About isomorphism of rings and fields If $A,B$ are rings and $A$ is a field. If $A$ is a field and $A\cong B$ so $B$ is a field too?
Thank you!
 A: Suppose $\phi: A \leftrightarrow B$ is a ring isomorphism. $B$ is a field if it is commutative, and if every non-zero element of $B$ has a multiplicative inverse.
Let $a,b$ be two elements of $B$. Then
$$ab = \phi\left(\phi^{-1}(a)\right)\phi\left(\phi^{-1}(b)\right) = \phi\left(\phi^{-1}(a)\phi^{-1}(b)\right) = \ldots$$
can you finish the argument?
Similarly, let $b$ be a nonzero element of $B$. Can you prove $\phi^{-1}(b)$ is nonzero? How might you use $\phi^{-1}(b)$ to construct the multiplicative inverse of $b$?
A: To check that $B$ is a field, we must check that the field axioms hold for $B$. That is, $B$ is a commutative ring such that every non-zero element of $B$ is a unit. If $A \cong B$ as rings, then clearly $B$ is a commutative ring if and only if $A$ is. $A$ is commutative as it's a field, hence $B$ is also. Now for the second part. Let $b \in B, b \neq 0$. Let $\phi:A\rightarrow B$ be a ring isomorphism. There is $a \in A, a \neq 0$ such that $\phi(a)=b$. As $A$ is a field, there is $a' \in A$ such that $aa'=a'a=1_A$ in $A$. Let $b'=\phi(a')$. So $1_B=\phi(1_A)=\phi(aa')=\phi(a)\phi(a')=bb'=b'b$, so $b$ is a unit in $B$. QED
A: Hint $\ $ If a polynomial has a root in a ring, then it has a root in every image of the ring. Applying this to $\, ax - 1$ we deduce that images of units remain units.
Remark $ $ I elaborate, since the hint appears to be too concise for at least one reader.  In order not to completely spoil the hint I consider an analogous example. Consider the class of rings that are root-closed, i.e. every element has a square-root in the ring. I claim that this property is preserved  in every ring image $\,h(R)\,$ of a root-closed ring $\,R.$ Indeed, an element of $\,h(R)\,$ has form $\,h(a).\,$ But $\,a = x^2$ for some $\,x\in R.$ Its image $\,h(a) = h(x^2)=(h(x))^2$ shows that $\,h(x)\,$ remains a root of $\,h(a)\,$ in $\,h(R).\,$ So, as claimed, $\,h(R)\,$ is root-closed too. 
Similarly, any "root existence" property will be preserved in ring-images, i.e. any property of the form $\,\forall a\,\exists x\!:\ f(a,x) = 0,\,$ for some polynomial $\,f.\,$ Fields are of this type, being those rings such that $\,ax = 1\,$ has a root, i.e. $\,\forall a\ne 0,\,\exists x\!:\ ax = 1.$
Model theory studies in much greater details the forms of statements that are preserved under homomomorphisms, subalgebras, products, etc, as well as many other types of relationships between syntax and semantics in equational algebras.
