Galois Theory - Question about Radical Extensions

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Homework Statement



Suppose that I\subseteq J are subfields of \mathbb{C}(t_1,...,t_n) (that is, subsets closed under the operations +, - , \times, \div), and J is generated by J_1,...,J_r where I \subseteq J_j \subseteq J for each j and J_j:I is radical. By induction on r, prove that J:I is radical.

Homework Equations



A relevant definition: An extension L:K in \mathbb{C} is radical if L = K(\alpha_1, ... ,\alpha_m) where for each j = 1, ... , m there exists an integer n_j such that \alpha_j^{n_j} \in K(\alpha_1,..., \alpha_{j-1}) with j \geq 1.


The Attempt at a Solution



I don't think I am understanding what it means when the question says "J is generated by J_1,...,J_r". Is the question implying that the union of all of these subsets equal J? I think once this is clarified, the proof should be straightforward.

My progress so far:

J_1:I is radical. By definition, this implies that J_1 = I(\alpha_1, ... ,\alpha_m) where for each k = 1, ... , m there exists an integer n_k such that \alpha_k^{n_k} \in K(\alpha_1,..., \alpha_{k-1}) with k \geq 1.

Because I don't understand the question, I don't know where to go from here. Help please? Thank you!
 
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Every element of J may be expressed as a finite sum/product of elements from J1,…, Jr.

Also, fields are not closed under division. Fields excluding the 0 element do, however, form a multiplicative group, and thus are closed under division.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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