Prove that $U(S) \subseteq U( R)$ (edited with new attempt)!

Let $R$ be a ring with unity which is denoted by $1_R$, and $S$ be a subring of $R$ with unity $1_S = 1_R$

1. Prove that $U(S) \subseteq U( R)$
2. Prove that $U(R )$ is closed under multiplication
3. Prove or disprove $U(S) = U(R) ∩ S$.

My attempt:

1. Let, $x \in U(S)$. Since $S$ is a subring of $R$ with unity $1_S = 1_R$, $1_S \in R$; hence, $x \in U( R)$... (Is this too good to be true?)

2. Let $x,y \in U(R )$. Then, there exist $x^{-1}$ and $y^{-1} \in R$ such that $(xy)(y^{-1}x^{-1}) = x*1_R*x^{-1} = xx^{-1} = 1_R \in R$. Hence, $xy \in U(R )$, correct?

3. not sure how to proceed.

• 1 and 2 are really easy; for 3, consider $\mathbb{Z}$ and $\mathbb{Q}$. – egreg Feb 15 '15 at 23:21

(1) Let $x\in U(S)$; then $xy=yx=1_S=1_R$ for some $y\in S$. In particular $x\in U(R)$, because $y\in R$.
(2) Let $x,y\in U(R)$; then $(xy)(y^{-1}x^{-1})=x(yy^{-1})x^{-1}=x1_Rx^{-1}=1_R$ and also $(y^{-1}x^{-1})(xy)=1$. Therefore $xy\in U(R)$ because it has an inverse.
(3) Consider $S=\mathbb{Z}$ and $R=\mathbb{Q}$. Then $U(\mathbb{Q})\cap\mathbb{Z}=\mathbb{Z}\setminus\{0\}$, so…
• For # 1, are you telling me if $x \in U(S)$,then $x^{-1} = y$ should be $\in S$ and vice versa? And, then $x \in U(R )$ – Jellyfish Feb 15 '15 at 23:31
• @abcd1234 Yes, an invertible element of $S$ is also invertible in $R$. Your attempt at (2) is not really correct: you're writing $(xy)^{-1}$ before knowing it exists. – egreg Feb 15 '15 at 23:53