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

In summary, the conversation discusses Exercise 3.24 in Chapter 3 of R Y Sharp's book "Steps in Commutative Algebra". The exercise asks the reader to show that the residue class ring S, which is the quotient of the ring of polynomials \mathbb{R}[x_1, x_2, x_3] over the real field \mathbb{R} by the ideal (x_1^2 + x_2^2 + x_3^2), is an integral domain. The conversation then delves into relevant thoughts and considerations for solving the problem, including the use of the division algorithm and the nature of the remainder. Deveno provides a proof that S is an integral domain
  • #1
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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:

-----------------------------------------------------------------------------
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.

----------------------------------------------------------------------------

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
 
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  • #2
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:

$ab' = 0$
$ac' = 0$
$ad' = 0$, anyone 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.
 
  • #3
Deveno said:
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:

$ab' = 0$
$ac' = 0$
$ad' = 0$, anyone 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
 
  • #4
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: $\text{deg}(f) = \text{deg}(g) = 1$.
 
  • #5
Deveno said:
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
 

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

What is a residue class ring of polynomials?

A residue class ring of polynomials, also known as a factor ring, is a mathematical structure formed by taking a polynomial ring and quotienting out by a specific ideal. This results in a smaller ring containing polynomials with coefficients in the original ring, but with a reduced set of operations and properties.

How is a residue class ring of polynomials different from a polynomial ring?

A polynomial ring is formed by taking polynomials with coefficients in a specific ring and adding, subtracting, and multiplying them. However, a residue class ring is formed by taking a polynomial ring and quotienting out by an ideal, resulting in a smaller ring with reduced operations. Additionally, the elements of a residue class ring are equivalence classes of polynomials, rather than individual polynomials as in a polynomial ring.

What is the significance of residue class rings in mathematics?

Residue class rings play an important role in algebraic number theory, algebraic geometry, and modular arithmetic. They allow for the study of structures that are more complicated than traditional rings, and they have applications in fields such as coding theory and cryptography.

How are residue class rings of polynomials used in coding theory?

In coding theory, residue class rings of polynomials are used to construct error-correcting codes. These codes are essential for transmitting data reliably over noisy communication channels. The structure of residue class rings allows for efficient encoding and decoding algorithms, making them a useful tool in this field.

Can residue class rings of polynomials be used in other contexts besides mathematics?

Yes, residue class rings of polynomials have applications in computer science, specifically in the field of cryptography. They are used to construct cryptographic systems such as the RSA algorithm, which is based on the difficulty of factoring large integers in a specific residue class ring. Additionally, residue class rings have been used in physics to study symmetries and gauge theories.

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