Field Extensions and Linear Algebra
I hope to show the relationship between field extensions and linear algebra, and how a proof using the latter can extend to a useful property of the former.
Extension Fields are Just Vector Spaces
An extension field of a field $F$ is any field that contains $F$ as a subfield. There are many ways to extend a field. If you take an element $u \notin F$, and then let $u$ be in $F$, then it extends $F$. We denote this $F(u)$, and this is a field extension. Since we only added a single element, this is called a simple extension.
Turns out there is a fundamental connection between field extensions, and linear algebra, and can be used to prove some interesting theorems. Most importantly, If $F$ is a field extension of $K$ then $F$ is simply a vector space over $K$. Elements of $K$ are scalars, and elements of $F$ are vectors.
D(imensions)egrees
We can also recontextualize dimension in terms of our fields, since sticking with linear algebra terms can get confusing when dealing with groups because of the different contexts (sorry Riju). The degree of $F$ as a vector space over $K$ is denoted $[F:K]$. This has the exact same meaning as dimension, but can be used to come to different conclusions from the eyes of groups.
Theorem 1: Multiplicativity Formula for Degrees
Let $E$ be an field extension of $K$ and $F$ be a field extension of $E$. Then, \([F:K] = [F:E][E:K]\)
The real interesting part of this for me (and why I’m writing this in the first place) is the fact that the proof uses basic concepts from linear algebra to prove this.
Proof. Since we are dealing with field extensions, $F$ is just a vector space over $E$, and $E$ is just a vector space over $K$. Therefore we can come up with a basis ${u_1, u_2, \dots, u_n}$ for $F$ over $E$ (assuming $[F:E]=n$), and the basis ${v_1,v_2,\dots,v_m}$ for $E$ over $K$ (assuming $[E:K]=m$).
I claim that the set of pairwise products, $u_iv_j$, which I will call the set $\mathcal{B}$ is the basis for the vector space of $F$ over $K$. We can do this by showing that $\mathcal{B}$ both spans $F$ over $K$ and each basis is linearly independent. The number of ways to combine $u$’s and $v$’s is $nm$, which would complete the proof.
To show that $\mathcal{B}$ spans $F$ over $K$, note that for any element $u$ in $F$, we can have
\[u= \sum_{i=1}^{n}a_iu_i,\]where each $u_i$ is the basis for $F$ over $E$, and each $a_i$ is in $E$, since $F$ is a vector space over $E$. Similarly, for each element $a_i$ in $E$, we can write it as
\[a_i = \sum_{j=1}^mc_{ij}v_j,\]where each $v_j$ is the basis for $E$ over $K$, and each $c_{ij}$ is in $K$. Well then we can combine these sums to get
\[u = \sum_{i=1}^n\sum_{j=1}^mc_{ij}v_ju_i,\]showing us that any $u$ in $F$ can be found as the sum of its basis vectors times $c_{ij}$ in $K$.
The next step is to show that $\mathcal{B}$ is linearly independent. Recall that a set is linearly independent if any linear combination of the set that equals $0$ implies that all coefficients are $0$ as well. Suppose $ \sum_{i,j}c_{ij}v_ju_i=0$ for a linear combination of the set $\mathcal{B}$. This is the same as
\[0 =\sum_{i=1}^n\left(\sum_{j=1}^mc_{ij}v_j\right)u_i.\]Notice that the inner “$j$” sum is a coefficient for the $u_i$ basis. Now since each $u_i$ form a basis for $F$ over $E$, each of its coefficients $\sum_{j=1}^mc_{ij}v_j$ must be $0$. Similarly, each $v_j$ form a basis for $E$ over $K$, so each coefficient $c_{ij}$ must also be $0$. This completes the proof, because we have shown that a linear combination of $\mathcal{B}$ that equals $0$ must have all coefficients $c_{ij}$ equal $0$. $\blacksquare$
Now what use does this linear algebra proof have beyond linear algebra itself? Here’s a useful example for how we can use it to determine the degree of one field extension over the original field.
Example 2: $[\mathbb{Q}(\sqrt3,\sqrt2):\mathbb{Q}]$
From Theorem 1, we know that
\[[\mathbb{Q}(\sqrt3,\sqrt2):\mathbb{Q}] = [\mathbb{Q}(\sqrt3,\sqrt2):\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2):\mathbb{Q}].\]We know that $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$, because the minimal polynomial for $\sqrt2$ in $\mathbb{Q}$ is $p(x)=x^2-2$. Moreover, we can show that the minimal polynomial for $\sqrt3$ in $\mathbb{Q}(\sqrt2)$ is $x^2-3$. Therefore $[\mathbb{Q}(\sqrt3,\sqrt2):\mathbb{Q}(\sqrt2)]=2$. So
\[[\mathbb{Q}(\sqrt3,\sqrt2):\mathbb{Q}] = 2\cdot 2= \boxed{4}.\]What’s more, we can use this to find the basis for $\mathbb{Q}(\sqrt3,\sqrt2)$ over $\mathbb{Q}$, it is a field extension and therefore a vector space.
The basis $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$ is ${1,\sqrt2}$, and the basis for $\mathbb{Q}(\sqrt3,\sqrt2)$ over $\mathbb{Q}(\sqrt2)$ is ${1,\sqrt{3}}$, so taking all possible products, the basis is
\[\{ 1, \sqrt2, \sqrt3, \sqrt6\},\]perfectly matching up with our degree, $4$.
The idea of a basis, as shown by the last example, is very different from the basis used in a typical introduction class for Linear Algebra. We don’t use vectors as a basis, instead we just use numbers. It becomes much more abstract it that way.
It should be clear that this theorem is very useful for understanding how field extensions interact, and allow us to learn more about the degrees of certain field extensions. I hope to show the further use of this theorem in future posts.
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