PLC Control Logic Extraction

Example: Consider the irreducible polynomial $f(x) = x^2 + 1$ with coefficients in $\mathbb{Q}$. One of its roots is $i = \sqrt{-1}$. Even though $i \notin \mathbb{Q}$, we can find a bigger field which contains it, namely

$$ \mathbb{Q}(i) = {a + bi : a,b \in \mathbb{Q}}. $$

($\mathbb{Q}(i)$ is “bigger” since $\mathbb{Q} \subset \mathbb{Q}(i)$.)

In the example above, we say $\mathbb{Q}(i)$ is an extension field of $\mathbb{Q}$. And in general, any field $K$ which contains a smaller field $F$ is called an extension of $F$, and $F$ is referred to as the base (or ground) field. But so far we’ve only talked about a single root of our polynomial $f(x)$. What about all of its roots? Can we find an extension of $F$ which contains all of the roots of a general $f(x)$? Again the answer is yes, and that field is called the splitting field of $f(x)$. (Technically, the splitting field is the smallest extension of $F$ which contains all of the roots of $f(x)$.)

Aside: We could also turn this question on its head. Suppose we have a field $F$ and an extension $K$, and we pick a random element^** $\alpha \in K$. Can we find a polynomial in $F[x]$ which has $\alpha$ as a root? This time, the answer is not always yes! But in the cases when it is, we say $\alpha$ is an algebraic element. Moreover, if the answer is yes for every element of $K$, we say $K$ is an algebraic field. As we’ve discussed before, algebraic elements are sort of like limit points in topology/analysis.)

Now the crux of The Field Story is the construction of such splitting fields. This construction is analogous to building a tower from the ground up – one floor at a time. We begin with the ground field $F$, and one by one adjoin to $F$ the roots of $f(x)$ until we obtain the field $K$ (the splitting field) which contains all the roots of $f$. I’m doing a lot of hand-waving here, but we eventually obtain a tower of fields which looks something like this:

$$ F \subseteq F_1 \subseteq F_2 \subseteq \cdots \subseteq F_m := K. $$

where $F_{i+1}$ is bigger than $F_i$ because it contains (at least) one more root of $f(x)$. It’s the structure of this tower of fields which is mirrored in the structure of the Galois group associated with $f(x)$. And that group is our next topic of discussion.

** In a longer essay, this would be a footnote marker attached to the aside. TESTING LATEX KEYBOARD: F ⊆ F₁