1. Feb 25, 2012

### iironiic

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 $j \geq 1$.

3. 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!

2. Feb 25, 2012

### alanlu

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.