Let $A \subseteq B$ be an extension where $A,B$ are Noetherian, commutative rings. If $B$ is algebraic over $A$, can we say that $\dim B\leq\dim A$?
Just read the following paper "Constructive Krull Dimension. I : Integral Extensions" that proves $\dim B\leq\dim A$ for $A,B$ commutative rings but the definition of algebraic is different. It says that $x \in B$ is algebraic if there exist comaximal elements $a_0, \ldots, a_k \in A$ such that $\sum_{i=0}^k a_i x^i = 0$. I didn't understand the whole proof but is there an intuitive proof for why comaximal is required in the definition of algebraic?
As far as I thought algebraic meant for every $b \in B$ there exists a non zero $f \in A[x]$ such that $f(b) = 0$. There was no requirement that coefficients in $f$ needed to be comaximal.
If one uses the usual definition of algebraic then the claim is wrong: set $A=\mathbb Z_4$, and $B=\mathbb Z_4[X]/(2X+2)$. We have $\dim A=0$ and $\dim B=1$.
However, if one uses the definition of algebraic from the quoted paper, then it's straightforward to show that $\dim B\le\dim A$.
To begin with, we assume that $A$ and $B$ are integral domains. For $b\in B$ there are $a_n,\dots,a_0\in A$ ($a_n\ne0$) such that $$a_nb^n+\cdots+a_1b+a_0=0.$$ This shows that $S^{-1}A\subset S^{-1}B$ is an integral extension of integral domains. (Here $S=A-\{0\}$.) Since $S^{-1}A$ is a field it follows that $S^{-1}B$ is also a field, and therefore $(0)$ is the only prime ideal of $B$ lying over $(0)$.
Now we deduce that the ring extension $A\subset B$ has the incomparable property, that is, if $P_1\cap A=P_2\cap A$ then $P_1$ and $P_2$ are incomparable with respect to the inclusion, and this entails $\dim B\le\dim A$.