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Over $F_{2}$ , $x^{4} + x^{2} + 1 = (x^{2} + x + 1)^{2}$ has two distinct roots, each repeated, in its four-element splitting field

Example

Over $F_{2}$ , $x^{4} + x^{2} + 1 = (x^{2} + x + 1)^{2} .$ The polynomial $x^{2} + x + 1$ is irreducible. If $u$ is one of its roots, the four-element field $F_{2} (u) = {0, 1, u, u + 1}$ is the splitting field, and the two distinct roots $u$ and $u + 1$ each occur with multiplicity two in $x^{4} + x^{2} + 1$ , meaning that their linear factors have exponent two in its factorisation.

Facts & Assumptions

Given: The polynomial $x^{4} + x^{2} + 1 \in F_{2} [x]$ .

[F1]

The ring $Z / 2$ is the field $F_{2}$ (For every prime $p$ , the two operations on $Z / p$ make it a field).

[F2]

A quadratic over a field is irreducible exactly when it has no root in the field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[F3]

For monic irreducible $p$ of degree $2$ , the quotient $F [x] / (p)$ is a field whose elements have unique form $a + b u$ , and $u$ is a root of $p$ ( $F [x] / (p)$ for monic irreducible $p$ is a field extension containing the root $x + (p)$ with unique reduced representatives).

[F4]

A root $a$ is repeated when $(x - a)^{2}$ divides the polynomial (Repeated roots in extension fields and separable polynomials).

[F5]

A splitting field is generated by the roots of a polynomial that splits there (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[F6]

For every field $K$ , the polynomial ring $K [x]$ is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Verification

technique · direct calculation

1.1

In characteristic $2$ , $(x^{2} + x + 1)^{2} = x^{4} + x^{2} + 1$ . The polynomial $p = x^{2} + x + 1$ takes the value $1$ at both $0$ and $1$ , so [F2] makes it irreducible.

F1F2algebra

1.2

By [F3], adjoining $u = x + (p)$ gives a field with the four distinct elements $0, 1, u, u + 1$ and relation $u^{2} + u + 1 = 0$ . Substituting $u + 1$ into $p$ in characteristic $2$ also gives zero, so $p = (x - u) (x - (u + 1))$ .

F1F3algebra

2.1

Squaring the factorisation yields $x^{4} + x^{2} + 1 = (x - u)^{2} (x - (u + 1))^{2}$ . The two roots are distinct by step 1.2, and any root of the displayed product equals one of them because a field has no zero divisors. Uniqueness of factorisation in [F6] makes both displayed exponents exactly two; in particular [F4] makes both roots repeated. Finally [F5] identifies $F_{2} (u)$ as their splitting field.

F4F5F6step 1.1step 1.2algebra∎

Depends on

Used by

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Sources

T. Judson, Abstract Algebra: Theory and Applications, Section 21.2 (standard reference, not scraped)

Over F2F2​, x4+x2+1=(x2+x+1)2x4+x2+1=(x2+x+1)2 has two distinct roots, each repeated, in its four-element splitting field