Equal Field Extensions

Question

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

Answer 1

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 $K[a_1, a_2, a_3, a_4]=\mathbb Q(\omega)\ne \mathbb Q$.

Answer 2

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 $a_n=b-\sum_{i=1}^{n-1} a_i$.

Answer 3

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.