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I have to prove the following, but I have no idea were to start.

Let $K$ be a field and $y$ transcendental over $K$. Then:

  1. $y^n$ with $n\in \mathbb{N}$ is transcendental in $K$.
  2. $K(y)$ is an algebraic extension of $K(y^n)$ with degree $[K(y):K(y^n)]=n$.
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    1) Do you mean "let $K$ be a field"? 2) Welcome to M.SE. What have you tried? Where are you stuck? – 2017-01-18
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    1) yes, I mixed up the translation 2) thanks, I was missing some kind of ansatz to solve it. For some reason I didn't think of assuming $y^n$ is algebraic and disproving that. Seems rather obvious after the other answers. – 2017-01-18

2 Answers 2

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  1. If $y^n$ is algebraic over $K$, then there exists a monic polynomial $f(x) \in K[x]$ such that $f(y^n)=0$. But then $g(x):=f(x^n)\in K[x]$ is such that $g(y)=0$, which contradicts the fact that $y$ is transcendental over $K$.

  2. The polynomial $f(x):=x^n-y^n$ is irreducible over $ K(y^n)$, and $f(y)=0$. Thus, $y$ is algebraic over $ K(y^n)$ and $[K(y):K(y^n)]=\deg f=n$.

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Here is the answers:

  1. Suppose the contrary, that $y^n$ is not transcendental for some $n\in N$. Then, it must satisfy some algebraic equation, e.g. there is such k and such $a_0,...,a_k$ so that $a_0+a_1y^n+...+a_k(y^n)^k = a_0+a_1y^n + ...+ a_ky^{nk} =0$. Note that the latter equation is a polynomial of y being zero, i.e. a contradiction because y is transcendental Are you sure about the other thing?? I doubt this holds...