For Field extension, $[E :F]=1$ implies $E=F$ Suppose $E/F$ is a field extension then
$[E :F]=1$ implies $E=F$.
This sounds very trivial but i don't know how to formally write this. 
 A: Remember that $E$ is a vector space over $F$. What is its dimension? What can you conclude from it?
A: The expression “$E/F$ is a field extension” has some ambiguity.
Almost everybody (including you, I am sure) uses this expression to mean that $F$ and $E$ are fields with $F\subset E$. In this case, equality between $F$ and $E$ is equivalent to the degree being $1$, and with others’ hints, I’m sure you can prove it.
There is another interpretation, though, and that is that $F\to E$ is a morphism of fields. For instance, let $k$ be a field (“constant field”), $k=\Bbb C$ or $k=\Bbb Q$ will do, and consider the field $k(x)$, where $x$ is an indeterminate, and the map $F=k(x)\to k(x^2)=E$ that sends $x$ to $x^2$, in other words $f(x)\in k[x]$ is sent to $f(x^2)$. The concept of field extension degree applies here, and the degree is $1$, but $E\ne F$.
To show the significance of the second way of looking at things, let $\sigma:k(x)\to k(x)$ by the same rule, $\sigma(f(x))=f(x^2)$, a good field morphism. In this case, the field extension degree is two, and $E=F$. (This is not nitpicking—the above way of looking at a field morphism is essential in some parts of algebraic and arithmetic geometry).
