Show that $\mathcal{I}(V)$ is the product ideal of $k=\mathbb{F}_2$

Question

This question is from Dummit and Foote Chapter 15 (Exercise 15). I want to show the following:

If $k=\mathbb{F}_2$ and $V=\{0,0),(1,1)\}\subset\mathbb{A}^2$, show that $\mathcal{I}(V)$ is the product ideal $m_1m_2$ where $m_1=(x,y)$ and $m_2=(x-1,y-1)$.

I think that I am overthinking this.. it feels sort of intuitive perhaps, but I am new to affine algebraic sets, so I am unsure of how to prove this.

Answer 1

Let $V_{0}= (0,0)$ and $V_{1}= (1,1)$. Then $V = V_{0} \cup V_{1} $. Hence $I(V)= I(V_{0})\cap I(V_{1})$. Now can you see that $I(V_{0}) = (x,y)$ and $I(V_{1}) = (x-1, y-1)$, using the definition you gave in comment? And since the ideals $(x-1, y-1), (x,y)$ are coprime, $I(V_{0})\cap I(V_{1})=I(V_{0})I(V_{1})$.