About maximal ideal of local ring homomorphism

Question

Suppose $R$ is a local domain; $f$ is a map from $R$ to $F$, where $F$ is the field of fractions of $R$.

1. Is $f$ injective?
2. If $m$ is the maximal ideal of $R$, then is it true that $f(m)$ is a maximal ideal?

Answer 1

In order for your question to be able to be answered, you need to specify which map $f$ is, I assume it is te canonical embedding (which by cancelation is injective if and only if $R$ has no non trivial zero devisors). regarding your second question, this is most definetely not possible, since the only $2$ ideals in a field are $0$ and the ring itself (depending if you allow the latter case). so in particulare, since $m\subset R$ is a strict subset, its image can't be an ideal in $F$. However, if it is, it needs to be zero, and hence your maximal ideal would be the ideal of nilpotents. In particular, if your field was an integral domain, this would mean that $R$ is already a field. btw, I recommend as an exercise to prove the following:

1) the only ideals in a field are the trivial ones,

2) the canonical map into the field of fractions is injective if and only if $R$ was integral