The Minimal Polynomial

The minimal polynomial is the computational backbone of algebraic number theory: the minimal polynomial of an algebraic number determines its ring of integers' local behavior, ramification of primes, and discriminants. In computer algebra systems (Magma, Sage, Mathematica), computing minimal polynomials of algebraic numbers or matrices is a core routine, underlying algorithms for simplifying nested radicals and computing Galois groups. In linear algebra, the minimal polynomial of a matrix $A$ over a field $F$ (the monic generator of the ideal of polynomials annihilating $A$) is the direct linear-algebraic analogue of this construction, controlling diagonalizability and Jordan form — the same PID argument (over $F[x]$ acting on the vector space via $A$) proves its existence and uniqueness. In cryptography, finite fields $\mathbb{F}_{p^n} \cong \mathbb{F}_p[x]/(f(x))$ for irreducible $f$ of degree $n$ are constructed by choosing minimal polynomials of primitive elements; this underlies AES's finite-field arithmetic and elliptic curve cryptography over binary fields. In coding theory, BCH and Reed–Solomon codes are built using minimal polynomials of roots of unity over $\mathbb{F}_2$ or $\mathbb{F}_p$. In algebraic geometry and Galois theory proper, the minimal polynomial's other roots (its "conjugates") are exactly the images of $\alpha$ under embeddings of $F(\alpha)$, the starting point for defining Galois groups and studying Simple Extensions and F(alpha).

:::quiz[Which property of $F[x]$ is essential to guarantee that the ideal $I_\alpha$ is generated by a single polynomial?]

• $F[x]$ is a field
• $F[x]$ is a Euclidean domain (hence a PID) ✓
• $F[x]$ is algebraically closed
• $F[x]$ is a finite ring

:::quiz[If $\alpha = \sqrt[3]{2}+1$ has minimal polynomial $x^3 - 3x^2+3x-3$ over $\mathbb{Q}$, what is $[\mathbb{Q}(\alpha):\mathbb{Q}]$?]

• 1
• 2
• 3 ✓
• 6

:::faq[What is the relationship between the minimal polynomial and $[F(\alpha):F]$?] They are numerically identical: $\deg m_\alpha(x) = [F(\alpha):F]$. This is because $F(\alpha) \cong F[x]/(m_\alpha(x))$ as $F$-vector spaces (via the evaluation isomorphism), and the quotient by a degree-$n$ polynomial has dimension exactly $n$ over $F$, with basis given by the images of $1, x, \ldots, x^{n-1}$. This single equality is what makes the minimal polynomial the primary tool for computing degrees of simple extensions.

Ask the tutor: How does the minimal polynomial of $\alpha$ change when you pass from $F$ to a larger base field, and what does that reveal about $[F'(\alpha):F']$ versus $[F(\alpha):F]$?

Think of $F[x]$ as a warehouse of every conceivable polynomial relation you could impose on a symbol $x$. Plugging in a specific algebraic element $\alpha$ collapses this warehouse: the evaluation map $\operatorname{ev}_\alpha$ sends each polynomial $f$ to the number $f(\alpha) \in E$. Some polynomials survive as nonzero values, but a whole ideal of them collapse to zero — these are exactly the polynomial identities that $\alpha$ satisfies. Because $F[x]$ behaves like the integers (a Euclidean domain with division with remainder), every ideal is generated by one element, the way every subgroup of $\mathbb{Z}$ is generated by one integer. The minimal polynomial is that one generator, normalized to be monic so it is unambiguous — exactly as we normalize the generator of an ideal in $\mathbb{Z}$ to be the positive generator. Irreducibility of $m_\alpha$ is not an extra assumption; it is forced, because a nontrivial factorization of $m_\alpha$ would produce a smaller-degree monic polynomial killing $\alpha$, since $E$ has no zero divisors. The picture below shows the ideal $I_\alpha \subset F[x]$ as a "sieve": polynomials pass through as multiples of $m_\alpha$, and everything that survives evaluation at $\alpha$ nonzero sits outside the ideal.

