Ring Element is Unit iff Unit in Integral Extension

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Theorem

let $A$ be a commutative ring with unity.

Let $a \in A$.

Let $B$ be an integral ring extension of $A$.


The following are equivalent:

$(1): \quad a$ is a unit of $A$
$(2): \quad a$ is a unit of $B$


Proof

1 implies 2

Follows from Ring Homomorphism Preserves Units.

$\Box$


2 implies 1

Let $a$ be a unit of $B$.

Let $P \in A \left[{x}\right]$ be a monic polynomial with $P \left({1 / a}\right) = 0$.

Let $n$ be its degree and $P \left({x}\right) = x^n + Q \left({x}\right)$.

Then $1 + a^n Q \left({1 / a}\right) = 0$.

Note that $a^{n-1} Q \left({1 / a}\right) \in A$.

Thus $a$ is a unit of $A$, with inverse $-a^{n - 1} Q \left({1 / a}\right)$.

$\blacksquare$