Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Example 3 - pag

[Math Amateur]

I am reading Dummit and Foote Ch 15, Commutative Rings and Algebraic Geometry. In Section 15.1 Noetherian Rings and Affine Algebraic Sets, Example 3 on page 660 reads as follows: (see attachment)

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Let V = \mathcal{Z}(x^3 - y^2) in \ \ \mathbb{A}^2.

If (a, b) \in \mathbb{A}^2 is an element of V, then a^3 = b^2.

If a \ne 0, then also b \ne 0 and we can writea = (b/a)^2, \ b = (b/a)^3.

It follows that V is the set \{ (a^2, a^3) \ | \ a \in k \}.

For any polynomial f(x,y) \in k[x,y]. we can write f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y)

... ... ... etc etc

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I cannot follow the line of reasoning:

"For any polynomial f(x,y) \in k[x,y]. we can write f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y)"

Can anyone clarify why this is true and why D&F are taking this step?

Peter

 

[R136a1]

It suffices to prove this for $f(x,y) = x^n y^m$.

We prove it by induction on $m$:
For $m=0$, take $f_0(x) = x^n$ and the rest $0$.
For $m=1$, take $f_1(x) = x^n$ and the rest $0$.

If it holds for $m<m^\prime$, then write $x^n y^m = - x^ny^{m-2} (x^3 - y^2 ) - x^{n+3}y^{m-2}$.
By induction, express $x^{n+3}y^{m-2}$ in the required form, then you can also express $x^n y^m$ in such form.

 

[Math Amateur]

Thank you so much for your help with this problem

I had nearly given up on it, and with it my progress into algebraic geometry!

So thanks again!

Peter