Why is acl not modular in this case?

Question

Given $\cal{M}=$(M,I)$ \models ACF_{0}$, show that $acl$ is not modular in $\cal{M}$.

($ACF_{0}$ denotes the theory of algebraically closed fields of characteristic zero).

So M is the underlying set, I is the interpretation of the language. For modularity, given $A,B\subseteq $M and some finite $C\subseteq $M such that $A\subseteq acl(B\cup C)$, it is the case that $dim(A|B\cup(acl(A)\cap acl(B)))$ =$dim(A|(acl(A)\cap acl(B)))$.

I sought a contradiction, and proceeded by assuming that $acl$ was modular in the structure. I considered bases of each side of the equation (and assumed that they had the same dimension due to modularity), and tried to simplify each side, but ran into problems since the union with $B$ ends up not really adding any constraints to the LHS of the expression so all I could find was that no element in the bases could be in $acl(B)$.

Please advise as to whether this is the correct approach, or whether I need to view this differently. I feel that my current attempt is at a dead end. Thank you.

Answer 1

It is important to have in mind that in this situation, dimension means transcendence degree.

Let $a,b,c$ be three algebraically independent transcendentals, and consider $d=a-bc\Rightarrow a=bc+d$.

Pick $A=\{a,c\}$, $B=\{b,d\}$. Then we have:

• $acl(A)\cap acl(B)=\overline{\mathbb{Q}(a,c)}^{alg}\cap \overline{\mathbb{Q}(b,d)}^{alg}=\overline{\mathbb{Q}}^{alg}$, by the algebraic independence (notice that if $P(a,b,c,d)=0$ for some polynomial with coefficients in $\mathbb{Q}$, then $P(a,b,c,a-bc)=0$, contradicting the algebraic independence of $a,b,c$).
• $\dim(A/(B\cup (acl(A)\cap acl(B)))=\dim(\{a,c\}/\{b,d\}\cup \mathbb{Q})=1$, because once you have $b,d,c$ we can produce $a$.
• $\dim(A/(acl(A)\cap acl(B))=trans.deg(\mathbb{Q}(a,c)/\mathbb{Q})=2$, because $a$ and $c$ are algebraically independent.

So, $M$ is not modular.