I came across the following problems:
If $\varphi: F \to G$ is an isomorphism of fields show that $\varphi^{-1}: G \to F$ is also an isomorphism.
So $\varphi(a+b) = \varphi(a) + \varphi(b)$ and $\varphi(ab) = \varphi(a) \varphi(b)$ for all $a,b \in F$. In addition, $\varphi$ is bijective. So $\varphi^{-1}: G \to F$ exists and is also bijective. So then it follows that $\varphi^{-1}(\phi(a) + \phi(b)) = \varphi^{-1}(\phi(a)) + \varphi^{-1}(\phi(b)) = a+b$ and $\varphi^{-1}(\phi(a) \phi(b)) = \varphi^{-1}(\phi(a)) \varphi^{-1}(\varphi(b)) = ab$ for all $\varphi(a), \varphi(b) \in G$?
If $\varphi: F \to G$ is a monomorphism of fields, show that $\varphi(1) = 1$.
So $\varphi(1 \cdot 1) = \varphi(1) \varphi(1)$ so that $\varphi(1) = \varphi(1) \varphi(1)$ or $\varphi(1) = 1$.
Show that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}(t)$ where $\mathbb{Q}(t)$ contains rational polynomials in the indeterminate $t$.
Suppose there is a monomorphism $\varphi: \mathbb{Q} \to \mathbb{Q}(t)$. Then $\varphi(r) = r$ for all $r \in \mathbb{Q}$? This means that non-constant polynomials are not hit and henece $\varphi$ is not surjective?
Do these seem correct?