If $\sigma : k \rightarrow L$ is a non-zero ring homomorphism of fields, show that $\sigma$ can be extended to an isomorphic copy of L If $\sigma : k \rightarrow L$ is a non-zero homomorphism of fields, show that there is an extension field $E$ of $k$ that is isomorphic to $L$ by an isomorphism that extends $\sigma.$
 A: Field homomorphisms preserve $1$ by definition and therefore are non-zero.
The answer is very simple: $E=L$ and the extension of $\sigma$ is the identity.
This works when you accept the correct definition of an extension field (which is, unfortunately, not really accepted yet), namely as a homomorphism of fields (not necessarily an inclusion on the underlying sets). Notice that most extension fields arise this way. For example when we can construct $\mathbb{C}=\mathbb{R}[x]/(x^2+1)$ from $\mathbb{R}$, it is not true that the underlying set of $\mathbb{R}$ is a subset of $\mathbb{C}$. But rather there is a canonical homomorphism $\mathbb{R} \to \mathbb{C}$ and this makes $\mathbb{C}$ an extension of $\mathbb{R}$.
If you want an answer where $E$ is an extension of $k$ in the "evil" sense that the underlying set $k$ is a subset of of the underlying set $E$, define the underlying set of $E$ to be the disjoint union of the underlying set of $k$ and the underlying set of the complement of the image of the underlying map of $\sigma$. There is a bijection from this set to the underlying set of $L$. Since $L$ is a field, the set carries the structure of a field $E$ in such a way that the bijection becomes an isomorphism $E \cong L$. Since the bijection is $\sigma$ on $k$, the isomorphism extends $\sigma$.
A: Any homomorphism of fields $\sigma : k\rightarrow L$ has to be One of :


*

*Zero map 

*Injective map.


This is because kernel of any homomorphism is an ideal and I have oly two ideals in $k$.
This should tell you that.... ???
As our map $\sigma : k\rightarrow L$ is non zero, It has to be an injection 
i.e., $k\cong \sigma(k)\subseteq L $ 
So, this should tell you that you have a copy of $k$ in $L$ so you can extend that copy to $L$.
