b - b_j)\) for
all \(i \geq 1, j \gt 1\).
We claim \(F(a, b) = F(c)\). The inclusion \(F(c) \subseteq F(a, b)\) is immediate from \(c = a + db\). For
the reverse, it suffices to show \(b \in F(c)\), since then \(a = c - db \in F(c)\) as well.
Consider the polynomials \(q(x)\) and \(r(x) = p(c - dx)\) over \(F(c)\). Both vanish at \(x = b\), so
their gcd in \(F(c)[x]\) is a polynomial of positive degree dividing \(q(x)\). The choice of \(d\) forces
\(b\) to be the only common root of \(q\) and \(r\) in \(K\), so \(\gcd(q, r) = x - b\) up to a unit.
The Euclidean algorithm computes the monic gcd of \(q\) and \(r\) from their coefficients, so this gcd
\(x - b\) lies in \(F(c)[x]\). We conclude \(b \in F(c)\), and therefore \(F(a, b) \subseteq F(c)\).
Example: Generator Compression and Data Packing
To see how the Primitive Element Theorem works as "algebraic data compression," we look at a concrete
numerical example. Consider the field \(E = \mathbb{Q}(\sqrt{2}, \sqrt{3})\), which is formed by adjoining
both \(\sqrt{2}\) and \(\sqrt{3}\) to the rational numbers \(\mathbb{Q}\).
-
Multidimensional Structure:
Since \(\sqrt{3} \notin \mathbb{Q}(\sqrt{2})\), the second
step below has degree \(2\), and the Tower Rule applied to
\(\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2}, \sqrt{3})\) gives
\[
\begin{align*}
[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]
&= [\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] \cdot [\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] \\\\
&= 2 \cdot 2 = 4.
\end{align*}
\]
A basis is \(\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}\), so the field behaves like a 4-dimensional vector
space over \(\mathbb{Q}\).
-
Compression via a "Primitive Element":
We do not need to keep track of two separate generators. We can define a single, combined element
\(c = \sqrt{2} + \sqrt{3}\). This single element is sufficient to generate the entire field:
\[
\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3}).
\]
-
Decoding the Information:
We can see this as a "lossless" encoding by showing that the original generators can be recovered
entirely through polynomial operations on \(c\). We calculate the cube of \(c\):
\[
\begin{align*}
c^3 &= (\sqrt{2} + \sqrt{3})^3 \\\\
&= (\sqrt{2})^3 + 3(\sqrt{2})^2\sqrt{3} + 3\sqrt{2}(\sqrt{3})^2 + (\sqrt{3})^3 \\\\
&= 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3} \\\\
&= 11\sqrt{2} + 9\sqrt{3}.
\end{align*}
\]
Now, if we subtract \(9c\) from \(c^3\), the \(\sqrt{3}\) terms cancel:
\[
c^3 - 9c = (11\sqrt{2} + 9\sqrt{3}) - 9(\sqrt{2} + \sqrt{3}) = 2\sqrt{2}.
\]
Therefore, we can perfectly decode \(\sqrt{2}\) using the formula \(\sqrt{2} = \frac{1}{2}(c^3 - 9c)\).
Once we have \(\sqrt{2}\), we easily get \(\sqrt{3} = c - \sqrt{2}\).
The CS Perspective:
This is analogous to lossless data packing. Instead of maintaining two separate variables,
or two separate memory addresses for \(\sqrt{2}\) and \(\sqrt{3}\), we apply a specific weighting and
sum them into a single variable \(c\). No structural information is lost. The Primitive Element Theorem
guarantees that such a lossless flattening is always possible in characteristic \(0\) for two algebraic
generators, and hence, by repeating the theorem, for any finite number of them.
Properties of Algebraic Extensions
In algorithm design and mathematical modeling, closure is a critical property. It ensures that an operation
applied to valid inputs always produces a valid output. Before we can transition to studying smooth manifolds and continuous
groups, we must guarantee that our algebraic building blocks are perfectly stable.
Theorem: Subfield of Algebraic Elements
Let \(E\) be an extension field of the field \(F\). Then the set of all elements in \(E\) that are
algebraic over \(F\) is a
subfield of \(E\).
