Dominant Maps and Generic Points

Proposition. Let $\varphi: A \to B$ and $f: \text{Spec} B \to \text{Spec} A$ be the induced map. Then $\text{Im} f$ is dense in $\text{Spec} A$ iff $\varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$
Equivalently, $\ker \varphi \subset \sqrt{0_A}.$

Proof.  Note that $\text{Im} f = \{ \varphi^{-1}(q) \mid q \in \text{Spec} B \}.$ So
$$\overline{\text{Im} f} = \mathbb{V}\left( \bigcap_{q \in \text{Spec} B} \varphi^{-1} q \right)$$
But $$\bigcap_{q \in \text{Spec} B} \varphi^{-1} q = \varphi^{-1} \left( \bigcap_{q \in \text{Spec} B} q\right) = \varphi^{-1}(\sqrt{0_B}).$$
Thus $$\overline{\text{Im} f} = \text{Spec} A = \mathbb{V}(\sqrt{0_A}) \iff \varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$$
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Note that for a set $S \in \text{Spec} B$, $$\overline{S} = \bigcup_{s \in S} \overline{\{s\}} = \bigcup \mathbb{V}(s) = \mathbb{V}\left( \bigcap \mathfrak{p}_s\right).$$

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If $\text{Spec} A$ and $\text{Spec} B$ are both irreducible, then their nilradicals are their generic points, so the above statement is equivalent to saying that $\text{Im} f$ is dense iff $f$ maps generic pt to generic pt. There is a generalization of this.

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Proposition. A rational map $\pi: X \dashrightarrow Y$ of irreducible schemes is dominant iff $\pi$ sends the genetic point $\eta_X$ of $X$ to the generic point $\eta_Y$ of Y. 

Proof. Let $(U, \pi)$ be a representation of $\pi$. Note that every open set of $X$ contains $\eta_X$.

Suppose $\pi(\eta_X) = \eta_Y$. Then $\pi(U) \ni \eta_Y$ so must be dense in $Y$.

Conversely, suppose that $\pi(U)$ is dense in $Y$.  Let $V\cong \text{Spec} B$ be an affine open of $Y$.   Cover $\pi^{-1}(V)$ by (finitely many) affine opens $U_i\text{Spec} A_i$. Then $\pi$ gives a morphism
$$\pi': \bigsqcup U_i \to V.$$
As $\pi$ is dominant, we must have image of $\pi'$ (which is equal to image of $\pi|_{\pi^{-1}(V)}$) is dense in $V$. On the other hand
$$\bigsqcup U_i = \bigsqcup \text{Spec} A_i = \text{Spec} \left(\prod A_i\right).$$ Let $A = \prod A_i$. Suppose $\varphi: B \to A$ is the induced map on global sections.

As the image of $\pi'$ is dense we know that $\varphi^{-1}(\sqrt{0_A}) \supset \sqrt{0_B}$, which is $\eta_Y$. On the other hand $(\sqrt{0_A}) = \bigcap_{p \in A_1, A_2, \ldots, A_n} p$  as $\text{Spec} A =  \bigsqcup \text{Spec} A_i$. Thus for each $A_i$, as $\bigcap_{p \in A_i} p \supset \bigcap_{p \in A_1, A_2, \ldots, A_n} p$ we have $\varphi^{-1}(\bigcap_{p \in A_i}) \supset \sqrt{0_B})$. But the former is just $\pi(\eta_X).$
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Note: Recall that every open subset of an irreducible space is irreducible.

Note: If $\eta$ is the generic point of an irreducible scheme $X$ and $U = \text{Spec} A$ is a affine open of $X$ then $\eta$ is the generic point of $U$ (i.e., corresponding to the nilradical of $A$).
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Quicker method: Suppose $\pi(\eta_X) \neq \eta_Y$. Then $U = Y - \overline{\pi(\eta_X)} \neq \emptyset$. As $\pi(X)$ is dense in $Y$, $U \cap \pi(X) \neq \emptyset$ so $\pi^{-1}(U)$ is not empty and thus must contain $\eta_X$, a contradiction.