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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?

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2 Answers 2

up vote 7 down vote accepted

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 $r\in\mathbb{Q}$.

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@Zev Chonoles: So $\phi^{-1}(\phi(a)+ \phi(b))= \phi^{-1}(\phi(a+b)) = a+b$ and $\varphi^{-1}(\phi (a) \phi(b)) = \varphi^{-1}(\varphi(ab)) = ab$ which shows that it is a homomorphism? –  Damien Jun 22 '11 at 3:12
    
@Damien: Precisely! Because $a=\phi^{-1}(\phi(a))$ and $b=\phi^{-1}(\phi(b))$, this shows that $\phi^{-1}$ is a homomorphism. –  Zev Chonoles Jun 22 '11 at 3:17
    
@Zev Chonoles: For (2), couldn't you multiply by $\varphi^{-1}$ on both sides to get the result? And also $0 = 0 \cdot 0$? –  Damien Jun 22 '11 at 3:22
    
@Damien: We have only assumed that $\phi$ is a monomorphism (i.e. an injective homomorphism); if $\phi$ were an isomorphism, so that there actually existed a map $\phi^{-1}$, then your argument would be correct. The equation in your question is valid; in fact, for any homomorphism $\phi$ between any two rings, $\phi(1)=\phi(1)\phi(1)$. Thus, $\phi(1)=x$ where $x\in F$ is some element for which $x^2=x$. Show that in a field, the only two such elements are $0$ and $1$, and show that we cannot have $\phi(1)=0$ when $\phi$ is a monomorphism. –  Zev Chonoles Jun 22 '11 at 3:25
    
@Zev Chonoles: Because if it was the case then $\varphi(0) = 0$ and $\varphi(1) = 0$ which contradicts injectivity? –  Damien Jun 22 '11 at 3:30

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 ${\cal M}_{2\times 2}(\mathbb{R})$.

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