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    Integral ring extensions

    D-DEFINTEG

    (integral ring morphism)

    A ring morphism \(\phi: B \to A\) is integral if every element of \(A\) is a root of a monic polynomial with coefficients in \(\phi(B)\). When \(\phi\) is an inclusion \(B \subseteq A\), we call \(A\) an integral extension of \(B\).

    \(A\) is a finite \(B\)-algebra if and only if it is integral and finitely generated as a \(B\)-algebra. For example, \(\bar{\QQ}\) is integral over \(\QQ\) and not finite.

    In \(k[t] \subseteq k[t,\inverseof{t}]\), the element \(\inverseof{t}\) is a root of the non-monic polynomial \(tx - 1\) and is not integral over \(k[t]\): multiplying a relation \(t^{-n}+b_1t^{-(n-1)}+\cdots+b_n=0\) by \(t^n\) gives \(1\in tk[t]\). So the open immersion \(\Spec k[t,\inverseof{t}]\to\Spec k[t]\) is not finite; its image \(\AA^1\sm\ts{0}\) is not closed.

    For an integral extension \(B\subseteq A\), lying over and going up hold, and no two distinct primes of \(A\) over the same prime of \(B\) are comparable [@AM18, Chapter 5]. Hence \(\Spec A\to\Spec B\) is surjective and closed, and its fibres have dimension \(0\); the fibres need not be finite, since the fibre of \(\Spec\bar\ZZ\to\Spec\ZZ\) over \((p)\) is infinite, where \(\bar\ZZ\) is the ring of algebraic integers.

    When is a ring map integral, and what is an integral extension?
    How does integral compare with finite as a condition on $B \to A$?
    Show that $k[t] \injects k[t, \inverseof{t}]$ is not integral.
    For a subring $R \subseteq F$ and $a \in F$, show that the following are equivalent: $a$ is integral over $R$; $R[a]$ is a finitely generated $R$-module; there is a ring $R[a] \subseteq L \subseteq F$ that is a finitely generated $R$-module.
    If $F$ is a field integral over a subring $R$, show that $R$ is a field.
    Show that $k[x][f^{-1}]$ is not a field for any $f \in k[x]$.

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