Currently, I am trying to understand Example 1.3.1. from Silverman's Arithmetic of Elliptic Curves:
Let $K$ be a perfect field and $V$ be an algebraic set in $\mathbb{A}^2 = \mathbb{A}^2(\bar{K})$ given by $X^2 - Y^2 = 1$.
The book says that $V$ is clearly defined over any (perfect) field $K$. However, my main issue is understanding the definition.
My own efforts:
The book defined an algebraic set $V$ to be defined over $K$ if its ideal $$I(V) = \{ f \in \bar{K}[X,Y] \, : \, f(P)=0 \: \forall P \in V \}$$ is generated by polynomials in $K[X,Y]$.
Also, $V$ which appears in the description of the set $I(V)$ is by definition given by $$ V = \{ (x,y) \in \mathbb{A}(\bar{K}) \: : \: x^2 - y^2 = 1 \}. $$
If the claim is true, this must mean that $I(V)$ is generated by polynomials in $K[X,Y]$. I also think that $g = X^2-Y^2-1$ generates it which would mean $I(V) = (g) \subseteq \bar{K}[X,Y]$.
I think this is true but I cannot find a way to convince myself why this equation is true (although it might seem trivial).
Another thing which might be helpful to convince myself through antoher way is the following remark in the book: It says that $V$ is defined over $K$ if and only if $I(V) = I(V/K)\bar{K}[X,Y]$ where $$ I(V/K) = \{ f \in K[X,Y] \: : \: f(P) = 0 \: \forall P \in V \} = I(V) \cap K[X,Y]. $$
However, I cannot use this observation to my advantage as well.
It would be really nice if could take your time to explain this probably trivial question.
It is true that $I(V) = (g)$. There are more elementary arguments, but this is an immediate consequence of Hilbert's Nullstellensatz, which says that there is a bijective correspondence between affine subvarieties and radical ideals. $(g)$ is radical, and its vanishing set is $V$, so $I(V) = (g)$.