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In my abstract algebra class, we have been tasked to find all monomorphisms $\mathbb{Q} \to \mathbb{C}$. The book (Stewart's Galois Theory) gives an example for $\mathbb{Q}(\sqrt[3]{2}) \to \mathbb{C}$, where $\mathbb{Q}(\sqrt[3]{2})$ is a field extension of $\mathbb{Q}$, which I mostly understand. There are 3 such monomorphisms because you can essentially permute the roots of the minimal polynomial of $\sqrt[3]{2}$ three ways in $\mathbb{C}$ (a better explanation would be welcome). However, for my problem the degree of the extension $\mathbb{Q}:\mathbb{Q}$ is 1, so my thoughts are that there is only the identity monomorphism. Is this true? A hint would be wonderful.

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up vote 4 down vote accepted

Hint: Let $f:\mathbb{Q}\to\mathbb{C}$ be a ring monomorphism. What can $f(1)$ be? Show that the value of $f(1)$ determines $f(n)$ for all $n\in\mathbb{Z}$, and hence $f(t)$ for all $t\in\mathbb{Q}$.

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More generally let $k$ be a prime field and $\varphi: k \rightarrow K$ a ring monomorphism. Then $\varphi(k)$ is the prime subfield of $K$ – JSchlather Feb 25 '13 at 5:24
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$f(1)$ has to be 1 right? I see now how $f(1)$ determines $f(n)$. – Jeremy Feb 25 '13 at 5:35
    
@Jeremy: Yup, that's right. – Zev Chonoles Feb 25 '13 at 5:48
    
@Zev great, thank you! – Jeremy Feb 25 '13 at 5:52
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@hhh: What you've written is incorrect. For any ring homomorphism $$f(a+b)=f(a)+f(b)\qquad\qquad f(a\cdot b)=f(a)\cdot f(b)$$ so for example, $$f(3)=f(1+1+1)=f(1)+f(1)+f(1)$$ – Zev Chonoles Mar 20 at 19:49

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