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The monic polynomial of least degree with a given root: why it is irreducible, why it divides everything else with that root, and the basis of powers that makes its degree the degree of the extension.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find a minimal polynomial by eliminating radicals, prove it irreducible by Eisenstein, by reduction or by a degree count, read the degree of a simple extension off it, write powers of the generator in the basis it supplies, recognise $K(\alpha)$ as a quotient of a polynomial ring, and prove that a minimal polynomial is always irreducible.
The last lesson measured every step of a tower by a degree and said the degree comes from a minimal polynomial, without saying what one is. It also relied on $x^{2} - 3$ staying irreducible over $\mathbb{Q}(\sqrt2)$. Both debts are paid here, and the tool is the quotient $K[x]/(f)$ the first course built.
$\alpha$ is algebraic over $K$ when some non-zero polynomial over $K$ has it as a root. Its minimal polynomial $m_\alpha$ is the monic polynomial over $K$ of least degree with $m_\alpha(\alpha) = 0$; monic means leading coefficient $1$, which is what makes it unique. A simple extension $K(\alpha)$ is one generated by a single element.
The definition. Let $\alpha$ be algebraic over $K$. Among all non-zero polynomials over $K$ with $\alpha$ as a root, take one of least degree and scale it to be monic. That is the minimal polynomial $m_\alpha$.
Three properties, each a one-liner.
The structure theorem. If $\deg m_\alpha = n$ then
$$K(\alpha) \cong K[x]/(m_\alpha), \qquad [K(\alpha) : K] = n,$$
with basis $\{1, \alpha, \alpha^{2}, \ldots, \alpha^{n-1}\}$. The isomorphism sends $x$ to $\alpha$; its kernel is exactly $(m_\alpha)$ by property 2, and the quotient is a field because $m_\alpha$ is irreducible and $K[x]$ is a principal ideal domain.
So adjoining a root and quotienting by an irreducible polynomial are the same construction from two sides. Kronecker's observation — that $K[x]/(f)$ contains a root of $f$ for any irreducible $f$ — is what lets a root be manufactured when none is lying around, and it builds the finite fields two lessons from now.
Arithmetic in $K(\alpha)$. Multiply as polynomials, then reduce modulo $m_\alpha$. For $\alpha = \sqrt[3]{2}$, $\alpha^{3} = 2$ turns every power back into the span of $1, \alpha, \alpha^{2}$. It is arithmetic modulo $n$ with a polynomial as the modulus.
It depends on the base field. $m_i = x^{2} + 1$ over $\mathbb{R}$ and $x - i$ over $\mathbb{C}$. Enlarging the base can only lower the degree, since more polynomials become available.
Finding one. Eliminate radicals to get a polynomial, make it monic, then prove irreducibility — by Eisenstein, by reduction modulo a prime, or by the absence of a rational root when the degree is $2$ or $3$. Skipping the last step is the standard error: $x^{4} - 4$ has $\sqrt2$ as a root and is not minimal.
Another way: picture
The minimal polynomial is the single relation $\alpha$ satisfies, from which every other relation follows. Think of $\alpha$ as an unknown that obeys one law: $m_\alpha(\alpha) = 0$. Everything in $K(\alpha)$ is a polynomial in $\alpha$, and that one law is enough to fold every high power back down into the first $n$. The field is the polynomial ring with that one relation imposed, and nothing else.
Another way: steps
To find a minimal polynomial over $\mathbb{Q}$: 1. Set $\alpha$ equal to the expression and clear the radicals by raising to powers and rearranging. 2. Make the result monic. 3. Test irreducibility: Eisenstein at a prime; reduce modulo a small prime; or, in degree $2$ or $3$, check for rational roots. 4. If it factors, keep the factor with $\alpha$ as a root. 5. The degree of the survivor is $[\mathbb{Q}(\alpha) : \mathbb{Q}]$; if that conflicts with a tower computation, the irreducibility check was wrong.
Producing a polynomial with the right root is mechanical. Proving it cannot be factored is not, and there are four standard tools.
Rational root test, degrees $2$ and $3$ only. A cubic or quadratic over $\mathbb{Q}$ factors exactly when it has a rational root, since one factor would have degree $1$. The candidates are $\pm p/q$ with $p$ dividing the constant term and $q$ the leading coefficient. This fails from degree $4$: $x^{4} + 4 = (x^{2} - 2x + 2)(x^{2} + 2x + 2)$ has no rational root and factors.
Eisenstein's criterion. If a prime $p$ divides every coefficient but the leading one, and $p^{2}$ does not divide the constant term, the polynomial is irreducible over $\mathbb{Q}$. $x^{n} - 2$ passes at $p = 2$ for every $n$, which is how $\sqrt[n]{2}$ is shown to have degree exactly $n$.
Reduction modulo a prime. If the reduction modulo $p$ is irreducible and the degree is unchanged, the original is irreducible over $\mathbb{Q}$. $x^{3} - x - 1$ modulo $2$ has no root in $\mathbb{F}_2$ and degree $3$, so it is irreducible there and hence over $\mathbb{Q}$.
The degree argument. Often the cleanest. For $\alpha = \sqrt2 + \sqrt3$, the quartic $x^{4} - 10x^{2} + 1$ has $\alpha$ as a root; and separately $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt3)$ has degree $4$ by the tower law. Since the minimal polynomial has degree $4$ and divides the quartic, it is the quartic — no irreducibility test needed.
