Proof of Field Embedding which does not restrict identity Suppose $b$ is algebraic over $F$, not in $F$, and lies in some algebraic closure $A$ of $F$. Show that there exists an embedding, $\varphi :F(b)\to A$ for which $\varphi(b)\neq b$. 
Would anyone know how to begin to tackle this problem/solve it?
EDIT: All fields assumed to have characteristic 0.
 A: In the 3rd edition of Lang's book, this is exercise 2 from section VII.§2, at page 274. Notice that at the beginning of this problem set it is explicitly stated that all fields are assumed to have characteristic $0$.
If $b$ is algebraic over $F$ then let $f \in F[X]$ be its minimal polynomial. If $b$ is the only root of $f$ in $A$, then $f = (X-b)^m$ for some $m \ge 1$, i.e. $f = X^m -mb X^{m-1} + \dots$ . Since $f \in F[X]$, it follows that $-mb \in F$, and by dividing by $-m$ (we may do it since $F$ has characteristic $0$) we get $b \in F$, which is forbidden by the hypothesis - which shows that $f$ has at least one other root $c \ne b$ in $A$ (a Galois conjugate of $b$). Notice that $f$ is the minimal polynomial of $c$, too.
Since $b$ is algebraic over $F$, it is known that $F(b) = F[b]$; if $n = [F(b) : F]$, then every element of $F(b)$ is of the form $\sum _{i=0} ^{n-1} a_i b^i$. Define then the morphism $\phi : F(b) = F[b] \to A$ by $\phi (1) = 1$ and $\phi(b) = c$, extended by multiplicativity and additivity, i.e. $\phi (\sum _{i=0} ^{n-1} a_i b^i) = \sum _{i=0} ^{n-1} a_i c^i$.
By construction, $\phi$ is a field morphism with $\phi(b) = c \ne b$. Since $\phi(1) = 1$ by construction, it follows that $\phi$ is injective, therefore an embedding.

Just for the fun of it, here is another proof of the injectivity of $\phi$. Notice that $\phi (\sum _{i=0} ^{n-1} a_i b^i) = 0$ is equivalent to $\sum _{i=0} ^{n-1} a_i c^i = 0$, so $c$ is a root of $\sum _{i=0} ^{n-1} a_i X^i \in F[X]$. Since $f$ is the minimal polynomial of $c$, it follows that $f \mid \sum _{i=0} ^{n-1} a_i X^i$. Since $b$ is a root of $f$, it follows that $b$ will also be a root of $\sum _{i=0} ^{n-1} a_i X^i$, i.e. $\sum _{i=0} ^{n-1} a_i b^i = 0$, showing that $\phi$ is injective.
