Extension of a field.. I am new in field theory. If we have to show that $E$ is a filed extension of $F$. Then we define a map from $F$ to $E$ and shows that map is injective and ring homomorphism. I found it  in proof of some theorems. What is connection between  injective ring homomorphism and extension?

Definition : If $K$ is a field     containing the subfield $F$, then $K$ is said to be an extension field of $F$.

What is intuitive idea of extension of a field?
Thanks!
 A: Let $E$ and $F$ be fields. Suppose there exists an injective ring homomorphism $\phi:E \to F$. Then by isomorphism theorem, $\phi(E) \cong E/\ker(\phi) = E$. Thus, $\phi(E)$ is a subfield of $F$ that is isomorphic to $E$. We sometimes abuse notation a little bit by saying that $E$ is a subfield of $F$.
Note: if $E$ is a field, then any ring homomorphism $\psi: E \to R$ is either injective or the zero map. This is due to the fact that $\ker(E)$ is an ideal of $E$ and the only ideals of $E$ are $(0)$ and $E$.
A: At the most basic level, an extension field is no more than simply a bigger field that contains all the stuff in the smaller field. It's just taking all of the things in your box and moving them into a bigger box (while preserving all the structure that goes along with them). So, it's like an enhanced version of superset that preserves addition, subtraction, multiplication, and division.
At a more useful level, however, I like to think of most field extensions by what "problem" they fix in our smaller field. Let's look at some basic canonical examples that are well studied at even a high school level:
$$\mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}\text{,}$$
the rationals, reals, and complex numbers.
The rationals $\mathbb{Q}$ can exist on their own right as a field. All of our operations work wonderfully. But we quickly find that we have no solutions to the polynomial $x^2-2=0$, since $\sqrt{2}$ does not exist in $\mathbb{Q}$. But if we instead imagine our universe to be $\mathbb{R}$, we can do exactly the same things we can do in $\mathbb{Q}$, but now have additional numbers to play with and can solve other problems, like find solutions to the above polynomial.
(Technical Note: $\mathbb{R}$ is not the smallest field extension that would let us solve that polynomial. That would be $\mathbb{Q}(\sqrt2)$, or what you get when you take the rationals, $\sqrt2$, and all other numbers for our field operations to still be defined. But $\mathbb{R}$ is more easily understood at first.)
Similarly, the polynomial $x^2+1=0$ has no solutions in $\mathbb{R}$, but if we expand our universe to be $\mathbb{C}$, we have enough numbers to solve that polynomial. (In fact, since $\mathbb{C}$ is the completion of $\mathbb{R}$, we can show that we have enough numbers to solve any polynomial with coefficients in $\mathbb{R}$!)
Now, back to the original question: "What is connection between injective ring homomorphism and extension?"
Recall, we said we want a bigger universe where all of our numbers can still live and where all or our field structure (specifically addition, subtraction, multiplication, and division) is preserved. So to prove that another field is an extension of of our smaller field, we would need to find a map that meets both of those requirements. A ring homomorphism is an obvious choice, since by definition it preserves structure. But the mapping would also need to be injective, since each distinct number in our smaller field would need to be mapped to a distinct number that acts the same in the bigger field. Since we're mapping distinct numbers to distinct numbers (i.e. it's one-to-one) in a way that preserves our field operations, we would thus need to find an injective ring homomorphism.
Note: there is a little more subtlety to that map, which tfp gets into some, but since the question focuses on intuition, I find it helpful to use some well-known examples as an introduction to the topic.
A: @tfp's answer is very nice, but I want to say something about the connection between subobjects/extensions and injective morphisms in general.
Since field theory is new to you, let's go back to something simpler, like vector spaces. Every subspace $W$ of a vector space $V$ comes with an inclusion map $W \to V$, which is of course both linear and injective. Moreover, every injective linear map $A \to B$ specifies a subspace of the codomain $B$, namely the image of the map. If you think about it, describing an injective linear map $f : A \to B$ is the same as describing a subspace of $B$ (the image of the map), along with a new name for each element of $B$ (that is, each element of the image comes from a unique element of the domain, which I'm calling its "new name"). In this sense, the inclusions of subspaces are the "simplest" linear maps: we specify a subspace to include, and don't change the names of the elements of that subspace. This description also makes it clear that two injective linear maps $f : A \to B$ and $f' : A' \to B$ having the same image don't need to be equal – however, $A$ and $A'$ must be isomorphic vector spaces, since both are just "renamed versions of $\operatorname{img}(f) = \operatorname{img}(f')$". Let me be more precise:
Claim Fix a vector space $V$. Then:

*

*For each subspace $W$ of $V$, the inclusion map $W \to V$ is an injective linear map whose image is $W$.

*Let $f_1 : W_1 \to V$ and $f_2 : W_2 \to V$ be injective linear maps. $f_1$ and $f_2$ have the same image if and only if there exists an isomorphism $g : W_1 \to W_2$ such that $f_1 = f_2 \circ g$.

Ignoring size issues, "having the same image" is an equivalence relation on the class of injective linear maps with codomain $V$. Part 1 of the claim tells us that sending each equivalence class to the common image of its elements gives a bijective correspondence between these equivalence classes and subspaces of $V$. Part 2 gives us a way to describe when two injections are equivalent without mentioning images, which can be helpful. And overall, we can say that subspaces of a vector space are in bijective correspondence with equivalence classes of injections to that vector space.
In fact, the same exact thing is true for subsets of sets, or subgroups of groups, or subposets of posets, etc. For practically any algebraic object you can think of, subobjects are the same as equivalence classes of injective homomorphisms.
In particular, subfields of a field $F$ are in bijective correspondence with equivalence classes of injective homomorphisms $E \to F$, where two injections $\phi_1 : E_1 \to F$ and $\phi_2 : E_2 \to F$ are deemed equivalent iff there exists an isomorphism $\alpha : E_1 \to E_2$ such that $\phi_1 = \phi_2 \circ \alpha$.
This motivates the general definition of a subobject in category theory:
Definition. Let $C$ be a category with an object $x$. A subobject of $x$ is an equivalence class of monomorphisms $y \to x$, where two monomorphisms $f_1 : y_1 \to x$ and $f_2 : y_2 \to x$ are deemed equivalent iff there exists an isomorphism $g : y_1 \to y_2$ such that $f_1 = g \circ f_2$.
Even if you don't know any category theory, this definition should look very similar to the previous discussion about subobjects in various settings, and indeed the notions of a sub-vector space/group/set/field/ring/module/etc. are all special cases of this definition.
Anyway, the idea of an extension is just to think about subobjects from the other perspective. Instead of thinking about $F$ as the "original field" and $E \subseteq F$ as a subfield, if we think about $E$ as the "original field" then suddenly $F \supseteq E$ becomes an extension of $E$. Since subobjects and extensions are just two perspectives on the same idea, and subobjects correspond to (equivalence classes of) injections, extensions also correspond to (equivalence classes of) injections, only this time it's the domain that's fixed. You can define the notion of an extension in any category in the exact same way: an extension of an object $X$ is just an equivalence class of monomorphisms from $X$.
