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Definition 9.1 (ring extensions)

[definition] Let $R \subseteq R’$ be rings. Then $R’$ is called an extension ring of $R$. We also say that $R \subseteq R’$ is a ring extension. [/definition]

[definition] Let $R \subseteq R’$ be a ring extension. We say that $a \in R’$ is integral over $R$ if there is monic polynomial $f \in R[x]$ with $f(a) = 0$. [/definition]

[definition] A ring extension $R \subseteq R’$ is finite if $R’$ is module-finite as an $R$-module. [/definition]

As an example, notice that $R\subseteq R’$ in the case of $R$ and $R’$ fields just amounts to field extensions and integral=algebraic and finite=finite as adjectives.

Example 9.3

[example] Let $R$ be a UFD and $R’ = \operatorname{Quot}(R)$ be the quotient field of $R$. Then $a \in R’$ is integral over $R$ if and only if $a \in R$. [/example]

[proof] ($\implies$) Let $a \in R’$ be integral over $R$. Then there exists a monic $f \in R[x]$ such that $f(a) = 0$. Since $R’$ is the quotient field of $R$, we can write $a = p/q$ for $p$ and $q$ relatively prime. If we write

then

We clear denominators by multiplying through by $q^n$ and move $p^n$ over to one side to get

This equation holds in $R$ and $R$ is a UFD, so it follows that $q \mid p$ or $q$ is a unit. The former is impossible by assumption so $q$ must be a unit. So $p/q \in R$.

($\impliedby$) This direction is obvious. The polynomial $x - a$ does the job. [/proof]

Example 9.4

[example] Let $R = \CC[x]$ and $R’ = R[y]/(f)$ where $f \in R[y]$ is a (non-constant) polynomial. There is an obvious extension of rings $R \subseteq R’$ and such a map is injective as $f$ is non-constant.

Geometrically, $R = A(X)$ and $R’ = A(X’)$ where $X = \AA_{\CC^1}$ (because $I(X)$ is the zero ideal here) and $X’ = V(f)$, correspond to the morphism $\pi:X’ \to X$, $(x, y) \mapsto x$ of varieties. Indeed, let

The latter map can be elaborated upon as

[/example]

Proposition 9.5

[proposition] The extension $R \subseteq R’$ of rings is finite if and only if $R’ = R[a_1, \ldots, a_n]$ for integral elements $a_1, \ldots, a_n \in R’$ over $R$.

Moreoever, when this case occurs, the extension $R \subseteq R’$ is integral. [/proposition]

[proof] ($\implies$) Let $R’ = \sum_{i=1}^n Ra_i$ as an $R$-module. Then, of course, $R’ = R[a_1, \ldots, a_n]$. Thus, it remains to show that $R\subseteq R’$ is an integral extension (which shows that the $a_i$ are integral over $R$). Let $a \in R’$. Since $a = \sum c_ia_i$ for some $c_i \in R$. So, by definition, $a \in RR’$. So the Cayley-Hamilton theorem implies there is a monic polynomial

for which $\chi(\phi) = 0$. So if we evaluate this operator on $x = 1$, we obtain

So $a$ is integral over $R$.

($\impliedby$) Let $R’ = R[a_1, \ldots, a_n]$. Then every element of $R’$ is a polynomial in $R[a_1, \ldots, a_n]$. Furthermore, is the root of a monic polynomial $f \in R$, we can use this to rewrite the elements of $R’$ so that the degree’s stay smaller than some bound (namely, sum of the degrees of the monics that the $a_i$ satisfy).
[/proof]

Lemma 9.6

[lemma] Let $R \subseteq R’ \subseteq R’’$ be ring extensions. Then

(a) If $R \subseteq R’$ and $R’ \subseteq R’’$ are finite, then so is $R\subseteq R’’$.

(b) If $R \subseteq R”$ and $R’ \subseteq R’’$ are integral, then so is $R \subseteq R’’$. [/lemma]

We’ll omit a proof of this lemma since the intended proof is pretty obvious.

Lemma 9.7

[lemma] Let $R’$ be an integral extension ring of $R$.

(a) If $I$ is an ideal of $R’$, then $R’/I$ is an integral extension ring of $R/(I\cap R)$.

(b) If $S \subseteq R$ is multiplicatively closed, then $S^{-1}R’$ is an integral extension ring of $S^{-1}R$.

(c) $R’[x]$ is an integral extension ring of $R[x]$. [/lemma]

[proof] (a) Notice that there is an obvious choice of extension $R/(I\cap R) \to R’/I$ which is injective and well-defined. For the integrality claism, just lift up to $R \subseteq R’$ and then pass back down.

(b) Same thing.

(c) Coefficients are all integral so this is immediate. [/proof]

Exercise 9.8

[exercise] (a) Is $\sqrt{2 + \sqrt{2}} + \frac{1}{2}\sqrt[3]{3} \in \RR$ integral over $\ZZ$?

(b) Let $R’ = \RR[x]$, $R = \RR[x^2 - 1] \subseteq R’$, $P’ = R’(x - 1)$, and $P = P’ \cap R$. Show that $R’$ is an integral extension of $R$ but the localization $R’_{P’}$ is an integral extension of $R_P$. Is this a contradiction to Lemma 9.7(b)? [/exercise]

(a) We claim that this element is not integral over $\ZZ$. Note that $\frac{1}{2}\sqrt[3]{3}$ itself is not integral over $\ZZ$ since its minimal polynomial over $\QQ$ is $x^3 - 3/8$. If it was, then we’d have a factorization of such a monic over $\ZZ$ into

where $p$ is also monic. By Gauss’ Lemma, we can find $r, s \in \QQ$ so that

is a factorization over $\ZZ$ without disturbing the monicity. In particular, this says that $rs = 1$. But since this is a factorization over $\ZZ$, this forces $r = s = 1$ (up to sign) by monocity of $p(x)$ and $x^3 - 3/8$, contradiction. So it follows that $\frac{1}{2}\sqrt[3]{3}$ is not integral over $\ZZ$.

Now note that $\sqrt{2 + \sqrt{2}}$ is clearly integral over $\ZZ$. So $\sqrt{2 + \sqrt{2}} + \frac{1}{2}\sqrt[3]{3}$ cannot be integral over $\ZZ$ (otherwise this would imply that $\frac{1}{2}\sqrt[3]{3}$ is).

(b) todo

Corollary and Definition 9.9

[definition] Let $R \subseteq R’$ be a ring extension. The set $\overline{R}$ of all integral elements in $R’$ over $R$ is a ring with $R \subseteq \overline{R} \subseteq R’$. The ring $\overline{R}$ is the integral closure of $R$ in $R’$. We say that $R$ is integrally closed in $R’$ if $\overline{R} = R$.

In the special case where $R$ is an integral domain, we say $R$ is integrally closed or normal if $R$ is integrally closed in $\operatorname{Quot}(R)$. [/definition]