That last technique is worth internalising: a degree computed two ways settles irreducibility without any factoring. Note too that the shift is applied to $\Phi_p(x) = x^{p-1} + \cdots + 1$ to make Eisenstein apply — $\Phi_p(x+1)$ passes at $p$ — which is how the cyclotomic polynomials are shown irreducible.
Forgetting irreducibility. Any polynomial with $\alpha$ as a root is not the minimal one. $x^{4} - 4$ is not the minimal polynomial of $\sqrt2$.
Forgetting monic. $2x^{2} - 4$ has the right root and the right degree, and is not minimal. The convention is what makes the answer unique.
Ignoring the base field. The minimal polynomial of $\alpha$ is meaningless without saying over what. Over $\mathbb{C}$ everything has degree $1$.
Using the rational root test above degree three. It only detects linear factors.
Assuming a sum of two degree-$n$ elements has degree $n^{2}$. It divides $n^{2}$; $\sqrt2 + \sqrt8 = 3\sqrt2$ has degree $2$, not $4$.
Confusing the minimal polynomial of an element with the characteristic polynomial of a matrix. They are related — the minimal polynomial of $\alpha$ is the minimal polynomial of multiplication by $\alpha$ as a linear map on $K(\alpha)$ — but the characteristic polynomial has degree $[L:K]$, which may be larger.
Let $\alpha = \sqrt2 + \sqrt3$. Then $\alpha^{2} = 5 + 2\sqrt6$, so $\alpha^{2} - 5 = 2\sqrt6$ and squaring gives $\alpha^{4} - 10\alpha^{2} + 25 = 24$.
Eliminate the radicals.
So $\alpha$ satisfies $x^{4} - 10x^{2} + 1$, which is monic of degree $4$.
A monic candidate.
And $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt3)$ has degree $4$ by the tower law, so the minimal polynomial has degree $4$ and must be this one.
Minimal, by a degree count.
Over $\mathbb{F}_2$, the polynomial $x^{2} + x + 1$ has no root: substituting $0$ gives $1$ and substituting $1$ gives $1$.
Irreducible over $\mathbb{F}_2$.
So $\mathbb{F}_2[x]/(x^{2} + x + 1)$ is a field, and in it the class of $x$ satisfies $x^{2} + x + 1 = 0$.
A root has been manufactured.
The field has four elements $0, 1, \alpha, \alpha + 1$ with $\alpha^{2} = \alpha + 1$ — it is $\mathbb{F}_4$, and this is the only way to build it.
Kronecker's construction.
It satisfies $x^{4} - 2$, which is monic.
A candidate.
Eisenstein at $p = 2$: the prime divides the constant term $-2$ and every other non-leading coefficient (they are zero), and $4$ does not divide $-2$.
Irreducible over $\mathbb{Q}$.
So the minimal polynomial is $x^{4} - 2$ and $[\mathbb{Q}(\sqrt[4]{2}) : \mathbb{Q}] = 4$, with basis $\{1, \sqrt[4]{2}, \sqrt{2}, \sqrt[4]{8}\}$. Note the field contains $\sqrt2$ as the square of the generator — a degree-$2$ subfield, as the tower law allows.
Match each number to its minimal polynomial over $\mathbb{Q}$.
| $x^{2} - 2$ | $x^{3} - 2$ | $x^{2} + 1$ | $x^{4} - 10x^{2} + 1$ | $x^{4} - 5$ | |
|---|---|---|---|---|---|
| $\sqrt2$ | |||||
| $\sqrt[3]{2}$ | |||||
| $i$ | |||||
| $\sqrt2 + \sqrt3$ |
What is $[\mathbb{Q}(a primitive fifth root of unity) : \mathbb{Q}]$?
Answer:
Select every statement that is true of the minimal polynomial of an algebraic element $\alpha$ over a field $K$.
This task has no paper form; do it on a device.
Let $\alpha = \sqrt[3]{2}$, so $\alpha^{3} = 2$ and $\{1, \alpha, \alpha^{2}\}$ is a basis of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$. Write each power in that basis by filling in its three coefficients.
| Coefficient of $1$ | Coefficient of $\alpha$ | Coefficient of $\alpha^{2}$ | |
|---|---|---|---|
| $\alpha^{3}$ | |||
| $\alpha^{4}$ | |||
| $\alpha^{5}$ | |||
| $\alpha^{6}$ |
Put in order the steps of finding the minimal polynomial of an algebraic number over $\mathbb{Q}$.
Number the steps in order (write the number in the box):
Build the proof that the minimal polynomial of an algebraic element is irreducible over the base field.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each number to its minimal polynomial over $\mathbb{Q}$.
| $x^{2} - 2$ | $x^{3} - 2$ | $x^{2} + 1$ | $x^{4} - 10x^{2} + 1$ | $x^{4} - 5$ | |
|---|---|---|---|---|---|
| $\sqrt2$ | |||||
| $\sqrt[3]{2}$ | |||||
| $i$ | |||||
| $\sqrt2 + \sqrt3$ |
You can find the minimal polynomial of an algebraic number and justify that it is minimal. Say in your own words why irreducibility is the step that cannot be skipped. Next: adjoining not one root but all of them.
9. Your turn: the minimal polynomial of $\sqrt[4]{2}$ over $\mathbb{Q}$, step 3