In Dummit & Foote (example on page $661$), it is stated that $\mathbb R[V] =\mathbb R[x,y]/ (xy-1)$. So, I'm assuming they are using the fact that $(xy-1)=\mathcal I (V)$ since by definition $$\mathbb R[V]=\mathbb R[x,y]/\mathcal I (V).$$
However, I do not see why $(xy-1)=\mathcal I(V=Z(xy-1))$. How is this true?
HINT:
Solution 1:
Let $P\in \mathbb{R}[x,y]$. We have $$P(t,\frac{1}{t}) = \frac{Q(t)}{t^m}$$ for some $Q\in \mathbb{R}[t]$ and $m\ge 0$. Now, if $P$ is zero on the real hyperbola then $$P(t,\frac{1}{t})= 0$$ for all $t\in \mathbb{R}\backslash\{0\}$, so $Q(t)= 0$ for all $t\in \mathbb{R}\backslash\{0\}$. This implies $Q$ is the zero polynomial. This implies that $P(t, \frac{1}{t})= 0$ for all $t \in \mathbb{C}\backslash\{0\}$, so $P$ is zero on the complex hyperbola. Now use the Nullstellensatz.
Note: In other cases, we could use analytic local parametrizations of varieties, as soon as the real locus contains nonsingular points.
Solution 2:
(only use basic algebra)
Consider $P\in \mathbb{R}[x,y]$. "Divide" $P$ by $(y x -1)$ with remainder. Since $y$ is not invertible in $\mathbb{R}[y]$ we will get an equality $$y^n P(x,y) = (y x-1) Q(x) + R(y)$$
Assume now $P$ is zero on the real hyperbola. Then for all $y\in \mathbb{R}\backslash\{0\}$ we get $R(t)= 0$, so $R$ is the zero polynomial. We get $y^n P(x,y) $ is divisible by $(x y -1)$. Since $y^n$ is relative prime to $(x y -1)$ we conclude $P(x,y)$ is divisible by $x y -1$.