A.5 · Polynomials and field extensions

A polynomial is irreducible over a field if it has positive degree and no factorisation into polynomials of smaller positive degrees there. A quadratic or cubic is irreducible exactly when it has no root in that field. This root test alone fails for degree four or higher; for instance (x²+1)(x²+2) has no real root but is reducible over R. Always name the base field.

The quotient F[x]/(p) is a field when p is irreducible. Reduce powers using p(alpha)=0, where alpha is the residue class of x. If F has q elements and p has degree d, the quotient has q^d elements represented by polynomials of degree below d. Z/4Z has four elements but has zero divisors, so it is not the field with four elements.

Over F2, p(x)=x²+x+1 has values 1 at both 0 and 1 and is irreducible. In its quotient alpha²=alpha+1 because subtraction equals addition in characteristic two. The four elements are 0,1,alpha,alpha+1. All three nonzero elements must be units; compute their products rather than treating alpha as an ordinary real number.

For nested finite-degree fields K inside L inside M, the tower law gives [M:K]=[M:L][L:K]. The degree of an algebraic element is the degree of its minimal polynomial. A finite extension of degree two does not contain an element of degree three over the base field. Over Q, sqrt(2) has degree two; adjoining sqrt(3) as well produces a degree-four extension, since sqrt(3) is not in Q(sqrt(2)). Over C, primitive nth roots of unity have exact order n and are the roots of the cyclotomic polynomial Φ_n. For n=10, divide x⁵+1 by x+1 to exclude the order-two root −1: Φ_10=x⁴−x³+x²−x+1. Vieta’s formulas give sum 1 and product 1 of its four primitive roots. Do not sum all tenth roots, or assume every nontrivial tenth root is primitive; a root’s order must be checked. For a monic degree d polynomial, product of its roots is (−1)^d times the constant coefficient.

In F2[alpha] with alpha²+alpha+1=0, multiply alpha(alpha+1) = alpha² + alpha = (alpha+1) + alpha = 1. Thus alpha^−1=alpha+1. Also alpha³=1 and alpha is not 1, so its multiplicative order is 3. This produces a field of four elements. By contrast 2·2=0 in Z/4Z, proving that quotient is not a field.

Polynomial reducibility changes with the coefficient field. The polynomial x²−2 is irreducible over Q but splits over R.

Construct $\mathbb F_2[x]/(x^2+x+1)$. List its elements and find the inverse of the residue $\alpha=[x]$.

Worked solution

The quadratic takes value one at both elements of $\mathbb F_2$, so it has no root and is irreducible. Its quotient is a field. Division leaves a representative of degree below two: $0,1,\alpha,1+\alpha$. The relation is $\alpha^2=\alpha+1$ in characteristic two. Consequently $\alpha(\alpha+1)=\alpha^2+\alpha=1$, so $\alpha^{-1}=\alpha+1$. This is not $\mathbb Z/4\mathbb Z$, which has a nonzero zero divisor.

Is $x^4+4$ irreducible over $\mathbb Q$? Does the absence of rational roots answer that question? Can a degree-two extension of $\mathbb Q$ contain a root of an irreducible cubic over $\mathbb Q$?

Check after attempting

Direct multiplication gives $x^4+4=(x^2-2x+2)(x^2+2x+2)$, so it is reducible over $\mathbb Q$ despite having no real, hence no rational, roots. The no-root test characterises irreducibility only for degree two or three over a field. If an irreducible cubic root $\beta$ belonged to a degree-two extension L, then $\mathbb Q\subset\mathbb Q(\beta)\subset L$ and the tower law would make 3 divide 2. This is impossible. The conclusion uses irreducibility, not merely a cubic equation.