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Prove function is in kernel of a morphism

Given $S,T$ algebraic varieties (over a field $\mathbb{K}$), $f : S\rightarrow T$ a morphism and $g : S\rightarrow \mathbb{K}$ a function.
Let $i$ be the kernel of $f$ and $j$ the kernel of $g$. Prove that $i\cap j$ is a kernel of $f\circ g$.

A:

I want to first start with a more general question: What’s the relationship between kernels of morphisms of varieties over a field $K$? If $f: X \rightarrow Y$, then $ker(f)$ can be identified with the variety of $K$-rational points of $X$ that map to the generic point of $Y$ under $f$. The usual argument can be used to show that this is a kernel of $f$ as a morphism of varieties over $K$. Similarly, we can define the kernel of $f$ to be the variety of $K$-rational points of $X$ that map to a point fixed by $Y$ (under $f$).
Now let’s get back to your question. Consider the following commutative diagram:
$$\require{AMScd} \begin{CD} i\cap j @>{}>> i @>{}>> i + j @>{}>> \mathbb{K}\\ @VV{}V @VV{}V @VV{}V @VV{}V\\ i @>{}>> j @>{}>> j + i @>{}>> \mathbb{K}\\ \end{CD}$$
This diagram represents a morphism of varieties over $K$ where $X$ and $Y$ are the varieties $i\cap j$ and $i + j$ respectively.
By the general results above, we have that $ker(f \circ g) = ker(g) = i \cap j$. By definition, this must also equal $ker(f) = i$. I’m having trouble proving that \$i \cap j = ker(

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