This means that the sum, difference, product, and quotient (by a nonzero element) of two algebraic numbers
are again algebraic numbers. The theorem after this one shows that the closure applies not just to
elements within a single extension, but to the extensions themselves.
Proof:
Let \(a, b \in E\) be algebraic over \(F\) with \(b \neq 0\). The set of algebraic elements contains
\(F\), since each \(f \in F\) is a zero of \(x - f\), so it is nonempty and contains \(1\). By the
Subring Test it therefore
suffices to show that \(a+b\), \(a-b\), \(ab\), and \(a/b\) are algebraic over \(F\). For this it suffices
to show that the extension degree \([F(a, b):F]\) is finite.
Since \(b\) is algebraic over \(F\), the
simple extension \(F(b)\) has finite degree over \(F\).
By the
Characterization of Extensions and the
structure theorem for \(F(b)\),
\([F(b) : F]\) equals the degree of the minimal polynomial of \(b\).
Since \(a\) is algebraic over \(F\), it satisfies the same nonzero polynomial when viewed over the larger field \(F(b)\),
so \(a\) is algebraic over \(F(b)\) as well. Applying the same argument to the simple extension \(F(b)(a) = F(a, b)\),
we see that \([F(a, b) : F(b)]\) is finite.
Here, by the Tower Rule, we know that
\[
[F(a, b):F] = [F(a, b):F(b)][F(b):F].
\]
Since both terms on the right are finite, \([F(a,b) : F]\) is finite. Because Finite Implies Algebraic,
every element inside \(F(a,b)\) must be algebraic over \(F\).
Since field operations are closed, \(a+b\), \(a-b\), \(ab\), and \(a/b\) are all in \(F(a,b)\). Therefore, they are all
guaranteed to be algebraic over \(F\). The algebraic elements form a subfield of \(E\).
Theorem: Algebraic over Algebraic is Algebraic
If \(K\) is an algebraic extension of \(E\), and \(E\) is an algebraic extension of \(F\), then
\(K\) is an algebraic extension of \(F\).
Proof:
Let \(\alpha \in K\). Since \(K\) is algebraic over \(E\), \(\alpha\) satisfies a nonzero polynomial
\(f(x) = e_n x^n + e_{n-1} x^{n-1} + \cdots + e_0\) with coefficients \(e_0, e_1, \ldots, e_n \in E\).
Consider the intermediate field
\[
E_0 = F(e_0, e_1, \ldots, e_n).
\]
Each \(e_i\) is algebraic over \(F\) (because \(E\) is algebraic over \(F\)), hence over every intermediate
field containing \(F\). Adjoining the \(e_i\) one at a time therefore produces simple algebraic extensions,
each of finite degree as in the previous proof, so the chain
\[
F \subset F(e_0) \subset F(e_0, e_1) \subset \cdots \subset F(e_0, e_1, \ldots, e_n) = E_0
\]
consists of finite extensions. Hence \([E_0 : F]\) is finite by repeated use of the
Tower Rule.
Now \(\alpha\) is algebraic over \(E_0\), since \(f(x)\) lies in \(E_0[x]\) by construction. Consequently
\([E_0(\alpha) : E_0]\) is also finite. Applying the Tower Rule, we obtain
\[
[E_0(\alpha) : F] = [E_0(\alpha) : E_0] \cdot [E_0 : F] \lt \infty.
\]
Since \(\alpha \in E_0(\alpha)\) and \([E_0(\alpha) : F]\) is finite,
Finite Implies Algebraic yields
that \(\alpha\) is algebraic over \(F\). As \(\alpha \in K\) was arbitrary, \(K\) is algebraic over \(F\).
These theorems prove that the "algebraic universe" is closed. We can stack algebraic extensions on top of
each other safely without ever producing a transcendental element, and when finitely many layers of finite
degree are stacked, the Tower Rule keeps the total degree finite. With this secure, discrete foundation of
dimensions and extensions established, we are ready to transition to continuous
geometry. The path starts with the discrete symmetries of polygons and leads
to the continuous Lie groups of 3D space, \(SO(3)\) and \(SE(3)\).