Let $F/K$ be a field extension of transcendence degree $r$ and let $\{x_1,\ldots,x_n\}$ be a set of elements which are algebraically independent over $K$. Then is it true that $tr.deg. F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)=r$?

If $\{t_1,\ldots,t_r\}$ is a transcendence base of the extension $F/K$ then we know $F/K(t_1,\ldots,t_r)$ is algebraic and $K(t_1,\ldots,t_r)/K$ is purely transcendental. Thus $F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)(t_1,\ldots,t_r)$ is also algebraic. Now it will be enough to show that $\{t_1,\ldots,t_r\}$ is algebraically independent over $K(x_1,\ldots,x_n)$. How should I approach?

My above queries are from the book of Lang's Algebra book in the context of being free from one field to another. I have added a photo from Lang's Algebra book: I want to understand why $tr. deg. F(y)/K(y)=r$ as shown in the diagram.

enter image description here

  • 1
    $\begingroup$ As written, $\{x_1,\ldots,x_n\}$ are merely algebraically independent over $K$, hence they might be the same as $\{t_1,\ldots,t_r\}$, making the claim false. $\endgroup$ Jul 29, 2021 at 4:26
  • $\begingroup$ What Hagen said. You need $\{x_1,\ldots,x_n\}$ to be algebraically independent over $F$ to guarantee this. $\endgroup$ Jul 29, 2021 at 4:33
  • $\begingroup$ @JyrkiLahtonen I have added a screenshot of the context of my queries. $\endgroup$
    – user371231
    Jul 29, 2021 at 5:30
  • $\begingroup$ The screenshot is missing something essential. What are $K$ and $k$ there? Also, apparently $F$ is not a random extension field but something dependent on the context. The original version of your question made it look like the claim is made for an arbitrary extension $F$, which is why Hagen von Eitzen and I protested. $\endgroup$ Jul 29, 2021 at 5:54
  • $\begingroup$ For example, the excerpt says that $F$ is a subfield of $K$, and the claim is about $F(y)/k(y)$ rather than $F(y)/K(y)$. In other words, your question, as it is currently phrased, does not match the excerpt at all. This may be the source of your misunderstanding of the presentation. $\endgroup$ Jul 29, 2021 at 5:56

1 Answer 1


If I understood the question correctly, it is about why the extension $F(y)/k(y)$ described in the excerpt has transcendence degree $r$.

This follows from the assumptions made in that claim. Let $(t):=\{t_1,t_2,\ldots,t_r\}$ be a transcendence basis of $F/k$. Then $(t)$ is a subset of $K$ that is algebraically independent over the base field $k$. Because $K$ was assumed to be free from $L$ over $k$, it follows that $(t)$ remains algebraically independent over $L$. Therefore $(t)$ is also algebraically independent over the subfield $k(y)\subseteq L$. Therefore the transcendence degree of $F(y)/k(y)$ is at least $r$. Because $F(y)$ is algebraic over $k(y)(t)$ the transcendence degree cannot be higher. Hence the transcendence degree of $F(y)/k(y)$ is exactly $r$ as the diagram claims.

The assumption of freeness of $K$ from $L$ is absolutely essential in this argument, and the claim may be false otherwise.

In the prescribed context $F$ is a subfield of $K$, so the title question is a bit strange.

  • $\begingroup$ Thanks for the explanation. I can’t make a general statement like what i made earlier. One has to use the fact $K$ is free from $L$ over $k$. $\endgroup$
    – user371231
    Jul 29, 2021 at 6:40
  • $\begingroup$ @user371231 Yeah. I recall details like this keeping me on my toes the first time I read about it. Lang makes the analogy with linearly disjoint extension. IIRC I only learned about those later, but it makes great pedagogical sense to draw attention to the analogy. Easier not to drop the ball when keeping that in mind. $\endgroup$ Jul 29, 2021 at 7:43
  • $\begingroup$ I am curious to know what is the meaning of IIRC? $\endgroup$
    – user371231
    Jul 29, 2021 at 12:23
  • $\begingroup$ @user371231: If I Recall / Remember Correctly. $\endgroup$
    – Paramanand Singh
    Jul 29, 2021 at 12:46

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