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While reading Kunz's commutative algebra book, I came across a statement I can't understand. First, let me define the notations.

Let $L/K$ be extension of fields, and let $\mathbb{P}^n (L)$ denote the projective n-space over $L$.

  • A Projective coordinate transformation is a mapping $\mathbb{P}^n (L) \rightarrow \mathbb{P}^n (L)$ given by a matrix $A\in GL(n+1,L)$ through the equation $(Y_0,...,Y_n) = (X_0,...,X_n).A$ $\space$ $\space$ $\space$ (1)
  • A Subset $V\subset \mathbb{P}^n (L)$ is called a projective $K$-variety if there are homogeneous polynomials $F_1,...,F_m\in K[X_0,...,X_n]$ such that $V$ is the set of all common zeros of the $F_i$ in $\mathbb{P}^n (L)$.

The author next says that "This concept is invariant under coordinate transformations (1) as long as A has coefficients in K". I have no idea what he means by this. I have tried to apply the transformation both on the n-tuples and on the variables, but in either case $V$ seems to be changing (If I'm not wrong). Can somebody explain what these mean?

Thanks in advance.

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$V$ is certainly changing in the sense that after applying a coordinate transformation $A$ to $\mathbb{P}^n$ you will find that $A(V)$ is not (in general) the zero locus of the same homogeneous polynomials $F_1,\ldots,F_m$ that $V$ was, but the author is saying that there still exist other homogeneous polynomials $G_1,\ldots,G_m$ that cut out $A(V)$ as their zero locus. So $A(V)$ is still a projective variety - i.e. the concept of being a projective variety is invariant under coordinate transformations. Indeed, you can check that letting $G_i:=F_i\circ A^{−1}$ suffices.

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