Fix a homomorphism $f:A\rightarrow B$.
Choose $\{b\_1,\dots,b\_n\}$, $\{b'\_1,\dots,b'\_m\}$ subsets of elements in $B$. Suppose that $B$ is algebraic over $f(A)[b\_1,\dots,b\_n]$ and $\{b\_1,\dots,b\_n\}$ are algebraically independent over $f(A)$. Suppose that $\{b'\_1,\dots,b'\_m\}$ satisfies the same condition. Does it imply $m=n$?


I checked the proof for an abstract dependence relation in Jacobson's Basic Algebra II (see 3.6). The condition $x<S<T\Rightarrow x<T$ is not satisfied.


But I cannot find a counter-example. Can anyone help me with that?



Can transcedence degree be defined for arbitrary ring homomorphism?
The problem is actually the following. Let $A\subset B$ be a subring and suppose that the subrings $C:=A[b\_1,\dots,b\_n]$ and $C':=A[b'\_1,\dots,b'\_m]$ of $B$ are the rings of polynomials in the indicated variables such that $B$ is algebraic over both. Does it follow $m=n$ ?


If "$B$ is algebraic over $C$" means
$\forall b\in B\ \ \exists 0\ne g(x)\in C[x]\quad g(b)=0$, the answer is **no**. Just take $B:=A[b\_1,b\_2,b']/\text{Ideal}(b\_1b',b\_2b')$, where $A[b\_1,b\_2,b']$ is the ring of polynomials in $b\_1,b\_2,b'$. It is easy to see that the images $C:=A[\overline b\_1,\overline b\_2]$ and $C':=A[\overline{b'}]$ are rings of polynomials respectively in $\overline b\_1,\overline b\_2$ and $\overline{b'}$ and that $\overline b\_1\overline b\_2B\subset C$ and $\overline{b'}B\subset C'$, implying that $B$ is algebraic over both $C$ and $C'$.


If "$B$ is algebraic over $C$" means $B$ is integer over $C$ (i.e., the above polynomial $g$ is always monic), the answer is **yes**. Here, you will need a bit of knowledge in commutative algebra. (Consult please some book, if necessary, say, "Commutative algebra" by M.F.Atiyah and I.G.Macdonald, the chapter about integer dependence.) You can relpace $B$ by the subring generated by $A$ and $\{b\_1,\dots,b\_n,b'\_1,\dots,b'\_m\}$. Let $M\triangleleft\_mA$ be a maximal ideal. Then $MC\triangleleft\_pC$ is a prime ideal in $C$ and $MC'\triangleleft\_pC'$ is a prime ideal in $C'$. By the "going up" theorem, there are prime ideals $P,P'\triangleleft\_pB$ such that $C\cap P=MC$ and
$C'\cap P'=MC'$. Taking the quotient $B/(P\cap P')$ in place of $B$, you can assume that $A$ is a field. As $B,C,C'$ are noetherian now, it remains to observe that the Krull dimensions of $B$, $C$, and $C'$ coincide, whereas that of $C$ and $C'$ equals respectively $n$ and $m$.


