Integral element

Let $𝐵 : 𝐴$ be a (commutative) ring extension.

An element $𝑏 \in 𝐵$ is called integral over $𝐴$ iff there exists a nonzero monic polynomial $𝑚 𝑏 (𝑥) \in 𝐴 [𝑥]$ such that $𝑚 𝑏 (𝑏) = 0$ . #m/def/ring We denote the set of all such elements as $O 𝐵 : 𝐴$ , and by ^P1 this is a ring.

If every $𝑏 \in 𝐵$ is integral over $𝐴$ , then $𝐵$ is called integral over $𝐴$ , and the extension $𝐵 : 𝐴$ is called integral.

An integral element over $ℤ$ is called an Algebraic integer.

Equivalent conditions

Let $𝑏 \in 𝐵$ . The following are equivalent¹

$𝑏$ is integral over $𝐴$ ;
$𝐴 [𝑏]$ is a module-finite $𝐴$ -ring;
$𝐴 \leq 𝐴 [𝑏] \leq 𝐶 \leq 𝐵$ where $𝐶 : 𝐴$ is some finitely generated extension

Proof

Suppose ^I1 holds with minimal polynomial

𝑚 𝑏 (𝑥) = 𝑥 𝑛 + 𝛼 𝑛 - 1 𝑥 𝑛 - 1 + \dots + 𝛼 0 .

Then $𝑏 𝑛 = - (𝛼 𝑛 - 1 𝑏 𝑛 - 1 + \dots + 𝛼 0)$ so $𝐴 [𝑏]$ is generated by ${𝑏 𝑗} 𝑛 - 1 𝑗 = 0$ .

If ^I2 is the case, then we get ^I3 trivially with $𝐶 = 𝐴 [𝑏]$ .

Now supposing ^I3 holds, let ${𝑐 𝑖} 𝑛 𝑖 = 1$ generate $𝐶$ as an $𝐴$ -module. Out task is to show that $𝑏$ satisfies a monic polynomial $𝑚 𝑏 (𝑥) \in 𝐴 [𝑥]$ . Since $𝑏 \in 𝐶$ , $𝑏 𝑐 𝑖 \in 𝐶$ . Letting $𝐜 = (𝑐 𝑖) 𝑛 𝑖 = 1 \in M 𝑛, 1 (𝑐)$ we have

𝑏 𝐜 = 𝐓 𝐜

for some $𝐓 \in 𝑀 𝑛, 𝑛 (𝐴)$ , and therefore

(𝑏 𝟏 𝑛 - 𝐓) 𝐜 = 0

whence by Zero of a matrix over a ring we have $d e t (𝑏 𝟏 𝑛 - 𝐓) 𝑐 𝑖 = 0$ for all $𝑖$ , implying $d e t (𝑏 𝟏 𝑛 - 𝐓) = 0$ . Now since $𝑚 𝑏 (𝑥) = d e t (𝑥 𝟏 𝑛 + 𝐓) = 𝑥 𝑛 + 𝑄 (𝑥)$ where $d e g 𝑄 (𝑥) < 𝑛$ , it follows $𝑚 𝑏 (𝑥)$ is the required monic polynomial.

Properties

The set of all integral elements over $𝐴$ forms a ring, namely the sum and product of integral elements is again integral.
Integrality is transitive, namely if $𝐶 : 𝐵$ is integral and $𝐵 : 𝐴$ is integral, then $𝐶 : 𝐴$ is integral.

Proof of 1–2

Suppose $𝑥, 𝑦 \in 𝐵$ are integral over $𝐴$ . By ^I2 it follows $𝐴 [𝑥] : 𝐴$ is finitely generated, and since $𝑦$ is also integral over $𝐴 [𝑥]$ it follows $𝐴 [𝑥, 𝑦] : 𝐴 [𝑥]$ is finitely generated. By Finitely generated module over a module-finite $K$ -monoid, it follows $𝐴 [𝑥, 𝑦] : 𝐴$ is finitely generated. Since both

𝐴 [𝑥 𝑦], 𝐴 [𝑥 + 𝑦] \leq 𝐴 [𝑥, 𝑦],

^P1 follows from ^I3.

Suppose $𝐶 : 𝐵$ is integral and $𝐵 : 𝐴$ is integral. Then by transitivity of finitely generatedness $𝐶 : 𝐴$ is integral, proving ^P2.

Integral element

Equivalent conditions

Properties

See also

Footnotes