Adjoining a number to a field

In the case of $\mathbb{Q}[\sqrt{2}]$, we have not only $\sqrt{2}$ and its multiplicative inverse, but everything necessary to retain closure under the operations. There is a notational point to make here:

  • $F[a]$ is defined to be the set $\{f(a) \ | \ f(x) \in F[x]\}$.
  • $F(a)$ is defined to be the "smallest" extension field of $F$ that contains $a$.

But notice that we are talking about $\mathbb{Q}[\sqrt{2}]$ as a field as in the second point! What gives?


Theorem: When $a$ is algebraic over a field $F$, then $F[a] = F(a)$.

Proof:

Since $F[a]$ is a ring, most field properties already hold. What is left is to demonstrate the existence of multiplicative inverses. To do this, we take advantage of the Euclidean algorithm:

Let $f(x) \in F[x]$ be the minimal polynomial for $a$. Every polynomial without $a$ as a root will correspond to a nonzero element in $F[a]$, and moreover, every such polynomial will be relatively prime to $f(x)$. That is, given such a $g(x)$, then there exists polynomials $h(x)$ and $k(x)$ such that:

$$f(x)h(x) + g(x)k(x) = 1$$

Since $a$ is a root of $f(x)$, evaluating the above at $a$ gives:

$$g(a)k(a) = 1$$

So given any nonzero $g(a) \in F[a]$, there exists some $k(a)$ that serves as its multiplicative inverse. This is to say: every nonzero element in $F[a]$ has a multiplicative inverse. We can conclude that, if $a$ is algebraic over $F$, then $F[a]$ is a field and $F[a] = F(a)$.


Final comments:

What makes an algebraic adjunction to a field special? Unlike transcendental adjunctions, algebraic ones are finite. That is, if $a$ is algebraic over $F$, then $F[a]$ can be viewed as a vector space over $F$ spanned by finitely many basis "vectors".

For example, $\mathbb{Q}[\sqrt{2}]$ is a finite extension of degree $2$, meaning any basis contains $2$ basis vectors. One possible basis is $\{1, \sqrt{2}\}$, and so $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.