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Let $B$ be an integral ring extension of $A$ and $B$ is an integral domain. Let $f:B \to C$ be a ring homomorphism to any commutative ring $C$ such that the restriction of $f$ to $A$ is injective. Then I have to show that $f:B \to C$ is also injective.

It is clear from the diagram that $\ker(f) \cap A= (0).$ How can I show that $\ker(f)=(0)$ ? Help me.

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Suppose $b\in \mathrm{ker}f$. $b$ is integral over $A$, so $$b^n + \alpha_{n-1}b^{n-1}+\dots+\alpha_0=0$$ for $\alpha_i\in A$. Assume $b\neq0$. Then we can assume $\alpha_0\neq 0$. But $f(b)=0$, so by applying $f$ to the above equation we find $f(\alpha_0)=0$, so $\alpha_0=0$ by injectivity of the composite map. This yields a contradiction, hence $b$ must be zero.

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