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Let $u \in \mathbb{K}(X), \ u = \frac{X^3}{X+1}$. Prove that $\mathbb{K}(X)\supset \mathbb{K}(u)$ is an algebraic extension and find $[\mathbb{K}(X):\mathbb{K}(u)]$.

My attemps were trying to show that is a finite extension. If I can prove that then automatically is an algebraic extension. What I have tried to do is to write every element of $K(X)$ as lineal combinations of $\{1,u,u^2,u^3,..,u^n\}$ but it seems impossible.

HFKy
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The minimal polynomial of $X$ over $\mathbb K(u)$ is seen by inspection to be $ z^3-uz-u, $ whereupon $\mathbb K(X)$ is a degree $3$ algebraic extension of $\mathbb K(u)$.

If this is confusing to you due to the fields containing indeterminates, think by analogy with a field extension of $\mathbb Q$ by a root of $z^3-z-1$.

pre-kidney
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  • I'm sorry I had a typo in ny answer. Anyways, you find an element that is algebraic, that does not implies that every element is algebraic, not in this case, because $\mathbb {K}(X) $ is the set of rational functions! Or Am i wrong? – HFKy Jun 20 '19 at 11:12
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    I think I get it! My problems was that I had never interpreted that $\mathbb{K}(X)$ is the minimal field containing $\mathbb{K}(u)$ and $X$. Knowing this, is clear that if you find a polinomial in $\mathbb{K}(u)[X]$ whose root is $X$ then you finish. – HFKy Jun 20 '19 at 22:43
  • However, how do you see that $z^3 -uz -u$ is irreducible over $\mathbb{K}(u)$? – HFKy Jun 21 '19 at 02:22
  • Every reducible cubic has a linear factor, say $z-t$ for some $t\in \mathbb K(u)$. Then $t^3=ut+u$, but this is impossible by degree considerations (the degree of the left side is $3\deg_u(t)$ and the degree of the right side is $\deg_u(t)+1$, which can never be equal to each other. – pre-kidney Jun 21 '19 at 03:41
  • Technically the degree of the right side is $\max(\deg_u(t)+1,1)$, since $\deg_u(t)$ could be negative, but the same argument goes through. – pre-kidney Jun 21 '19 at 03:47