The following is the problem:
Let $K$ be a field, and $A$ is a commutative ring containing $K$. $\phi:A \rightarrow K[X]$ is a ring homomorphism which is the identity on $K$. If $M$ is a maximal ideal of $K[X]$, show that $\phi^{-1}(M)$ is a maximal ideal of $A$.
I tried to prove that $A/\phi^{-1}(M)$ is a field. Since $K[X]$ is a PID, $M$ should be generated by some irreducible polynomial $g(X)\in K[X].$ Because the indeterminate $X$ is transcendental over $K$, $\phi^{-1}(X)$ should be another indeterminate, say $Y$ in $A$. Hence $\phi^{-1}(M)$ is generated by the irreducible polynomial $g(Y)\in K[Y]$, which makes $A/\phi^{-1}(M)$ isomorphic to a simple extension $K(\alpha),$ where $\alpha$ is a root of $g(Y).$ So $A/\phi^{-1}(M)$ is a field and $\phi^{-1}(M)$ is a maximal ideal.
Are there any mistakes in my argument? Please let me know if there are any problems in it. Thanks a lot.
If $A$ is a commutative ring and $f\in A[X]$ is a monic polynomial, then the ring extension $A\subset A[X]/(f)$ is integral.
In your case $K\subset K[X]/M$ is integral (algebraic) and then every intermediate ring is a field, so $A/M\cap A$ is a field.