Let $K$ be an integral domain and let $R$ an unit ring such that $K \cong R$. Can I say that $R$ is another integral domain? (can I omit that $1_R$ belongs to $R$?). Show that $R$ is conmutative is easy. Let $f$ be an isomorphism from $K$ to $R$. In addition, if $y , y' \in R \setminus \{0\}$, there exist $x , x' \in K$ such that $f(x) = y$ and $f(x') = y'$ because $f$ is surjective and we have also that $x \neq 0 \neq x'$ because $\ker f = \{0\}$ ($f$ is injective) and $y \neq 0 \neq y'$. Now, since $K$ is an integral domain, $$ x x' \neq 0 \qquad \Longrightarrow \qquad 0 \neq f(x x') = f(x) f(x') = y y' $$ and we conclude that $R$ is an integral domain too. Is it correct?
On the other hand, can we state that $R$ is a field if $K$ is? My thought is the next: if $y \in D \setminus \{0\}$, there exists $x \in K$ such that $f(x) = y$, because $f$ is surjective. As $K$ is a field, $x^{- 1} \in K$ and $$ 1 = f(1) = f(x x^{- 1}) = y f(x^{- 1})\mbox{,} $$ so $y^{- 1} = f(x^{- 1})$ is an element in $R$ such that $y y^{- 1} = 1$ (and $y^{- 1} y = 1$ because $R$ is a conmutative ring), so $R$ would be a field. Is this argument correct?
Thank you very much in advance.