# Algebraic extension and Krull dimension

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.

• The definition of algebraicity with comaximal elements is more general than the usual definition. So the answer should be “yes”. Aug 17, 2015 at 8:46
• so the answer to the question $\mathrm{dim}(B) \leq \mathrm{dim}(A)$ is yes if $B$ is algebraic over $A$?
– Zoey
Aug 17, 2015 at 8:54
• Yes, if $B$ is algebraic over $A$, then $B$ is also algebraic over $A$ in the sense of the paper. Aug 17, 2015 at 9:02
• @user218931 I'm afraid I didn't get it: why "The definition of algebraicity with comaximal elements is more general than the usual definition."? Aug 17, 2015 at 15:37
• For an element $x$ to be algebraic over $A$ in the “comaximal“ sense, means that there exist elements $a_0,\dotsc,a_n\in A$ such that $(a_0,\dotsc,a_n) = (1) = A$ (the comaximality condition) and $a_nx^n + a_{n-1}x^{n-1}+\dotsb+a_1x+a_0=0$. If an element $x$ is algebraic (in the usual sense), then there exist $a_0,\dotsc,a_{n-1}\in A$ such that $x^n + a_{n-1}x^{n-1} + \dotsb+ a_1x+a_0 = 0$. But then $1,a_{n-1},\dotsc,a_0$ are obviously comaximal. Aug 17, 2015 at 18:44

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 $\dim B\le\dim A$.

• Here we don't need comaximality?
– Zoey
Aug 18, 2015 at 2:27
• @Zoey Yes, we do need comaximality because we need the extension $A/p\subset B/P$ ($P\cap A=p$) to be algebraic, so not all coefficients of a polynomial $f\in A[X]$ which has a root from $B$ should become zero in $A/p$. Aug 18, 2015 at 6:36
• @Zoey Suppose $P_1\cap A=P_2\cap A=p$, and $P_1\subset P_2$. Now use the extension of integral domains $A/p\subset B/P_1$ and deduce that $A\subset B$ satisfies INC. (This also shows why we consider integral domains as a first step.) Aug 18, 2015 at 6:39
• @Zoey If $A\subset B$ has INC, then a chain of primes in $B$ contracts to a chain of primes in $A$ and no equality occurs. Aug 18, 2015 at 6:44
• @Zoey 1. I've used the extension $S^{-1}A\subset S^{-1}$ since I've noticed that it is integral which is more convenient. 2. I've also wondered if the inequality $\dim B\le\dim A$ holds for algebraic extensions of integral domains, but I don't have a clear answer. (I suspect the answer is still negative.) Aug 20, 2015 at 12:25