Many polynomials have $\sqrt{2}$ as a root: $x^{2} - 2$, $x^{4} - 4$, $x^{3} - 2x$. One of them is special: the smallest, and it tells you the size of $\mathbb{Q}(\sqrt{2})$. This lesson finds it in general and connects it to the degree of an extension.
Definitions. $\alpha$ (in an extension of $F$) is algebraic over $F$ if it is a root of some nonzero polynomial in $F[x]$. Its minimal polynomial is the monic polynomial of least degree with $\alpha$ as a root.
Why it works
Theorem. The minimal polynomial $m(x)$ of $\alpha$ is irreducible, it divides every polynomial in $F[x]$ with root $\alpha$, and $[F(\alpha) : F] = \deg m$.
Proof. Divide any $f$ with $f(\alpha) = 0$ by $m$: $f = qm + r$ with $\deg r \lt \deg m$. Then $r(\alpha) = 0$, so $r = 0$ by minimality: $m \mid f$. If $m = gh$ with smaller degrees, then $g(\alpha)h(\alpha) = 0$ in a field, so one factor has root $\alpha$, contradicting minimality: $m$ is irreducible. Finally, $f \mapsto f(\alpha)$ sends $F[x]$ onto $F(\alpha)$ with kernel $(m)$, so $F(\alpha) \cong F[x]/(m)$, which has degree $\deg m$. $\square$ So: to find $[F(\alpha) : F]$, find any monic polynomial with root $\alpha$ and prove it irreducible.
Proof. Divide any $f$ with $f(\alpha) = 0$ by $m$: $f = qm + r$ with $\deg r \lt \deg m$. Then $r(\alpha) = 0$, so $r = 0$ by minimality: $m \mid f$. If $m = gh$ with smaller degrees, then $g(\alpha)h(\alpha) = 0$ in a field, so one factor has root $\alpha$, contradicting minimality: $m$ is irreducible. Finally, $f \mapsto f(\alpha)$ sends $F[x]$ onto $F(\alpha)$ with kernel $(m)$, so $F(\alpha) \cong F[x]/(m)$, which has degree $\deg m$. $\square$ So: to find $[F(\alpha) : F]$, find any monic polynomial with root $\alpha$ and prove it irreducible.
Worked example: A quadratic number
Find the minimal polynomial of $1 + \sqrt{3}$ over $\mathbb{Q}$.
- $\alpha - 1 = \sqrt{3}$, so $(\alpha - 1)^{2} = 3$, that is, $\alpha^{2} - 2\alpha - 2 = 0$.
- $x^{2} - 2x - 2$ has no rational root ($\alpha$ is irrational), so it is irreducible.
Answer: $x^{2} - 2x - 2$, degree $2$.
Worked example: A sum of square roots
Find the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3}$.
- $\alpha^{2} = 5 + 2\sqrt{6}$, so $(\alpha^{2} - 5)^{2} = 24$, that is, $\alpha^{4} - 10\alpha^{2} + 1 = 0$.
- It is irreducible: $\mathbb{Q}(\alpha)$ contains $\alpha^{2} = 5 + 2\sqrt{6}$, and also $\sqrt{2} = \frac{\alpha^{2} - 1}{2\alpha}$, so it is all of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$, which has degree $4$ (next lesson). A minimal polynomial of degree less than $4$ is impossible.
Answer: $x^{4} - 10x^{2} + 1$.
Worked example: A cube root
Find the minimal polynomial of $\sqrt[3]{2} + 1$.
- $(\alpha - 1)^{3} = 2$: $\alpha^{3} - 3\alpha^{2} + 3\alpha - 3 = 0$.
- It is irreducible: Eisenstein at $3$ applies ($3$ divides $-3, 3, -3$, not $1$, and $9 \nmid 3$).
Answer: $x^{3} - 3x^{2} + 3x - 3$.
The minimal polynomial depends on the base field. Over $\mathbb{Q}$, $\sqrt{2}$ has minimal polynomial $x^{2} - 2$; over $\mathbb{R}$, or over $\mathbb{Q}(\sqrt{2})$ itself, it is $x - \sqrt{2}$, of degree $1$.
Common mistake
Giving a polynomial with root $\alpha$ that factors. $x^{4} - 4$ has root $\sqrt{2}$, but $x^{4} - 4 = (x^{2} - 2)(x^{2} + 2)$, and the minimal polynomial is the factor $x^{2} - 2$. Before answering, check that your polynomial is monic and irreducible; its degree is then $[F(\alpha) : F]$.
Key idea
The minimal polynomial of $\alpha$ is the monic irreducible polynomial with root $\alpha$; it divides every polynomial with root $\alpha$, and $F(\alpha) \cong F[x]/(m)$, so $[F(\alpha) : F] = \deg m$.