*(This is part of my series *Field theory for high-schoolers*.)*

This post describes how to construct simple algebraic extensions of some field $F$, by adjoining to it an element of a known extension $E$ of $F$ that is algebraic over $F$. It is also shown that $\mathbf{C}$ can be constructed from $\mathbf{R}$ in that way.

Let $F$ be a field, and $f(x)$ be a non-constant polynomial with coefficients in $F$. Kronecker’s theorem states that there must exist some extension $E$ of $F$ and some element $\alpha$ of $E$ such that $\alpha$ is a root of $f(x)$. Thus $\alpha$ is algebraic over $F$, and there is an algebraic extension $F(\alpha)$ of $F$, which is the smallest field containing $F$ and $\alpha$.

Let $n$ be the degree of $\alpha$ over $F$. It can be shown that the elements of $F(\alpha)$ are precisely the elements

\[ c_0 + c_1\alpha + c_2\alpha^2 + \dots + c_{n-1}\alpha^{n-1} \]

with all the coefficients $c_i$ being in $F$. Indeed, it is tedious but straightforward to verify that those elements form a field, and it is readily seen that any field containing both $F$ and $\alpha$ must contain all the elements of the above form.

If the degree of $\alpha$ over $F$ is $n$, we say that $F(\alpha)$ is an *algebraic extension of $F$ of degree $n$*, and we note $[F(\alpha):F] = n$.

*Examples:*

- We have seen that the degree of $i$ over $\mathbf{R}$ is $2$, and so $\mathbf{R}(i)$ is the set of all elements of the form $c_0 + c_1i$ with $c_0,c_1 \in \mathbf{R}$. This is the definition of $\mathbf{C}$ that we have always worked with, so we see that $\mathbf{R}(i) = \mathbf{C}$, and that $[\mathbf{C}:\mathbf{R}] = 2$.
- The degree of $\sqrt[3]{2}$ over $\mathbf{Q}$ is $3$, since it is a root of the irreducible polynomial $x^3-2$. Thus $\mathbf{Q}(\sqrt[3]{2})$ is the set of all elements of the form $c_0 + c_1\sqrt[3]{2} + c_2(\sqrt[3]{2})^2$, and we have $[\mathbf{Q}(\sqrt[3]{2}):\mathbf{Q}] = 3$.
- The degree of $\pi$ over $\mathbf{R}$ is $1$, so the elements of $\mathbf{R}(\pi)$ are of the form $c_0$ with $c_0 \in \mathbf{R}$. This is exactly the elements of $\mathbf{R}$, so we see that $\mathbf{R}(\pi) = \mathbf{R}$. In general, if $\alpha \in F$, then $F(\alpha) = F$.

**Next:** Summary of the construction of **C**

**Questions:**

- Describe the elements of $\mathbf{R}(i+\sqrt{2})$. Do we have $\mathbf{R}(i+\sqrt{2}) = \mathbf{C}$?
- Describe the elements of $\mathbf{Q}(\sqrt{2}+\sqrt{3})$.
- Describe the elements of $\mathbf{Q}(\sqrt{2}\cdot\sqrt{3})$.
- Since $\mathbf{Q}(\sqrt{2})$ is a field, we can adjoin to it the element $\sqrt{3}$, to form $(\mathbf{Q}(\sqrt{2}))(\sqrt{3})$, which is noted simply $\mathbf{Q}(\sqrt{2},\sqrt{3})$. Describe the elements of $\mathbf{Q}(\sqrt{2},\sqrt{3})$.
- Do we have $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{3},\sqrt{2})$?