Units in integral extensions of rings

Question

I want prove the following;

Let $X$ be a ring extension of $Y$, then $u\in Y$ is a invertible in $Y$ if and only if $u$ is invertible in $X$.

Answer 1

Provided that you want the extension to be integral, this can be proven as follows:

If $v\in Y$ denotes the inverse of $u$, we have an integral relation $v^n +a_1 v^{n-1} +\dots + a_n =0$ with coefficients in $X$. Multiplying both sides by $u^{n-1}$ yields $v=-(a_1 + \dots + a_n u^{n-1})$, hence $v\in X$ as the right hand side is contained in $X$.

Answer 2

This is not true; the ring extension $$ 0 \to \mathbb Z \xrightarrow{\cdot 5} \mathbb Z \to \mathbb Z/(5) \to 0 $$ has $2$ invertible in $\mathbb Z/(5)$, even though 2 (or any other preimage of $2 \in \mathbb Z/(5)$) is not invertible in $\mathbb Z$.