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Assume that:

(1) $A \subsetneq B$ are integral domains and finitely generated algebras over a field $k$ ($k$ is algebraically closed of characteristic zero, if this helps).

(2) $A$ is algebraically closed in $B$.

(3) The field of fractions of $A$, $Q(A)$, has transcendence degree over $k$ one less than the transcendence degree of $Q(B)$ over $k$, that is, $\dim A=\dim B-1$.

My question: Is it true that $Q(A)$ is algebraically closed in $Q(B)$? If not, it would be nice to have a counterexample. (If I am not wrong, $k[x] \subsetneq k[x,y]$ is an example to my question; just use Exercise 1.3.)

I would appreciate any help in solving my question.

Edit: Perhaps Exercise 1.4 is a counterexample to my question; I am not sure if it satisfies my assumption (3) or not.

user26857
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user237522
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  • $A$ and $B$ are $k$-algebras? Of finite type? – paf Aug 12 '16 at 00:51
  • Yes, $A$ and $B$ are $k$-algebras. Yes, $A$ and $B$ are finitely generated as $k$-algebras. – user237522 Aug 12 '16 at 00:54
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    What do you mean by "$A$ is algebraically closed in $B$"? – Rob Arthan Aug 12 '16 at 01:08
  • If $b \in B$ is algebraic over $A$ (namely, $a_n b^n+ a_{n-1} b^{n-1} + \ldots + a_1 b + a_0 = 0$ for some $a_0,\ldots,a_n \in A$), then $b \in A$. – user237522 Aug 12 '16 at 01:12
  • Can you explain your example? I believe that in your definition the product $a_1 \cdots a_n \neq 0$. Why is $y$ "algebraic" over $k[x]$? Also, what is your reference for two rings to be algebraic and what is some basic properties of such extension? In general, there is something called an integral extension whose definition is the same of the one you gave with a_n = 1. – Youngsu Aug 12 '16 at 01:22
  • I will try to explain my example: $k[x] \subsetneq k[x,y]=k[x][y]$. $y$ is transcendental over $k[x]$, so it is not relevant to assumption (2). What I claimed is that if $h(x,y) \in k[x,y]$ is algebraic over $k[x]$, then it is already in $k[x]$ (I think I can prove this). Generally, $A \subset B$ is an algebraic ring extension if every element $b \in B$ is algebraic over $A$. – user237522 Aug 12 '16 at 01:32
  • @Youngsu: Please note that I did not assume that $A \subsetneq B$ is an algebraic ring extension. I assumed that if an element in $B$ is algebraic over $A$, then it is already in $A$; there may exist in $B$ transcendental elements over $A$. – user237522 Aug 12 '16 at 01:41
  • Thank you. I get it now. And thanks for the reference. – Youngsu Aug 12 '16 at 03:11
  • :-) and thank you for trying to help. – user237522 Aug 12 '16 at 03:19
  • The example 1.4 in the link is a counter example to your question. For instance, look at Proposition 11.31 in http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/chapter-11.pdf. – Youngsu Aug 15 '16 at 05:16
  • Thanks! I like the idea of showing that the (Krull) dimension of $R=\mathbb{Q}[x,y]/(x^2+y^2)$ is $1$ (one has to show that $(x^2+y^2)$ is a height $1$ prime), hence the transcendence degree of $Q(R)$ over $\mathbb{Q}$ is also $1$. This is indeed a counterexample, with $k=\mathbb{Q}$. Can you find a counterexample with $k$ algebraically closed? If we change $\mathbb{Q}$ to $\mathbb{C}$ we do NOT get a counterexample, since $\mathbb{C}$ is algebraically closed in $\mathbb{C}(t)$. http://math.stackexchange.com/questions/1889765/what-is-the-field-of-fractions-of-mathbbqx-y-x2y2 – user237522 Aug 15 '16 at 09:28
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    $B$ is no longer a domain. – Youngsu Aug 15 '16 at 21:51
  • One more try: $A = \mathbb{C}[x,y]$, $B = \mathbb{C}[x,y,z,w]/(xz^2 - yw^2)$ and take $\frac{z}{w}$ algebraic over $A$ but not in $\operatorname{Frac}(A)$ – Jay Aug 15 '16 at 22:15
  • @Youngsu Oops, my mistake, you are right $B=\mathbb{C}[x,y]/(x^2+y^2)$ is not an integral domain, since $(x^2+y^2)$ is not a prime ideal (it contains $(x-iy)(x+iy)$, but none of $x-iy, x+iy$). – user237522 Aug 16 '16 at 08:09
  • @Jay I will have to think about your counterexample, namely, to check if I understand why assumptions (2) and (3) are satisfied (at the moment, I only see that $z/w$ is algebraic over $\mathbb{C}(x,y)$). If and when you have time to elaborate a little in an answer- that would be nice. – user237522 Aug 16 '16 at 08:25

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The example I gave above is $A = \mathbb{C}[x,y]$, $B = A[z,w]/(xz^2 - yw^2)$. $xz^2 - yw^2$ is irreducible so $B$ is an $\mathbb{N}$-graded domain with $A$ as the degree $0$ part. So $A$ is algebraically closed in $B$ by Exercise 1.3 in your link (basically anything in $B \setminus A$ can't satisfy a polynomial over $A$ since the highest degree term always survives as $B$ is a domain). Also your assumption (3) is just that $\dim A = \dim B - 1$ which holds here. But $Q(A)$ is not algebraically closed in $Q(B)$ e.g. $(z/w)^2 = y/x$ is in $Q(A)$ but $z/w$ is not in $Q(A)$.

Jay
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  • Thank you very much! Please: (1) Do you have a reference to the general claim you applied to get that $B$ is an $\mathbb{N}$-graded domain? (2) Why $xz^2-yw^2$ is a height one prime in $\mathbb{C}[x,y,z,w]$? (in order to get that $dimB=3$). Sorry if my questions are trivial. – user237522 Aug 17 '16 at 02:06
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    Happy to help! I think a general statement is that if $S$ is an $\mathbb{N}$-graded ring with degree $0$ part $A$ and $I$ is a graded $S$-ideal such that $I \cap A = 0$ then $S/I$ is also an $\mathbb{N}$-graded ring with degree $0$ part $A$. You can apply this to $S = A[z,w]$ and $I = (xz^2 - yw^2)$ in this case. Next, $xz^2 - yw^2$ is irreducible so $I$ is a prime ideal. Then the principal ideal theorem tells you $I$ has height at most one. But the height can't be zero because $I$ lives in a domain – Jay Aug 17 '16 at 04:51
  • Thanks for the additional helpful explanations! – user237522 Aug 17 '16 at 12:52