Monomorphisms and Isomorphisms 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?
 A: For the first problem, you correctly used the fact that every element of $F$ is equal to $\phi$ of some other element, but you have not actually argued why $\phi^{-1}$ is a homomorphism; you need to show that
$$\phi^{-1}(x+y)=\phi^{-1}(x)+\phi^{-1}(y)$$
and
$$\phi^{-1}(xy)=\phi^{-1}(x)\phi^{-1}(y)$$
for all $x,y\in F$. Use the fact that every element of $F$ is equal to $\phi$ of some other element, and the fact that $\phi$ is a homomorphism.
For the second problem, your conclusion is incorrect; $\phi(1)=\phi(1)\phi(1)$ does not imply that $\phi(1)=1$. There is one other element of a field that has this property. Show that, because $\phi$ is a monomorphism (i.e., it is injective), this other element cannot be the value of $\phi(1)$.
For the third problem, your argument has exactly the right idea, but you need to include a proof of the fact that $\phi(r)=r$ for all $r\in\mathbb{Q}$. Start by showing that, if $\phi:\mathbb{Q}\rightarrow\mathbb{Q}(t)$ were an isomorphism, then $\phi(1)=1$ (this is true by the second problem), then use that to show that $\phi(n)=n$ for all $n\in\mathbb{Z}$, then use that to show that $\phi(r)=r$ for all $r\in\mathbb{Q}$.
A: The following (interesting) exercise is relevant to (2) of your question and the general fact that in the context of arbitrary rings and non-unital rings homomorphisms, one cannot conclude that $\phi(1)=1$ from $\phi(1)\phi(1)=\phi(1)$:
Exercise:
Let $\phi:\mathbb{R}\to {\cal M}_{2\times 2}(\mathbb{R})$ (where ${\cal M}_{2\times 2}(\mathbb{R})$ is the ring of $2\times 2$ matrices with real entries) be the map defined by the rule:
$\phi(a)=\begin{bmatrix}
0 & a\\
0 & 0\\
\end{bmatrix}$
for $a\in\mathbb{R}$. 
(a) Prove that $\phi$ is $\textit{additive}$, that is, prove that $\phi(a+b)=\phi(a)+\phi(b)$ for all $a,b\in\mathbb{R}$. 
(b) Prove that $\phi$ is $\textit{multiplicative}$, that is, prove that $\phi(ab)=\phi(a)\phi(b)$ for all $a,b\in\mathbb{R}$. 
(c) Deduce that $\phi(1)\phi(1)=\phi(1)$. However, prove that $\phi(1)\neq 1$, that is, $\phi$ does not map the multiplicative identity of $\mathbb{R}$ to the multiplicative identity of ${\cal M}_{2\times 2}(\mathbb{R})$.
