Q:

If a and b are transcendent and algebraically dependent

If $a$ and $b$ are transcendental numbers and algebraically dependent, then for any $\alpha$ and $\beta$ algebraic, it follows that the linear combination of $a$ with $\alpha$, and the linear combination of $b$ with $\beta$ are also algebraically dependent.This is true?If this is true, can  you show me the proof?

A:

Suppose we have a base field $K$, with $\alpha,\beta$ algebraic over $K$ and $a,b$ transcendental over $K$ but algebraically dependent over $K$. This means that the field extension $K(a,b)/K$ has transcendence degree $1$ and that $K(a,b,\alpha,\beta)/K$ also has transcendence degree $1$ since $K(a,b,\alpha,\beta)=K(a,b)(\alpha,\beta)$. Therefore
any two elements of $K(a,b,\alpha,\beta)$ are algebraically dependent over $K$ (since otherwise the extension would have transcendence degree at least 2), in particular given $\lambda_{1,2},\mu_{1,2}\in K$ there is some polynomial $p\in K[X,Y]$ such that $p(\lambda_1 a+\mu_1 \alpha, \lambda_2 b+\mu_2 \beta)=0$.

