# Can transcedence degree be defined for arbitrary ring homomorphism?

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?

• I'd say it makes sense. – Fernando Muro Oct 11 '13 at 11:23
• But I am not sure because I did not see such a definition anywhere which does not require either to be field or domain. – Qixiao Oct 11 '13 at 11:29
• You should be more confident in yourself! – Fernando Muro Oct 11 '13 at 11:32

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 $m$.

• Thanks for your clarification on the definition of algebraic.I think this is the point. Also I think we may use Jacobson's Proof with your clarification. The use of krull dim is unexpected. And thank you for your nice example! – Qixiao Oct 11 '13 at 23:32
• @mqx I am afraid of that the property 4 (Steinitz exchange axiom) does not hold for integral dependence in spite of the fact that $the\ best\ dependence\ is\ the\ integral\ one$ (used to say so to students). – Sasha Anan'in Oct 12 '13 at 0:45
• You're right the exchange axiom just hold for the algebraic case. Originally I was unable to check axiom 3 for algebraic case.. – Qixiao Oct 12 '13 at 14:44
• A question is I don't know why $MC$ is a prime ideal of $C$. I can show the quotient ring is a domain if ${b_1,.,b_n}$ are algebraic independent over A. But does it still hold if they are just integral independent? Could you explain it a little bit? – Qixiao Oct 12 '13 at 14:51
• $MC$ consist of all polynomials with coefficients in $M$ (remember, $C=A[b_1,\dots,b_n]$ is the ring of polynomials in $b_1,\dots,b_n$). So, $C/MC\cong(A/M)[b_1,\dots,b_n]$. As $A/M$ is a domain, a polynomial ring over it is also a domain, implying that $MC$ is prime. – Sasha Anan'in Oct 12 '13 at 15:05

This is a sidenote to Sasha's answer. The "yes" part can be proven in a completely elementary way without prime ideals and Krull dimension. Here is a sketch, as I have to prepare a talk for Monday and finish a paper for very soon:

Theorem 1. Let $$A$$ be a subring of a commutative ring $$B$$. Let $$n$$ and $$m$$ be two distinct nonnegative integers. Assume that $$C = A\left[b_1, b_2, ..., b_n\right]$$ and $$C' = A\left[b'_1, b'_2, ..., b'_m\right]$$ be two subrings of $$B$$ (with all $$b_i$$ and all $$b'_j$$ lying in $$B$$, obviously) such that $$b_1$$, $$b_2$$, ..., $$b_n$$ are algebraically independent over $$A$$ (that is, $$C$$ is the polynomial ring in $$b_1$$, $$b_2$$, ..., $$b_n$$ up to isomorphism) and such that $$b'_1$$, $$b'_2$$, ..., $$b'_m$$ are algebraically independent over $$A$$. Assume that both ring extensions $$B / C$$ and $$B / C'$$ are integral. Then, $$A$$ is the trivial ring (that is, $$A=0$$).

(This is stated in an unusual form for reasons of constructivism.)

Proof sketch for Theorem 1. Assume WLOG that $$n > m$$.

Let $$B'$$ be the $$A$$-subalgebra of $$B$$ generated by the $$b_i$$ and the $$b_j'$$. Then, $$A$$, $$C$$ and $$C'$$ are subrings of $$B'$$. Moreover, the ring extension $$B' / C$$ is integral (since a subring of a ring integral over $$C$$ is still integral over $$C$$), and similarly the ring extension $$B' / C'$$ is integral. Hence, we can replace the ring $$B$$ by $$B'$$ without anything else changing. So let us WLOG assume that $$B = B'$$. Thus, $$B$$ is the $$A$$-subalgebra of $$B$$ generated by the $$b_i$$ and the $$b_j'$$. Hence, $$B$$ is a finitely generated $$C'$$-algebra and integral over $$C'$$, therefore a finitely generated $$C'$$-module. Choose any finite generating set of the $$C'$$-module $$B$$, and throw in the element $$1$$. The resulting set is finite and generates $$B$$ as a $$C'$$-module. Denote this set by $$S$$. Let $$U$$ be the $$A$$-submodule of $$B$$ spanned by $$S$$.

For every polynomial algebra $$\mathfrak C$$ over $$A$$ (such as $$C$$ and $$C'$$) and any nonnegative integer $$i$$, let $$\mathfrak C_{\leq i}$$ denote the $$A$$-submodule of $$\mathfrak C$$ consisting of polynomials of degree $$\leq i$$. Recall that

(1) $$\mathfrak C_{\leq i}$$ is a free $$A$$-module of rank $$\dbinom{i+r}{r}$$, where $$r$$ is the number of indeterminates of the polynomial algebra $$\mathfrak C$$.

