Transcendence Degree of a field extension over $\mathbb C$ Consider the $2 \times n$ matrix  $\begin{bmatrix}
    x_{11}       & x_{12} & x_{13} & \dots & x_{1n} \\
    x_{21}       & x_{22} & x_{23} & \dots & x_{2n}
\end{bmatrix}$. Form all the products by taking exactly one entry from each column. So there are $2^n$ monomials in the matrix entries of degree $n$. What is the transcendence degree over $\mathbb C$ of the field generated by these $2^n$ monomials ?
 A: Let $S=K[X_1,\dots,X_n,Y_1,\dots,Y_n]$. Let $R$ be the $K$-subalgebra of $S$ generated by the set $M$ of all degree $n$ monomials consisting of the products obtained by taking exactly one entry from each column of the matrix
$$\begin{bmatrix}
    X_{1} & X_{2} & \dots & X_{n} \\
    Y_{1} & Y_{2} & \dots & Y_{n}
\end{bmatrix}.$$
Then $R=k[M]$ can be viewed as an affine semigroup ring. The Krull dimension of $R$ equals the transcendence degree of $Q(R)$ (the field of fractions of $R$) over $K$, and at its turn this equals the rank of (the submonoid generated by) $M$ in $\mathbb N^{2n}$. The elements of $M$ are of the form $$e_{i_1}+\cdots+e_{i_k}+e_{n+j_1}+\cdots+e_{n+j_l},$$ where $(e_i)_{1\le i\le 2n}$ is the canonical basis of $\mathbb Z^{2n}$, $1\le i_1<\cdots<i_k\le n$, $1\le j_1<\cdots<j_l\le n$, $k+l=n$, and $\{i_1,\dots,i_k\}\cup\{j_1,\dots,j_l\}=\{1,\dots,n\}$. It remains to show that the rank of $M$ is $n+1$. One can view $M$ as an $2^n\times 2n$ matrix. Adding the first column to the $(n+1)$th, and so on, we get a matrix having the last $n$ columns equal to a $2^n\times 1$ vector with all entries equal to $1$, so $n-1$ of them can be replaced by $0$. Now consider any $n+1$ rows and the first $n+1$ columns in the last matrix and note these are linearly independent. (A simpler case is to consider the vectors $e_1+e_{n+1},\dots,e_n+e_{n+1},e_{n+1}$.)
To conclude, the answer to the question is $n+1$. 
