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Residue Class Rings (Factor Rings) of Polynomials _ R Y Sharp

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading R Y Sharp: Steps in Commutative Algebra.

In Chapter 3 (Prime Ideals and Maximal Ideals) on page 44 we find Exercise 3.24 which reads as follows:

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Show that the residue class ring \(\displaystyle S \) of the ring of polynomials \(\displaystyle \mathbb{R}[x_1, x_2, x_3] \) over the real field \(\displaystyle \mathbb{R} \) in indeterminates \(\displaystyle x_1, x_2, x_3 \) given by

\(\displaystyle S = \mathbb{R}[x_1, x_2, x_3]/(x_1^2 + x_2^2 + x_3^2) \) is an integral domain.

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Can someone please help me make a significant start n this problem.

Some thoughts that I think are relevant to the problem follow:

First I tried to get an idea of the nature of \(\displaystyle (x_1^2 + x_2^2 + x_3^2) \) and \(\displaystyle S = \mathbb{R}[x_1, x_2, x_3]/(x_1^2 + x_2^2 + x_3^2) \) and their elements.

Thus ...

\(\displaystyle (x_1^2 + x_2^2 + x_3^2) = \{ f(x_1, x_2, x_3)(x_1^2 + x_2^2 + x_3^2) \ | \ f(x_1, x_2, x_3) \in \mathbb{R}[x_1, x_2, x_3] \} \)

\(\displaystyle \mathbb{R}[x_1, x_2, x_3]/(x_1^2 + x_2^2 + x_3^2) = \{ g(x_1, x_2, x_3) + (x_1^2 + x_2^2 + x_3^2) \ | \ g(x_1, x_2, x_3) \in \mathbb{R}[x_1, x_2, x_3] \} \)

Now I suspect that one now uses the division algorithm to determine a remainder which will be of lower degree than \(\displaystyle x_1^2 + x_2^2 + x_3^2 \) - that is lower than deg 2 (??? is that right ?? )

BUT, what exactly would be the nature of the remainder - ie how does the division algorithm work for polynomials of several variables?

And then ... where to go from there ...

Can someone assist me in this ...

Further does anyone know of a text that gives an example of the division algorithm, applied to polynomials in several variables ...

An elementary or undergraduate text dealing in all aspects of the theory of polynomials in several variables would be helpful in getting a sense of what is happening in the abstract theorems ...

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I don't think the division algorithm generalizes well to multivariate polynomials.

For example, we have:

$\text{gcd}(x_1,x_2) = 1$

but I daresay we cannot find $f(x_1,x_2),g(x_1,x_2) \in \Bbb R[x_1,x_2]$ such that:

$x_1f(x_1,x_2) + x_2g(x_1,x_2) = 1$

for, if we could, we would have, at $(x_1,x_2) = (0,0)$:

$0 = 1$.

But all is not lost:

What we need to show is that $x_1^2 + x_2^2 + x_3^2$ is irreducible (= prime, since we have a UFD) over $\Bbb R$ (we will use an important property of the real numbers in the process). This will show that our ideal is a prime ideal, which will mean the quotient is an integral domain.

First of all, it should be clear that if:

$x_1^2 + x_2^2 + x_3^2 = f(x_1,x_2,x_3)g(x_1,x_2,x_3)$ where these are non-units, then neither $f$ nor $g$ can contain terms of degree higher than 1, so we have:

$x_1^2 + x_2^2 + x_3^2 = (a + bx_1 + cx_2 + dx_3)(a' + b'x_1 + c'x_2 + d'x_3)$.

Now, what do we ALWAYS DO FIRST, when we consider polynomials? If you answered: "look at the constant term", you win today's prize.

We thus get: $aa' = 0$, so without loss of generality, we can take: $a' = 0$.

Looking at the $x_i^2$ terms, we get:

$bb' = 1$
$cc' = 1$
$dd' = 1$, so none of these can be 0.

Since the $x_i$ terms must all be 0, we get:

$ab' = 0$
$ac' = 0$
$ad' = 0$, any one of which implies $a = 0$.

Now let's consider the $x_1x_2$ term, which is:

$bc' + cb' = \dfrac{b}{c} + \dfrac{c}{b} = 0$.

This gives:

$\dfrac{b^2 + c^2}{bc} = 0$

and since $bc \neq 0$, we must have $b^2 + c^2 = 0$.

This in turn implies $b = c = 0$ (this is where we use the properties of real numbers), a contradiction, so no such $b,c$ can exist, and we are done.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I don't think the division algorithm generalizes well to multivariate polynomials.

