# I Affine Algebraic Sets - Dummit and Foote, page 660, Ex. 3

1. Jun 17, 2016

### Math Amateur

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need someone to help me to fully understand the reasoning/analysis behind one of the statements in Example (3) on Page 660 of D&F ...

On page 660 (in Section 15.1) of D&F we find the following text and examples (I am specifically focused on Example (3)):

In the above text, in Example (3), we find the following:

"... ... 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 someone explain ( slowly and carefully) exactly how/why this is true ... ...

Peter

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In order for readers of the above post to understand the context of the question and the notation employed I am providing the introductory pages on affine algebraic sets in the D&F text ... ... as follows:

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2. Jun 17, 2016

### Staff: Mentor

Hi Peter,

let us start and consider $f(x,y)$ as polynomial in $y$. Then it can be written:
$f(x,y) = c_0 \cdot f_0(x) + c_1 \cdot y \cdot f_1(x) + c_2 \cdot y^2 \cdot f_2(x) + \dots$
What we are doing next is to gather everything, that "disturbs" us and putting it into the terms $f_0$ and $f_1$.
There are no restrictions on those, beside that they aren't allowed to contain $y$'s. so our first substitution will be $f_0^{new} = c_0 f_0$ and $f_1^{new} = c_1 f_1$. For the next steps I will drop the $^{new}$ because it is not necessary to keep a record on them.

Now $c_2 \cdot y^2 \cdot f_2(x) = c_2 \cdot y^2 \cdot (d_0 + d_1 \cdot x + d_2 \cdot x^2 + \dots) = c_2d_0 y^2 + c_2d_1xy^2+c_2d_2x^2y^2 + \dots$ and every term is of the form $e_n \cdot x^n \cdot y^2.$
This can be written as $e_n \cdot x^n \cdot y^2 = (x^3-y^2) \cdot (-e_n x^n) + e_n \cdot x^{n+3}.$
The first factor at $(x^3-y^2)$ will be used to build up $g(x,y)$ and the second will be added to $f_0.$

You can proceed this way with every following term $c_n y^n f_n(x) \; (n>2)$, i.e. perform the long division on polynomials.

3. Jun 17, 2016

### mathwonk

To summarize fresh34's nice answer, for any domain R, we can divide by any polynomial in the ring R[Y] whose lead coefficient is a unit. E.g. we can always divide by Y^2 + aY + b. Moreover, as is usual in long division, the remainder has lower degree than the divisor.

Thus if we take R = k[X], then in the ring R[Y] = k[X][Y] ≈ k[X,Y], we can divide any polynomial by Y^2 - X^3, and get a remainder of degree lower than 2 (in Y). I.e. the remainder will have form f1Y + f0 where f0 and f1 lie in the coefficient ring k[X].

The point of this somewhat tedious example is to illustrate that over a finite field, one does not keep much information about a polynomial from looking only where it vanishes, since over Z/2Z, any multiple of the polynomial x(x-1) vanishes everywhere. to insure that the zero locus has as much information as possible, one usually assumes the field algebraically closed. see e.g. hilbert's nullstellensatz in DF, p.675.

Last edited: Jun 17, 2016
4. Jun 17, 2016

### Math Amateur

Thanks fresh_42 ... that post was most helpful ...

You really helped me when you wrote:

"This can be written as $e_n \cdot x^n \cdot y^2 = (x^3-y^2) \cdot (-e_n x^n) + e_n \cdot x^{n+3}.$
The first factor at $(x^3-y^2)$ will be used to build up $g(x,y)$ and the second will be added to $f_0.$"

I was really perplexed at how to deal with this situation ...

Thanks again for the help,

Peter

5. Jun 17, 2016

### Math Amateur

Thanks mathwonk ... really helpful to get a general perspective on this matter ...

Thank you for moving my understanding forward ... really appreciate it ...

Peter

6. Jun 17, 2016

### Staff: Mentor

You're welcome. It's always a pleasure to read your carefully prepared and precise questions.