The Minimal Polynomial of an Algebraic Element in a Field Extension

# The Minimal Polynomial of an Algebraic Element in a Field Extension

Recall from the Algebraic and Transcendental Elements in a Field Extension page that if $(F, +, *)$ is a field and $(K, +, *)$ is a field extension then an element $u \in K$ is said to be algebraic over $F$ if there exists a nonzero polynomial $f \in F[x]$ such that:

(1)
\begin{align} \quad f(u) = 0 \end{align}

And if no such nonzero polynomial exists, then $u$ is said to be transcendental over $F$.

We will now look at some basic results regarding algebraic and transcendental elements in a field extension.

 Theorem 1: Let $(F, +, *)$ be a field and let $(K, +, *)$ be a field extension. If $u \in K$ is algebraic over $F$ then there exists a unique irreducible monic polynomial $p \in F[x]$ such that $p(u) = 0$, and furthermore, if $f \in F[x]$ is such that $f(u) = 0$ then $p | f$.
• Proof: Let $u \in K$ be algebraic over $F$. Consider the set of all polynomials with coefficients in $F$ that have $u$ as a root, and let this set be denoted by $I$:
(2)
\begin{align} \quad I = \{ f \in F[x] : f(u) = 0 \} \end{align}
• Let $f, g \in I$. Then:
(3)
\begin{align} \quad (f + g)(u) = f(u) + g(u) = 0 \end{align}
• So $I$ is closed under $+$. Moreover, if $f \in I$ then $-f(u) = -0 = 0$, so $-f \in I$. Lastly, $0 \in I$. Hence $(I, +)$ is a subgroup of $(F[x], +)$.
• Now let $f \in I$ and let $g \in F[x]$. Then:
(4)
\begin{align} \quad (fg)(u) = f(u)g(u) = 0g(u) = 0 \end{align}
• The polynomial $p$ has minimal degree in $I$.
• Now let $f, g \in F[x]$ be such that $fg \in I$. Then $f(u)g(u) = 0$. So this implies that either $f(u) = 0$ or $g(u) = 0$. So either $f \in I$ or $g \in I$. So $I$ is a prime ideal. Since $I$ is a prime ideal, its generator $p$ must be irreducible.
• Now let $f \in F[x]$ be such that $f(u) = 0$. Then $f \in I = p(x)F[x]$ and so there exists an element $g \in F[x]$ such that $f = pg$. So $p | f$. $\blacksquare$

The polynomials obtained from the above theorem are given a special name which we define below.

 Theorem 1: Let $(F, +, *)$ be a field and let $(K, +, *)$ be a field extension. If $u \in K$ is algebraic over $F$ then the Minimal Polynomial of $u$ over $F$ is defined to be the unique irreducible monic polynomial $p \in F[x]$ such that $p(u) = 0$. The Degree of $u$ over $F$ is defined to be the degree of $p$.

For example, one again we consider the field $F = \mathbb{Q}$ and the field extension $K = \mathbb{R}$. The element $\sqrt{3} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$ since for the polynomial $f \in \mathbb{Q}[x]$ given by $f(x) = x^3 + x^2 - 3x - 3$ we have that:

(5)
\begin{align} \quad f(\sqrt{3}) = (\sqrt{3})^3 + (\sqrt{3})^2 - 3\sqrt{3} - 3 = 3\sqrt{3} + 3 - 3 \sqrt{3} - 3 = 0 \end{align}

It is very easy to check that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$ is:

(6)
\begin{align} \quad p(x) = x^2 - 3 \end{align}

And so the degree of $\sqrt{3}$ over $\mathbb{Q}$ is $2$ since $\deg p = 2$.