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Let $A \le B$ be rings. Suppose that there exists a unique ring homomorphism $ \phi : B \rightarrow A$ such that $ \phi (a) = a$ for all $a \in A$. Does it follow that $A=B$?

I proved that if $B$ is an integral domain and $A \neq B$ then every element in $B \setminus A$ is transcendental over $A$:
Let $x \in B$ be integral over $A$. Let $y=x- \phi (x)$. Then $y$ is integral over $A$ and $\phi (y) =0$. Let $f \in A[T]$ be monic polynomial of minimal degree such that $f(y)=0$. Then $f$ has constant term $0$, so $f(T)=T$. Hence $x=\phi (x) \in A$.

user1142333
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    In this forum you should post your attempts to solve a problem or a problem. You can't just give a problem and ask to solve it! – Dmitry Apr 07 '23 at 20:34
  • @Dmitry The OP added a valuable attempt. – Anne Bauval Apr 07 '23 at 22:24
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    One counterexample would be A reduced (e.g., a field k) and B = A[nilpotent things] (e.g., k[x]/(x^2)) with the obvious inclusion of A. I can’t say much about the case where B is specifically an integral domain. – Rafi Apr 08 '23 at 00:30
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    Another genre of counterexample comes from playing off of separability issues by considering things like A = ₚ(t)[x]/(x^p-t), B = ₚ(t)[x, y]/(x^p - t, y^p - t), again with the obvious inclusion. – Rafi Apr 08 '23 at 00:41

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