If $R$ and $S$ are isomorphic rings, is $R$ an ordered ring iff $S$ is an ordered ring? Yesterday I had my final for my introduction to abstract algebra course. One of the questions on the final asked you to prove that the field $<\mathbb{R},+, \cdot>$ was not isomorphic to $<\mathbb{C},+,\cdot>$. I didn't have time to write out a good proof, but I basically stated that $\mathbb{C}$ can not be isomorphic to $\mathbb{R}$ since $\mathbb{R}$ is an ordered field while $\mathbb{C}$ is not. At the time I was not 100% positive that this was true, but since I was having trouble coming up with anything else I figured that this was my best shot. After the exam, I went home and looked at some proofs that $\mathbb{R}$ is not isomorphic to $\mathbb{C}$, and realized that showing that $\mathbb{C}$ has more elements than $\mathbb{R}$ that are of finite order would have been sufficient to prove that they are not isomorphic, but my original answer to this problem got me curious about proving the following:
Let $R$ and $S$ be isomorphic rings. $R$ is an ordered ring if and only if $S$ is an ordered ring.
I could not find any such proof on the internet, although perhaps there is some text out there that contains a proof, although it would surprise me to say the least.
In our course notes the definition of an ordered ring is as follows:

An ordered ring is a commutative ring $R$ with a
  subset $R^+$ satisfying:
(a) $R^+$ is closed under addition,
(b) $R^+$ is closed under multiplication,
(c) if $a$ is any element of $R$, then exactly one of the following hold: $a$
  is in $R^+$, $−a$ is in $R^+$, $a = 0$. (Trichotomy Law)

This is my proof attempt:
Let $R$ and $S$ be isomorphic rings, hence there exists an isomorphism $f: R \rightarrow S$.
$(=>)$ Suppose $R$ is an ordered ring, that is there exists $R^+ \subseteq R$ that holds the properties displayed above. Let $S^+=f(R^+)$. Since $f$ is surjective, we have that $\forall a \in S$, $\exists b \in R$ such that $a=f(b)$. Since $b \in R$, we have that exactly one of the following holds:


*

*($b \in R^+$) Suppose $b \in R^+$, then since $a=f(b)$, $a \in f(R^+)$ and $\therefore a \in S^+$.

*($-b \in R^+$) Suppose $-b \in R^+$, then we have that $a=f(b) \iff -a=-f(b) \iff -a=f(-b)$, hence $-a \in f(R^+)$ and $\therefore -a \in S^+$.

*($b=0$) Suppose $b=0$, then we have that $a=f(b)=f(0)=0$, $\therefore a=0$.


So we have that the trichotomy law holds for $S^+$. Now, let $x,y \in S^+$, hence $\exists w, \exists z \in R^+$ such that $x=f(w)$ and $y=f(z)$. We have that
$$
xy=f(w)f(z)=f(wz)
$$
Since $R^+$ is closed under multiplication, $wz \in R^+$ so $xy \in S^+$, hence $S^+$ is closed under multiplication. Also,
$$
x+y=f(w)+f(y)=f(w+y)
$$
Since $R^+$ is closed under addition, $w+y \in R^+$ so $x+y \in S^+$, hence $S^+$ is closed under addition. We have that $R$ is a commutative ring since it is ordered, so by isomorphism $S$ must be commutative as well. $\therefore$ Since there exists $S^+ \subseteq S$ such that $S^+$ follows the properties displayed above, and $S$ is commutative, we have that $S$ is an ordered ring.
$(<=)$ The converse seems trivial to prove, following essentially the same method of proof as above, so I won't write it here.
My questions are: Is my proof correct? If it is correct, is there a shorter method of proving this? Is this a valid method of showing that $<\mathbb{R},+, \cdot>$ and $<\mathbb{C},+, \cdot>$ are not isomorphic fields?
Thanks for taking the time to read this I know it's a long one.
 A: Two comments. First, you should talk about what you're trying to prove before you dive into the weeds. So instead of this:

$(=>)$ Suppose $R$ is an ordered ring, that is there exists $R^+ \subseteq R$ that holds the properties displayed above. Let $S^+=f(R^+)$. Since $f$ is surjective, we have that $\forall a \in S$, $\exists b \in R$ such that $a=f(b)$...

Write this:

$(=>)$ Suppose $R$ is an ordered ring, that is there exists $R^+ \subseteq R$ that holds the properties displayed above. Let $S^+=f(R^+)$. We claim that $S^+$ makes $S$ an ordered ring.
[new paragraph] First, we verify trichotomy. Since $f$ is surjective, we have that $\forall a \in S$, $\exists b \in R$ such that $a=f(b)$...

But more importantly: This is the sort of proof that nobody would actually write out! It's enough to observe that the definition of an ordered ring is stated entirely in terms of the ring axioms, so ring isomorphisms must preserve orderedness. That's it, you're done.
(There is general principle that isomorphic structures have the same properties. I just haven't studied enough model theory to state that as a precise theorem.)
A: The theorem is true when you state it in the form

Let $R$ and $S$ be isomorphic commutative rings. The $R$ can be given the structure of ordered ring if and only if $S$ can.

Suppose $R$ admits an ordering, defined by a subset $R^+$ (there can be several orders on $R$) and set $S^+=f(R^+)$, where $f\colon R\to S$ is a ring isomorphism (there can be several isomorphisms).
Then $S^+$ satisfies trichotomy. Set, for simplicity, $-R^+=\{-a:a\in R^+\}$ and $-S^+=\{-b:b\in S^+\}$. Then $f(-R^+)=-f(R^+)=-S^+$. Since $f$ is bijective, it maps unions of sets to union of sets, so
$$
S=f(R)=f(R^+\cup\{0\}\cup(-R^+))=S^+\cup\{0\}\cup(-S^+)
$$
Again, since $f$ is bijective, $S^+$, $\{0\}$ and $-S^+$ are pairwise disjoint, because so are $R^+$, $\{0\}$ and $-R^+$ in $R$.
Closure of $S^+$ under sums and products is easy (and you nailed it).
You don't need to prove the converse, because the inverse map of an isomorphism is an isomorphism as well.

However, you're looking too deep. In an ordered ring, $1\in R^+$, so $-1\in(-R^+)$. Suppose there is an ordering defined by $\mathbb{C}^+$. Then either $i\in\mathbb{C}^+$ or $-i\in\mathbb{C}^+$. However, this implies that $-1=i^2=(-i^2)\in\mathbb{C}^+$, which is a contradiction.
You can use the same idea to prove there is no ring homomorphism $f\colon\mathbb{C}\to\mathbb{R}$. Indeed, such a homomorphism must satisfy $f(1)=1$, so also $f(-1)=-1$. Consider $x=f(i)$; then
$$
x^2=f(i)^2=f(i^2)=f(-1)=-1
$$
and there's no real number with this property.
