# If $tr. deg. F/K=r$ then $tr.deg. F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)=r$.

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. • 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. Jul 29, 2021 at 4:26
• What Hagen said. You need $\{x_1,\ldots,x_n\}$ to be algebraically independent over $F$ to guarantee this. Jul 29, 2021 at 4:33
• @JyrkiLahtonen I have added a screenshot of the context of my queries. Jul 29, 2021 at 5:30
• 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. Jul 29, 2021 at 5:54
• 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. Jul 29, 2021 at 5:56

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.

• 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 $k$. Jul 29, 2021 at 6:40
• @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. Jul 29, 2021 at 7:43
• I am curious to know what is the meaning of IIRC? Jul 29, 2021 at 12:23
• @user371231: If I Recall / Remember Correctly. Jul 29, 2021 at 12:46