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    The rings and fields $K[x]$, $K(x)$, $K[\alpha]$, and $K(\alpha)$

    P-HI4U2

    Area
    Algebra
    Topics
    Field Extensions, Fields, Polynomials
    Status
    draft

    What are the following objects?

    • \(K(x)\)

    • \(K[x]\)

    • \(K( \alpha)\)

    • \(K[ \alpha]\)

    Solution

    For a field \(K\) and a variable \(x\), \(K[x]\) is the ring of polynomials in \(x\) with coefficients in \(K\), and \(K(x)=\operatorname{Frac}(K[x])\) is the field of rational functions \(p/q\) with \(p,q\in K[x]\), \(q\neq0\).

    For a field extension \(K\subseteq F\) and \(\alpha\in F\), \(K[\alpha]\) is the smallest subring of \(F\) containing \(K\) and \(\alpha\), and \(K(\alpha)\) is the smallest subfield of \(F\) containing \(K\) and \(\alpha\). Evaluation \(x\mapsto\alpha\) is the unique \(K\)-algebra homomorphism \(K[x]\to K[\alpha]\) sending \(x\) to \(\alpha\), and it is surjective. If \(\alpha\) is transcendental over \(K\), it is injective, so \(K[\alpha]\cong K[x]\) and \(K(\alpha)\cong K(x)\). If \(\alpha\) is algebraic with minimal polynomial \(m\), its kernel is \((m)\), and \(K[\alpha]\cong K[x]/(m)\) is already a field, so \(K[\alpha]=K(\alpha)\).