In abstract algebra, the transcendence degree of a field extension $L/K$ is a certain rather coarse measure of the “size” of the extension. Specifically, it is defined as the largest cardinality of an algebraically independent subset of $L$ over $K$.
Questions tagged [transcendence-degree]
59 questions
24
votes
1 answer
Are all transcendental numbers a zero of a power series?
So I came across the concept of extending the notion of irrationality to higher degree polynomials. The base case of this is standard irrationality. That is, a number is irrational if it cannot be expressed as the ratio of two integers. This is…
Ozaner Hansha
- 707
10
votes
2 answers
Showing $[K(x):K(\frac{x^5}{1+x})]=5$?
Let $K$ be a field and $x$ be trnacendental over $K$. Compute
$[K(x):K(\frac{x^5}{1+x})]$.
I've never came across questions like these. It's easy to see that this degree is at most $5$, since: $$x^5-\alpha(x+1)=0$$ ($\alpha$ being…
35T41
- 3,601
6
votes
2 answers
Is the subextension of a purely transcendental extension purely transcendental over the base field?
Let $K/E/F$ be extension of fields, where $K/F$ is purely transcendental.
It is generally not true that $K/E$ is purely transcendental. For example, take $F(x)/F(x^2)/F$. I wonder what is the situation for $E/F$. Specifically, is $E/F$ purely…
fantasie
- 1,851
6
votes
2 answers
Show that $[k(t): k(t^4 + t) ] = 4$
Let $k$ be the field with $4$ elements, $t$ a transcendental over $k$, $F = k(t^4 + t)$ and $K = k(t).$ Show that $[K : F] = 4.$
I think I have to use the following theorem, but I'm not quite putting it together.
If $P = P(t), Q = Q(t)$ are nonzero…
user525033
5
votes
1 answer
Algebraic independence of elementary symmetric polynomials
I am following the book Lectures on Algebra by Abhyankar, p. 638.
Let $k$ be a field, $x_1,x_2,\ldots,x_n$ be independent variables, and define
$$e_1=x_1+x_2+\cdots+x_n\\
e_2=\sum_{i
Maths Rahul
- 3,293
4
votes
1 answer
Proper subfields of $\mathbb{C}$ isomorphic to $\mathbb{C}$
It is known that $\mathbb{C}$ has proper subfields which are isomorphic to $\mathbb{C}$, see this question; let $K$ be such subfield of $\mathbb{C}$.
Let $\iota: K \to \mathbb{C}$, $\iota(k)=k$ for all $k \in K$.
Now, I am confused:
(1) If $K…
user237522
- 7,263
4
votes
0 answers
Concerning $k \subset L \subset k(x,y)$
The following is a known result in algebraic geometry:
Let $k$ be an algebraically closed field of characteristic zero (for example, $k=\mathbb{C}$).
Let $L$ be a field such that $k \subset L \subset k(x,y)$
and $L$ is of transcendence degree two…
user237522
- 7,263
4
votes
1 answer
Transcendence degree and Krull dimension of finitely generated algebras
Let $K$ be a field, and let $a_1,\dots,a_{n+1}$ be $n+1$ elements of a finitely generated $K$-algebra $A$ of Krull dimension $n$.
Are the elements $a_1,\dots,a_{n+1}$ always algebraically dependent over $K$?
I.e: Are the monomials…
Pierre-Yves Gaillard
- 20,446
3
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2 answers
Is $\mathbb{R}(x,\sqrt{1+x^2})$ a purely transcendental extension of $\mathbb{R}$?
I'm working on Exercise 9, Section 19 of Morandi's Field and Galois Theory:
If $K=\mathbb{R}(x,\sqrt{1+x^2})$, show that there is a $t\in K$ with $K=\mathbb{R}(t)$.
Trying to deduce a workable $t$, I assume if $K=\mathbb{R}(t)$ for some $t$, then…
Adelaide Dokras
- 1,703
3
votes
1 answer
$L/K$ is algebraic iff its transcendence degree is zero
On Wikipedia, I read that
A field extension $L/K$ is algebraic if and only if its transcendence degree is $0$.
$[\Rightarrow]$ Suppose $L/K$ is an algebraic field extension, i.e., for every $l \in L$, there exists a non-zero polynomial $g(X)$ in…
stoic-santiago
- 14,877
3
votes
2 answers
Proof verification: transcendence degree additive in towers
I am trying to prove that if $k\subseteq E\subseteq F$ are field extensions, then $$\text{tr.deg}_k F=\text{tr.deg}_k E+\text{tr.deg}_E F.$$
If $A=\{a_1,\ldots, a_n\}$ is a transcendence basis for $E$ over $k$ and $B=\{b_1,\ldots, b_m\}$ is a…
SVG
- 4,283
3
votes
1 answer
Can you determine the transcendence degree of an algebra by looking at a generating set?
Let $K$ be a field and $A$ be a $K$-algebra generated (as $K$-algebra) by a set $S$. The transcendence degree of $A$ is$$
\operatorname{trdeg}(A) = \sup\{|T| : T \subset A,\, T \text{ algebraically independent}\} \quad .
$$
Setting
$$
N = \sup\{|T|…
Carlos Esparza
- 3,747
3
votes
1 answer
Dimension of scheme of finite type over a field under base change (Hartshorne Ex. II.3.20)
Consider an integral scheme $X$ of finite type over a field $k$. If $k\subseteq k'$ is a field extension, then the scheme $X' = X\otimes_k k'$ is not necessarily integral. For instance, take $X = \operatorname{Spec} \mathbb{R}[x,y]/(x^2+y^2)$ over…
2
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Schanuel's Conjecture $\implies $ no surprises on the integers. $2^t+3^t=1 \implies t\notin\overline{\mathbb{Q}}$
Simple Question:
Is the proof given below correct?
Let $t$ satisfy $2^t+3^t=1$. We have
$t \approx -0.787884911025869783628555917298434738269083137354182194199 \dots$ according to wolfram alpha.
Then $t$ is demonstrably transcendental assuming…
Mason
- 4,489
2
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0 answers
Find a field extension $K=k(x,y)$ of transcendence degree 1 where $x\notin k(y)$, $y\notin k(x)$ and $K|k$ is purely transcendental
I saw this question online and it's actually a true or false question, the question is:
True of False: Let $K=k(x,y)$ be a field extension with transcendence degree 1. If $x\notin k(y)$ and $y\notin k(x)$, then $K|k$ isn't purely transcendental
I…
user8785084
- 347