:::svg[The kernel ideal $I_\alpha$ inside $F[x]$, generated by $m_\alpha$]

Example 1: $\alpha = \sqrt[3]{2} + 1$ over $\mathbb{Q}$. Let $\beta = \sqrt[3]{2}$, so $\alpha = \beta + 1$, i.e. $\beta = \alpha - 1$. We know $\beta^3 = 2$. Substituting:

Expand: $(\alpha-1)^3 = \alpha^3 - 3\alpha^2 + 3\alpha - 1$. So

Let $f(x) = x^3 - 3x^2 + 3x - 3 \in \mathbb{Q}[x]$; this is monic and $f(\alpha) = 0$. We claim $f$ is irreducible over $\mathbb{Q}$, which by the theorem's uniqueness statement will identify $f = m_\alpha$. By the Rational Root Theorem, any rational root $p/q$ (in lowest terms) of $f$ must have $p \mid 3$ and $q \mid 1$, so candidates are $\pm 1, \pm 3$. Check:

• $f(1) = 1 - 3 + 3 - 3 = -2 \neq 0$
• $f(-1) = -1 - 3 - 3 - 3 = -10 \neq 0$
• $f(3) = 27 - 27 + 9 - 3 = 6 \neq 0$
• $f(-3) = -27 - 27 - 9 - 3 = -66 \neq 0$

No rational roots, so $f$ has no linear factor over $\mathbb{Q}$. Since $\deg f = 3$, the only possible nontrivial factorization is (linear)(quadratic); the absence of a rational root rules this out. Hence $f$ is irreducible over $\mathbb{Q}$. By the theorem, $m_\alpha(x) = x^3 - 3x^2 + 3x - 3$, and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, with $\mathbb{Q}$-basis $\{1, \alpha, \alpha^2\}$ for $\mathbb{Q}(\alpha)$. (This also confirms $\mathbb{Q}(\sqrt[3]{2}+1) = \mathbb{Q}(\sqrt[3]{2})$, as expected since $\beta = \alpha - 1 \in \mathbb{Q}(\alpha)$.)

Example 2: $\alpha = \sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$, verified by the divisibility criterion. Compute powers of $\alpha$:

Square again:

So $f(x) = x^4 - 10x^2 + 1$ is monic in $\mathbb{Q}[x]$ with $f(\alpha) = 0$.

Claim: $f$ is irreducible over $\mathbb{Q}$. First, rational roots: by the Rational Root Theorem candidates are $\pm 1$; $f(1) = 1 - 10 + 1 = -8 \neq 0$, $f(-1) = -8 \neq 0$. No linear factor. Now rule out factorization into two monic quadratics over $\mathbb{Q}$ (the only remaining possibility since $f$ has no rational root and degree 4). Suppose

using that the $x^3$-coefficients cancel (0 total, forcing the linear terms to be negatives of each other, as one checks by matching the general product $(x^2+ax+b)(x^2+cx+d)$'s $x^3$-coefficient $a+c=0$). Expanding:

More carefully, expand fully:

Matching coefficients with $x^4 + 0\cdot x^3 -10x^2 + 0\cdot x + 1$:

Case $a = 0$: then $b + c = -10$ and $bc = 1$, so $b,c$ are roots of $t^2 + 10t + 1 = 0$, giving $t = -5 \pm \sqrt{24} = -5 \pm 2\sqrt6$, irrational. So $b, c \notin \mathbb{Q}$ — contradiction (we need integer/rational $b,c$).

Case $c = b$: then $b^2 = 1$ so $b = c = \pm 1$.

• If $b = c = 1$: $2 - a^2 = -10 \Rightarrow a^2 = 12$, not a perfect square of a rational (since $12$ is not a rational square), no rational $a$.
• If $b = c = -1$: $-2 - a^2 = -10 \Rightarrow a^2 = 8$, again not a rational square.

All cases fail to produce rational $a,b,c$, so $f$ has no factorization into two rational monic quadratics, and none into a rational linear times a cubic (checked above). Hence $f(x) = x^4 - 10x^2+1$ is irreducible over $\mathbb{Q}$, and by the theorem's uniqueness,

As a sanity check via the divisibility criterion (part 3 of the theorem): the obvious polynomial $g(x) = (x^2-5)^2 - 24 = x^4 - 10x^2+1$ that we derived is $f$ itself here, so divisibility is immediate; but if someone hands us a different annihilating polynomial, say $h(x) = x^8 - 20x^6 + 98x^4 - 40x^2 + 1$ (which one can check also vanishes at $\alpha$, e.g. by squaring $f(\alpha)=0$ combined with extra terms), the theorem guarantees $m_\alpha(x) \mid h(x)$ exactly — this is what makes minimal polynomials useful as a test, not just a description.