Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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 $t\in\mathbb{Q}$.

share|improve this answer
    
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
    
$f(1)$ 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.