Also, $$\mathfrak C_{\leq a+b} = \mathfrak C_{\leq a} \mathfrak C_{\leq b}$$ for all nonnegative integers $$a$$ and $$b$$.

We have $$C' = \bigcup\limits_{p\geq 0} C'_{\leq p}$$ and thus $$C' U = \bigcup\limits_{p\geq 0} C'_{\leq p} U$$ (since $$C'_{\leq 0} \subseteq C'_{\leq 1} \subseteq C'_{\leq 2} \subseteq \cdots$$).

Since $$B$$ is generated by $$S$$ as a $$C'$$-module, while $$U$$ is the $$A$$-linear span of $$S$$, we have $$B = C' U = \bigcup\limits_{p\geq 0} C'_{\leq p} U$$.

Let $$\ell = \left|S\right|$$ (we know that $$S$$ is finite) and $$S = \left\lbrace s_1, s_2, ..., s_{\ell} \right\rbrace$$. Then, the $$A$$-module $$U$$ is spanned by the $$s_j$$ with $$j$$ ranging over $$\left\lbrace 1, 2, ..., \ell \right\rbrace$$ (since the $$A$$-module $$U$$ is spanned by $$S$$).

There are only finitely many products $$b_i s_j$$ with $$i \in \left\lbrace 1, 2, ..., n\right\rbrace$$ and $$j \in \left\lbrace 1, 2, ..., \ell \right\rbrace$$, and thus there exists some nonnegative integer $$p$$ such that these products all lie in $$C'_{\leq p} U$$ (since $$B = \bigcup\limits_{p\geq 0} C'_{\leq p} U$$). Fix such a $$p$$. Then,

(2) $$b_i U \subseteq C'_{\leq p} U$$ for all $$i \in \left\lbrace 1, 2, ..., n\right\rbrace$$

(because the $$A$$-module $$U$$ is spanned by the $$s_j$$ with $$j$$ ranging over $$\left\lbrace 1, 2, ..., \ell \right\rbrace$$, so that the $$A$$-module $$b_i U$$ is spanned by the $$b_i s_j$$ with $$j$$ ranging over $$\left\lbrace 1, 2, ..., \ell \right\rbrace$$).

Now, every nonnegative integer $$N$$ satisfies

(3) $$C_{\leq N} U \subseteq C'_{\leq pN} U$$.

Why is this so? Indeed, in order to prove (3), we need to show that $$b_{i_1} b_{i_2} ... b_{i_k} U \subseteq C'_{\leq pN} U$$ for every $$k\leq N$$ and any $$k$$-tuple $$\left(i_1,i_2,...,i_k\right) \in \left\lbrace 1,2,...,n\right\rbrace^k$$ (because such products $$b_{i_1} b_{i_2} ... b_{i_k}$$ span $$C_{\leq N}$$ as an $$A$$-module). In order to prove this, it is clearly enough to show that $$b_{i_1} b_{i_2} ... b_{i_k} U \subseteq C'_{\leq pk} U$$ for every $$k\leq N$$ and any $$k$$-tuple $$\left(i_1,i_2,...,i_k\right) \in \left\lbrace 1,2,...,n\right\rbrace^k$$ (because if $$k\leq N$$, then $$pk \leq pN$$ an thus $$C'_{\leq pk} U \subseteq C'_{\leq pN} U$$). This is shown by induction over $$k$$, using the induction step

$$b_{i_1} b_{i_2} ... b_{i_k} U = b_{i_1} b_{i_2} ... b_{i_{k-1}} \underbrace{b_{i_k} U}_{\subseteq C'_{\leq p} U\ \text{(by (2))}} \subseteq b_{i_1} b_{i_2} ... b_{i_{k-1}} C'_{\leq p} U$$

$$= C'_{\leq p} \underbrace{b_{i_1} b_{i_2} ... b_{i_{k-1}} U}_{\subseteq C'_{\leq p(k-1)} U\ \text{(by induction hypothesis)}} \subseteq \underbrace{C'_{\leq p} C'_{\leq p(k-1)}}_{= C'_{\leq p + p(k-1)} = C'_{\leq pk}} U = C'_{\leq pk} U$$.

Armed with (3), we can close in for the kill. Recall that $$1 \in S \subseteq U$$. Hence, every nonnegative integer $$N$$ satisfies

(4) $$C_{\leq N} \subseteq C_{\leq N} U \subseteq C'_{\leq pN} U$$ (by (3)).

