Tag Archives: Eisenstein criterion

One example of inseparable polynomial

\mathbf{Theorem:} F is a field. If F[x] contains an inseparable polynomial, then F can not have characteristic 0 or F is some finite field with characteristic p.

In general, we do not need to worry about the separability of the extension field, because we often deal with finite fields or the one of characteristic 0.

An example of inseparable polynomial
Let K be a field of characteristic p, and F=K(x) is the field of rational polynomials, x is the indeterminate. Consider F[y], y is the indeterminate, x is an irreducible element in this UFD. So f(y)=y^p-x is irreducible in E by the Einsenstein criterion. There exists a splitting field of f(y) over F. Suppose \sigma is root of f(y) then \sigma^p=x. By freshman’s dream f(y)=(y-\sigma)^p. So f has multiple roots. It is not separable.

\mathbf{Remark:} Isaacs, p281

Splitting field of x^5-2 over rational field and its primitive element

\mathbf{Problem:} Consturct a splitting field over \mathbb{Q} of x^5-2. Find its dimensionality over \mathbb{Q}.
\mathbf{Proof:} Let \omega is the primitive 5th root of unity in \mathbb{C}, whose minimal polynomial is x^4+x^3+x^2+x+1.
\displaystyle x^5-2 has roots \sqrt[5]{2}, \sqrt[5]{2}\omega, \sqrt[5]{2}\omega^2,\sqrt[5]{2}\omega^3,\sqrt[5]{2}\omega^4 in \mathbb{C}. Then a splitting field of x^5-2 is \mathbb{Q}(\sqrt[5]{2},\omega). And [\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]=20.
To see this, \displaystyle [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5, because x^5-2 is irreducible by Eisenstein criterion. Furthermore f(x)=x^4+x^3+x^2+x+1 is irreducible in \mathbb{Q}(\sqrt[5]{2}). Suppose not, since \omega\not \in \mathbb{Q}(\sqrt[5]{2}), f must split into two quadratic polynomial, namely f(x)=(x^2-(\omega+\omega^4)x+1)(x^2-(\omega^2+\omega^3)x+1). So \displaystyle \omega+\omega^4\in \mathbb{Q}(\sqrt[5]{2}). While \omega+\omega^4 satisfies x^2-5x+2=0, which has roots \displaystyle\frac{5+\sqrt{17}}{2} and \displaystyle\frac{5-\sqrt{17}}{2}. So \sqrt{17}\in \mathbb{Q}(\sqrt[5]{2}). This is impossible because [\mathbb{Q}(\sqrt{17}):\mathbb{Q}]=2 does not divide 5.
We can consider this problem in another way. [\mathbb{Q}(\omega):\mathbb{Q}]=5, x^5-2 is irreducible in \mathbb{Q}(\omega). Otherwise \mathbb{Q}(\sqrt[5]{2}) can be imbedded in \mathbb{Q}(\omega). This is impossible because [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5>4=[\mathbb{Q}(\omega):\mathbb{Q}].
\mathbf{Problem:} Find a primitive element of a splitting field of x^5-2 over \mathbb{Q}.
\mathbf{Proof:} The Galois group of x^5-2 over \mathbb{Q} are permutations of roots. Every permutation \eta is uniquely determined by \eta(\sqrt{2}) and \eta(\omega). Let \alpha=\sqrt{2}+\omega, the only \eta fixes \alpha is \eta=id. So \mathbb{Q}(\alpha) is the splitting field of x^5-2 and \alpha is a primitive element.