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We do not impose homomorphism must map 1 to 1. The two rings are commutative, unital rings. How to show that the preimage of a prime ideal must not be the whole ring/domain? or show a counter example.

Everything single proof I saw impose $f(1)=1$ for ring homomorphisms but I'm yet to see a counter example where both rings are unital, commutative.

Any help appreciated!

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Consider $f:\mathbb{R} \to \mathbb{R}$ given by $f(x):=0$ for any $x \in \mathbb{R}$. Note that the zero ideal is a prime ideal in $\mathbb{R}$, but its preimage is the whole ring $\mathbb{R}$. This gives a counterexample when 1 is not required to be mapped to 1.

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Let $F_2$ be the field of two elements and consider $R=F_2\times F_2$. Then $x\mapsto (x,0)$ is such a homomorphism from $F_2\to R$.

It maps onto a prime ideal of $R$ (indeed, a maximal ideal) and its inverse image is clearly all of $F_2$.

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