Let $A \subseteq B$ be two commutative $\mathbb{C}$-algebras. Let $q$ be an ideal of $B$ and denote $p=q \cap A$.
By [Lemma 3][1], if $A \subseteq B$ is an integral extension, then: $q$ is maximal in $B$ if and only if $p$ is maximal in $A$.
If $A \subseteq B$ is not necessarily integral, when is it possible to deduce the following half claim: If $p$ is maximal (prime) in $A$, then $q$ is maximal (prime) in $B$.
I guess that this does not hold in general. Does one of the following additional conditions help: (i) flatness of $A \subseteq B$; (ii) faithful flatness of $A \subseteq B$, see [this][2]; (iii) separability of $A \subseteq B$; (iv) $A$ and $B$ being UFD's.
Thank you very much!
Edit: One direction of the proof of Lemma 3 says: "If $p$ is maximal then $A/p$ is a field and every element of $B/q$ is algebraic over $A/p$. Hence $B/q$ is a field" (so $q$ is maximal).
Therefore, if I am not wrong (but I may be wrong), the following claim is true: If $A \subseteq B$ is algebraic, then: If $p$ is maximal in $A$, then $q$ is maximal in $B$.
Reason: If I am not worng, exactly the same proof holds; the critical line is: and every element of $B/q$ is algebraic over $A/p$.
Am I right?
strong text [1]: http://math.stanford.edu/~conrad/210BPage/handouts/math210b-going-up.pdf [2]: Extensions and contractions of prime ideals under integral extensions
For $(1), (2), (4)$, take $k \subseteq k[x]$ for any field. This is a free (a fortiori faithfully flat) extension of UFDs in which the non-maximal ideal $0$ of $k[x]$ contracts to the maximal ideal $0$ of $k$.
I highly doubt that separability is even remotely related to the property in question, and will politely decline to think about it until you provide some motivation.
As for algebraic..... no, even if we restrict attention to $A, B$ domains this doesn't work out. The key property of integral extensions in your "lemma 3" is that they are stable under quotients, but such is not the case for algebraic extensions.
For example $\mathbb{C}[x,xy] \subseteq \mathbb{C}[x,y]$ is an algebraic extension but the height $1$ (non-maximal) prime $(x)\mathbb{C}[x,y]$ contracts to the maximal ideal $(x,xy) \mathbb{C}[x,xy]$.