Since $$C_{\leq N}$$ is a free $$A$$-module of rank $$\dbinom{N + n}{n}$$ (by (1)), there exists an $$A$$-module isomorphism from $$A^{\dbinom{N + n}{n}}$$ to $$C_{\leq N}$$. Due to (4), this yields that there exists an $$A$$-module injection from $$A^{\dbinom{N + n}{n}}$$ to $$C'_{\leq pN} U$$. Denote this injection by $$\phi$$.

But $$U$$ is generated by $$S = \left\lbrace s_1, s_2, ..., s_{\ell} \right\rbrace$$ as $$A$$-module. In other words, $$U = \sum\limits_{q=1}^{\ell} s_q A$$. Hence, $$C'_{\leq pN} U = \sum\limits_{q=1}^{\ell} C'_{\leq pN} s_q$$ is a quotient of the $$A$$-module $$\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$$. Since $$C'_{\leq pN}$$ is a free $$A$$-module of rank $$\dbinom{pN + m}{m}$$ (by (1)), the direct sum $$\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$$ is a free $$A$$-module of rank $$\ell \dbinom{pN + m}{m}$$. In other words, there exists an $$A$$-module isomorphism from $$A^{\ell \dbinom{pN + m}{m}}$$ to the direct sum $$\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$$. Hence, there exists an $$A$$-module surjection from $$A^{\ell \dbinom{pN + m}{m}}$$ to the $$A$$-module $$C'_{\leq pN} U$$ (because the $$A$$-module $$C'_{\leq pN} U$$ is a quotient of this direct sum). Denote this surjection by $$\psi$$.

The $$A$$-module $$A^{\dbinom{N + n}{n}}$$ is free and thus projective. Hence, the $$A$$-module injection $$\phi : A^{\dbinom{N + n}{n}} \to C'_{\leq pN} U$$ lifts (through the $$A$$-module surjection $$\psi : A^{\ell \dbinom{pN + m}{m}} \to C'_{\leq pN} U$$) to an $$A$$-module map $$\chi : A^{\dbinom{N + n}{n}} \to A^{\ell \dbinom{pN + m}{m}}$$ satisfying $$\phi = \psi \circ \chi$$. This $$A$$-module map $$\chi$$ is clearly injective again.

Since $$n > m$$, the polynomial $$\dbinom{x + n}{n} \in \mathbb Q\left[x\right]$$ has a higher degree than the polynomial $$\ell \dbinom{px + m}{m} \in \mathbb Q\left[x\right]$$. Hence, the former polynomial (having a positive leading coefficient) grows faster than the latter. Thus, there exists a nonnegative integer $$N$$ such that $$\dbinom{N + n}{n} > \ell \dbinom{pN + m}{m}$$. Fix this $$N$$.

But there is a well-known fact saying that if $$a$$ and $$b$$ are two nonnegative integers satisfying $$a > b$$, and if there is an injective $$A$$-module map $$\gamma : A^a \to A^b$$, then $$A$$ is the trivial ring. (This is equivalent to theorem (2) in Fred Richman's Nontrivial uses of trivial rings, Proceedings of the American Mathematical Society, vol. 103, no. 4, 1988, pp. 1012-1014, and part of Corollary 5.11 in Keith Conrad's Exterior powers.) Applying this fact to $$a = \dbinom{N + n}{n}$$, $$b = \ell \dbinom{pN + m}{m}$$ and $$\gamma = \chi$$, we conclude that $$A$$ is the trivial ring, qed. $$\blacksquare$$

Remark. We can use the above argument to prove a slightly stronger result:

Theorem 2. Let $$A$$ be a subring of a commutative ring $$B$$. Let $$n$$ and $$m$$ be two nonnegative integers such that $$n > m$$. Assume that $$C = A\left[b_1, b_2, ..., b_n\right]$$ and $$C' = A\left[b'_1, b'_2, ..., b'_m\right]$$ be two subrings of $$B$$ (with all $$b_i$$ and all $$b'_j$$ lying in $$B$$, obviously) such that $$b_1$$, $$b_2$$, ..., $$b_n$$ are algebraically independent over $$A$$ (that is, $$C$$ is the polynomial ring in $$b_1$$, $$b_2$$, ..., $$b_n$$ up to isomorphism). Assume that the ring extension $$B / C'$$ is integral. Then, $$A$$ is the trivial ring (that is, $$A=0$$).

Proof of Theorem 2. This proof is similar to the proof of Theorem 1 above, but the following changes need to be made:

• We no longer need to WLOG assume that $$n > m$$, because $$n > m$$ is already given by the assumptions.

