# If there are $k_1, k_2 \in K$ such that $K(\alpha + k_1\beta)=K(\alpha + k_2\beta)$ then $K(\alpha,\beta) = K(\alpha + c\beta)$ for some $c \in K$.

Here is the problem: "Let $K \subset M$ be a finite field extension, and $\alpha, \beta \in M$. Suppose there are $k_1, k_2 \in K$ are distinct and such that $K(\alpha + k_1\beta)=K(\alpha + k_2\beta)$. Then $K(\alpha,\beta) = K(\alpha + c\beta)$ for some $c \in K$."

Some help?

• For future reference, you can always edit a question in order to bump it back to the front page for more attention. You don't need to use up a quarter of your rep to do it! Jun 18, 2015 at 20:21
• oh, interesting! thanks! i don't really care about this reputation stuff, though. Jun 19, 2015 at 0:50
• But if you thought bounties were the only way to garner more attention, that would have left you with a very limited number of opportunities to bump a question. Jun 19, 2015 at 16:54

## 1 Answer

In fact if $K(\alpha+k_1\beta)$ and $K(\alpha+k_2\beta)$ are the same then they are both $K(\alpha,\beta)$.

Denote the given field by $L$. Hint: from $\alpha+k_1\beta,~\alpha+k_2\beta\in L\,$ deduce $\alpha,\beta\in L$.

• God, I'm so dumb... Thanks (: Jun 18, 2015 at 3:27