Elementary Algebraic Geometry - D&F Section 15.1 - Exercise 15

[Math Amateur]

Dummit and Foote (D&F), Ch15, Section 15.1, Exercise 15 reads as follows:

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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).

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I am having trouble getting started on this problem.

One issue/problem I have is - what is the exact nature of m_1, m_2 and m_1m_2. What (explicitly) are the nature of the elements of these ideals.

I would appreciate some help and guidance.

Peter



Note: D&F define \mathcal{I} (V) as follows:

\mathcal{I} (V) = \{ f \in k(x_1, x_2, ... , x_n) \ | \ f(a_1, a_2, ... , a_n) = 0 \ \ \forall \ \ (a_1, a_2, ... , a_n) \in V \}

 

[R136a1]

The ideal $m_1m_2$ is generated by $\{xy~\vert~x\in m_1,y\in m_2\}$. So $m_1m_2$ consists of sums of these elements.

Is this what you wanted to know?

 

[Math Amateur]

Thanks R136a1

Can you also specify what the elements of m_1 and m_2 look like?

Thanks Again,

Peter

 

[R136a1]

Elements of $m_1$ have the form

\alpha x + \beta y

for $\alpha,\beta\in \mathbb{F}_2$. Analogous for $m_2$.

 

[Math Amateur]

Thanks R136a1

Just reflecting on your reply.

Given the context of this exercise (algebraic geometry) and the definition of

\mathcal{I} (V) as follows:

\mathcal{I} (V) = \{ f \in k(x_1, x_2, ... , x_n) \ | \ f(a_1, a_2, ... , a_n) = 0 \ \ \forall \ \ (a_1, a_2, ... , a_n) \in V \}

would it be more accurate to (following our lead) to define the elements of (x,y) as

f_1x + f_2y

where f_1, f_2 \in \mathbb{F}_2[x,y]

What do you think?

Peter

 

[R136a1]

Yes, of course. What was I thinking...