• Our definition of $$\mathfrak{C}_{\leq i}$$ no longer gives us $$A$$-modules $$C'_{\leq i}$$, because $$C'$$ is not necessarily a polynomial algebra. We thus need to define $$C'_{\leq i}$$ differently. For any nonnegative integer $$i$$, we define $$C'_{\leq i}$$ to be the $$A$$-submodule of $$C'$$ consisting of all elements that can be written as polynomials of degree $$\leq i$$ in the generators $$b'_1, b'_2, \ldots, b'_m$$. This $$A$$-module $$C'_{\leq i}$$ might not be free, but

(5) it is finitely generated with $$\dbinom{i+m}{m}$$ generators

(namely, the $$\dbinom{i+m}{m}$$ monomials of degree $$\leq i$$ in the generators $$b'_1, b'_2, \ldots, b'_m$$). We can easily see that $$C'_{\leq a+b} = C'_{\leq a} C'_{\leq b}$$ for all nonnegative integers $$a$$ and $$b$$. Again, $$C' = \bigcup\limits_{p\geq 0} C'_{\leq p}$$ holds.

• Our definition of $$\psi$$ needs to be modified from the place on where I claim that $$C'_{\leq pN}$$ is a free $$A$$-module of rank $$\dbinom{pN + m}{m}$$. In fact, the $$A$$-module $$C'_{\leq pN}$$ is not necessarily free of rank $$\dbinom{pN + m}{m}$$ anymore. But it is finitely generated with $$\dbinom{i+m}{m}$$ generators (according to (5)), and thus is a quotient of a free $$A$$-module of rank $$\dbinom{pN + m}{m}$$. Hence, the direct sum $$\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$$ is a quotient of a free $$A$$-module of rank $$\ell \dbinom{pN + m}{m}$$. In other words, there exists an $$A$$-module surjection from $$A^{\ell \dbinom{pN + m}{m}}$$ to the direct sum $$\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$$. Hence, there exists an $$A$$-module surjection from $$A^{\ell \dbinom{pN + m}{m}}$$ to the $$A$$-module $$C'_{\leq pN} U$$ (because the $$A$$-module $$C'_{\leq pN} U$$ is a quotient of this direct sum). Denote this surjection by $$\psi$$. $$\blacksquare$$

Theorem 2 has a well-known corollary:

Corollary 3. Let $$A$$ be a subring of a commutative ring $$B$$. Let $$n$$ and $$m$$ be two nonnegative integers such that $$n > m$$. Let $$b_1, b_2, \ldots, b_n$$ be $$n$$ algebraically independent elements of $$B$$. Furthermore, let $$b'_1, b'_2, \ldots, b'_m$$ be $$m$$ elements of $$B$$ such that $$B = A\left[b'_1, b'_2, ..., b'_m\right]$$. Then, $$A$$ is the trivial ring (that is, $$A=0$$).

Proof of Corollary 3. Define a subring $$C$$ of $$B$$ by $$C = A\left[b_1, b_2, ..., b_n\right]$$. The ring extension $$B / B$$ is clearly integral. Hence, Theorem 2 (applied to $$C' = B$$) yields that $$A$$ is the trivial ring. This proves Corollary 3. $$\blacksquare$$

• Nice! I checked all the details (including those in the reference) and found that you successfully avoided the use of the axiom of choice. (What you wrote is not really a sketch. You might be much more sketchy and save time for preparing your talk.) Yes, most of the mathematics is constructive. (A sort of a counter-example is given by the answer $(\sqrt2^{\sqrt2})^{\sqrt2}=2$ to the question whether is it possible that $a^b$ is rational with $a,b$ irrational.) I have no option to accept your answer, so can only upvote. Have a nice talk! – Sasha Anan'in Oct 12 '13 at 6:12
• Yes, I started writing it as a sketch but then I ended up making so many mistakes that I could only convince myself of the proof being correct by extending it to this level of detail. Thanks for checking it! By "constructive" I mean not only avoiding the axiom of choice but also avoiding the tertium non datur, hence being entirely algorithmic. (Of course, the algorithmic value of Theorem 1 only becomes noticeable when, for instance, $A$ is constructed as a quotient ring of a ring $A'$ by some ideal $I$, in which case the triviality of $A$ gives an algorithm to write $1_{A'}$ as a linear ... – darij grinberg Oct 12 '13 at 15:58
• ... combination of the given generators of $I$.) – darij grinberg Oct 12 '13 at 15:58
• Sure, when checking, I had "constructive" in mind, "the axiom of choice" is just a bad wording. – Sasha Anan'in Oct 12 '13 at 16:05