For example, we have:

$\text{gcd}(x_1,x_2) = 1$

but I daresay we cannot find $f(x_1,x_2),g(x_1,x_2) \in \Bbb R[x_1,x_2]$ such that:

$x_1f(x_1,x_2) + x_2g(x_1,x_2) = 1$

for, if we could, we would have, at $(x_1,x_2) = (0,0)$:

$0 = 1$.

But all is not lost:

What we need to show is that $x_1^2 + x_2^2 + x_3^2$ is irreducible (= prime, since we have a UFD) over $\Bbb R$ (we will use an important property of the real numbers in the process). This will show that our ideal is a prime ideal, which will mean the quotient is an integral domain.

First of all, it should be clear that if:

$x_1^2 + x_2^2 + x_3^2 = f(x_1,x_2,x_3)g(x_1,x_2,x_3)$ where these are non-units, then neither $f$ nor $g$ can contain terms of degree higher than 1, so we have:

$x_1^2 + x_2^2 + x_3^2 = (a + bx_1 + cx_2 + dx_3)(a' + b'x_1 + c'x_2 + d'x_3)$.

Now, what do we ALWAYS DO FIRST, when we consider polynomials? If you answered: "look at the constant term", you win today's prize.

We thus get: $aa' = 0$, so without loss of generality, we can take: $a' = 0$.

Looking at the $x_i^2$ terms, we get:

$bb' = 1$
$cc' = 1$
$dd' = 1$, so none of these can be 0.

Since the $x_i$ terms must all be 0, we get:

$ab' = 0$
$ac' = 0$
$ad' = 0$, any one of which implies $a = 0$.

Now let's consider the $x_1x_2$ term, which is:

$bc' + cb' = \dfrac{b}{c} + \dfrac{c}{b} = 0$.

This gives:

$\dfrac{b^2 + c^2}{bc} = 0$

and since $bc \neq 0$, we must have $b^2 + c^2 = 0$.

This in turn implies $b = c = 0$ (this is where we use the properties of real numbers), a contradiction, so no such $b,c$ can exist, and we are done.
Thank you for the help, Deveno ... just starting to work through this now.

Just a preliminary question ...

You write:

"I don't think the division algorithm generalizes well to multivariate polynomials."

and then

"$x_1^2 + x_2^2 + x_3^2 = f(x_1,x_2,x_3)g(x_1,x_2,x_3)$ where these are non-units, then neither $f$ nor $g$ can contain terms of degree higher than 1"

but how does this second statement follow ... since I thought this type of statement was dependent for its validity on the division algorithm working and giving a remainder of lower degree than the dividend ...

How are you justifying this statement ...

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well we can still talk about the degree of a polynomial in more than one variable:

First we define the degree of the monomial:

$ax_1^{r_1}x_2^{r_2}\cdots x_n^{r_n}$ to be:

$\displaystyle \sum_{i = 1}^n r_i$

and then define the degree of the polynomial to be the maximum of the set of all the degrees of the monomials (we have a finite number of monomials, so we have a maximum, which may be shared by several monomials).

So we can still say that:

$\text{deg}(f(x_1,\dots,x_n)g(x_1,\dots,x_n)) = \text{deg}(f(x_1,\dots,x_n)) + \text{deg}(g(x_1,\dots,x_n))$ if $f,g \neq 0$, and that:

$f(x_1,\dots,x_n)$ is a constant polynomial if and only if $\text{deg}(f(x_1,\dots,x_n)) = 0$.

In particular, if $\text{deg}(fg) = 2$ and neither polynomial is constant (a unit) we must have: $\text{deg}(f) = \text{deg}(g) = 1$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Well we can still talk about the degree of a polynomial in more than one variable:

First we define the degree of the monomial:

$ax_1^{r_1}x_2^{r_2}\cdots x_n^{r_n}$ to be:

$\displaystyle \sum_{i = 1}^n r_i$

and then define the degree of the polynomial to be the maximum of the set of all the degrees of the monomials (we have a finite number of monomials, so we have a maximum, which may be shared by several monomials).

So we can still say that:

$\text{deg}(f(x_1,\dots,x_n)g(x_1,\dots,x_n)) = \text{deg}(f(x_1,\dots,x_n)) + \text{deg}(g(x_1,\dots,x_n))$ if $f,g \neq 0$, and that:

$f(x_1,\dots,x_n)$ is a constant polynomial if and only if $\text{deg}(f(x_1,\dots,x_n)) = 0$.

In particular, if $\text{deg}(fg) = 2$ and neither polynomial is constant (a unit) we must have: $\text{deg}(f) = \text{deg}(g) = 1$.
Thanks Deveno ....

Yes of course ... That cleared up that issue ...

Peter