Algebraically independent equivalent conditions I have some problems to understand the field extensions. Namely,
Let $K$ be a field and $E$ its extension. Let $x_1,\ldots ,x_n$ in $E$ and $0<k<n$. Show that TFAE


*

*Family $(x_1,...,x_n)$ is algebraically independent with respect to $K$.

*Family $(x_1,...,x_k)$ is algebraically independent with respect to $K$ and family $(x_ {k+1},...,x_n)$ is algebraically independent with respect to $K(x_1,...,x_k)$.
Where do this follows? Do I have to make polynomials and check if those has a root in $K$ and its extensions or what?
 A: We have $K[X_1, \ldots, X_n] \simeq (K[X_1, \ldots, X_k])[X_{k + 1}, \ldots, X_n]$.  Informally this means you can always think of a polynomial in the variables $X_1, \ldots, X_n$ with coefficients in $K$ as a polynomial in the variables $X_{k + 1}, \ldots, X_n$ with coefficients in $K[X_1, \ldots, X_k]$.
So if $x_1, \ldots, x_n$ is not algebraically free there's some polynomial in $K[X_1, \ldots, X_n]$ for which $X_i = x_i$ is a root.  But then consider this as a polynomial in $(K[X_1, \ldots, X_k])[X_{k + 1}, \ldots, X_n]$ and just plug in $x_1, \ldots, x_k$.  You get a polynomial in $K(x_1, \ldots, x_k)[X_{k + 1}, \ldots, X_n]$ for which $X_i = x_i$ ($i > k$) is a root.  If this is a nonzero polynomial then $x_{k + 1}, \ldots, x_n$ is not free over $K(x_1, \ldots, x_k)$.  If this is the zero polynomial then when you plugged in $x_1, \ldots, x_k$ you got zero coefficients so $x_1, \ldots, x_k$ are not free over $K$.
That's one direction of the if and only if, I'll leave the other to you.
