Algebraic independence

Question

Let $a_1,\dots,a_n$ be transcendental numbers. If the set $\{a_1,\dots,a_n\}$ is algebraically independent over $\mathbb{Q}$, then so is the set $\{a_1,\dots,a_n,1\}$?

Answer 1

Nope. Let $r\in K$ and $L/K$. Define $f(x,y)=0x+1y-r$ (it's nonzero). Then $f(x,r)=0$ for any value $x\in L$, so $\{x,r\}$ is algebraically dependent over $K$ for any $x\in L$. Thus so is any $\{\cdots,x,r\}$.