Equal Field Extensions Let $K$ be a field, $a_1, a_2, ..., a_n$ be algebraic over $K$, and $c_1, ..., c_n \in K$. My question is regarding the extension $K[a_1, ..., a_n]$. 
Is it possible to have an extension $K[c_1a_1+c_2a_2+...+c_na_n]$ which is not equal to $K[a_1, ..., a_n]$ if $a_1, a_2, ..., a_n$ are linearly independent, non zero and $c_i \neq 0$ for all $i$?
I have tried to but have yet to find anything.
Thanks
 A: Take $K=\mathbb Q$ and $n=4$ and let $a_i=\omega^i$, where $\omega$ is a primitive $5$-th root of unit.
Then the $a_i$ are linearly independent over $\mathbb Q$.
Indeed, 
$c_1 \omega + c_2 \omega^2 + c_3 \omega^3 + c_4 \omega^4 = 0$
implies
$c_1 \omega + c_2 \omega^2 + c_3 \omega^3 =  -c_4 \omega^4 = c_4 (1+\omega + \omega^2 + \omega^3)$ because $1+\omega + \omega^2 + \omega^3 +\omega^4=0$, and this implies that all $c_i=0$ because $1,\omega,\omega^2,\omega^3$ are linearly independent over $\mathbb Q$.
Take $c_1a_1+c_2a_2+c_3a_3+c_4a_4$ with all $c_i=1$.
Then $c_1a_1+c_2a_2+c_3a_3+c_4a_4=\omega + \omega^2 + \omega^3 +\omega^4=-1$ and so $K[c_1a_1+c_2a_2+c_3a_3+c_4a_4]=\mathbb Q$, but $K[a_1, a_2, a_3, a_4]=\mathbb Q(\omega)\ne \mathbb Q$.
A: Consider a field of finite characteristic $p >0$, for example $\mathbb F_p$.
The algebraic fraction field of two variables $K=\mathbb F_p(X,Y)$ is a finite field extension of degree $p^2$ of $k=\mathbb F_p(X^p,Y^p)$. $X$ and $Y$ are independent over $k$ and $k(Y^pX+Y) \subsetneq k(X,Y)$.
For more details on the proves, you can have a look here.
A: Consider any example of a finite extension $K\subset L$ (say, of degree $n$) and a nonzero element $b\in L$ such that $K[b]\neq L$.  Then we can easily construct a basis $\{a_1,\dots,a_n\}$ for $L$ over $K$ such that if we write $b=\sum c_i a_i$, every $c_i$ is nonzero.  For instance, you can take $\{a_1,\dots,a_{n-1}\}$ to be any set of $n-1$ linearly independent elements of $L$ whose span does not contain $b$, and then let $a_n=b-\sum_{i=1}^{n-1} a_i$.
