Proposition 5.21 in Atiyah-MacDonald There's just one step in this proof I can't see for the life of me.  
Set up: We have a field K and an algebraically closed field $\Omega$.  $(B, g)$ is maximal in the set $\Sigma$ of ordered pairs $(A, f)$ where $A$ is a subring of K and $f$ a homomorphism into $\Omega$, where $\Sigma$ has the partial order $(A, f) \leq (A', f')$ if $A$ is a subring of $A'$ and $f'|_{A} = f$.  The overall claim is that $(B, g)$ is a valuation of $K$.  We let $M$ be the unique maximal ideal of $B$ (which exists).  We take $x \in K$ with $x \neq 0$ and may assume that $M[x]$ is not the unit ideal of $B' = B[x]$(by a lemma) and so is contained in some maximal ideal $M'$.  Let $k = B/M$ and $k' = B'/M'$.  
The claim I don't understand: Since $k' = k[\bar{x}]$ for $\bar{x}$ the image of x in k' (which I see), $\bar{x}$ is algebraic over k.  

Suppose $\bar x$ is not algebraic over $k$.
Then $k[\bar x]$ is isomorphic to the polynomial ring $k[X]$.
However $k[X]$ is not a field(any polynomial of degree $\geq 1$ in $k[X]$ is not invertible in $k[X]$).
Hence $k[\bar x]$ is not a field.
This is a contradiction.
Hence $\bar x$ is algebraic over $k$.
