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Group of polynomials with coefficients from Z_10.

ZaidAlyafey

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Jan 17, 2013
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Contemporary Abstract Algebra by Gallian

This is Exercise 14 Chapter 3 Page 69

Question

Let $G$ be the group of polynomials under the addition with coefficients from $Z_{10}$. Find the order of $f=7x^2+5x+4$ .

Note: this is not the full question, I removed the remaining parts.

Attempt

For simplicity I represented the coefficients as ordered pair $(a,b,c)$

Then to compute $|f|$ , I add $f$ until I get a positive integer $n$ such that $f^n=(0,0,0)$

\(\displaystyle f^2=(7,5,4)+(7,5,4)=(4,0,8)\)
\(\displaystyle f^3=(4,0,8)+(7,5,4)=(1,5,2)\)
\(\displaystyle f^4=(1,5,2)+(7,5,4)=(8,0,6)\)
\(\displaystyle f^5=(8,0,6)+(7,5,4)=(5,5,0)\)
\(\displaystyle f^6=(5,5,0)+(7,5,4)=(2,0,4)\)
\(\displaystyle f^7=(2,0,4)+(7,5,4)=(9,5,8)\)

That approach seems wrong since that could go anywhere.

If we put $f=ax^2+bx+c$ then $f=(a,b,c)$ either circulates or approaches a finite order but what would be the method to prove it is of infinite or finite order ?
 

Evgeny.Makarov

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Jan 30, 2012
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\(\displaystyle f^2=(7,5,4)+(7,5,4)=(4,0,8)\)
\(\displaystyle f^3=(4,0,8)+(7,5,4)=(1,5,2)\)
\(\displaystyle f^4=(1,5,2)+(7,5,4)=(8,0,6)\)
\(\displaystyle f^5=(8,0,6)+(7,5,4)=(5,5,0)\)
\(\displaystyle f^6=(5,5,0)+(7,5,4)=(2,0,4)\)
\(\displaystyle f^7=(2,0,4)+(7,5,4)=(9,5,8)\)
Why did you stop??? Three more steps, and you would have reached enlightening! (Smile)
 

mathbalarka

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Mar 22, 2013
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Hint : $\Bbb Z/10 \Bbb Z$ is of order $10$.

See, Z, every member of I&S goes senile from time to time (not to mention me), as I suspected before.
 

ZaidAlyafey

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Jan 17, 2013
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Ok , I must be kidding. It is easy to see that the system of equations

\(\displaystyle 7n \equiv 0 (\text{mod} \, 10)\)

\(\displaystyle 5n \equiv 0 (\text{mod} \, 10)\)

\(\displaystyle 4n \equiv 0 (\text{mod} \, 10)\)

\(\displaystyle n=10\) is an immediate solution .
 

mathbalarka

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Mar 22, 2013
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ZaidAlyafey

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Jan 17, 2013
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Hint : $\Bbb Z/10 \Bbb Z$ is of order $10$.

See, Z, every member of I&S goes senile from time to time (not to mention me), as I suspected before.
I hope that one won't last for a long time!
 

mathbalarka

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Mar 22, 2013
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Zaid said:
I hope that one won't last for a long time!
That forum's cursed. Forget all hope of the light he who enters there!! :D
 

Deveno

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Feb 15, 2012
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Just a few comments:

1. Your observation that additively, $\Bbb Z_{10}[x]$ is isomorphic to $\Bbb Z_{10}^{\infty}$ is actually a very astute one. The set of $n$ degree polynomials is a subgroup, and isomorphic to $\Bbb Z_{10}^n$ (in this case, $n$ = 3).

2. It is customary to write $f^n$ in an abelian group as $nf$. Beginners in abstract algebra always seem to struggle with this largely notational issue.

3. Both of these observations come in handy much later, when one is studying $R$-modules (which act "sort of" like vector spaces, but with ring elements as coordinates, which introduces some peculiarities). Here the ring is $\Bbb Z_{10}$, and the fact that $\Bbb Z_{10}$ is a quotient ring of $\Bbb Z$ allows us to view it as a $\Bbb Z$-module with some interesting $\Bbb Z$-linear dependencies.

4. $\Bbb Z$-modules are just abelian groups, and this is why you are seeing this example at such an early stage. you have just shown that the integer 10 is the smallest positive $n$ for which $nI$ annihilates $(a,b,c)$ with a = 7, b = 5, c = 4. A little reflection shows that it annihilates all of $\Bbb Z_{10}^3$. This should suggest to you that matrices over certain rings (even when the ring is commutative) don't always act as nicely as we want them to.

5. mathblalarka's observation that none of the entries of (7,5,4) are 0 (mod 10) is somewhat incomplete, consider the order of $p(x) = 2 + 4x + 8x^2$ in the same group....
 

ZaidAlyafey

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Jan 17, 2013
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I think if we have \(\displaystyle f= \sum_{k=0}^n a_k x^k \) then if we restrict that $\text{gcd}(a_1,a_2, \cdots , a_n)=1$ we have the order as $10$.

If we do $f=(8,4,2)$ then we expect that $|f|=5$ since for the simultaneous equations

\(\displaystyle 8n \equiv 0 (\text{mod}\, 10)\)
\(\displaystyle 4n \equiv 0 (\text{mod}\, 10)\)
\(\displaystyle 2n \equiv 0 (\text{mod}\, 10)\)

sine $\text{gcd}(2,4,8)=2$ it suffies to put $n=5$.

To check we have

\(\displaystyle 2f= (8,4,2)+(8,4,2)=(6,8,4)\)
\(\displaystyle 3f= (6,8,4)+(8,4,2)=(4,2,6)\)
\(\displaystyle 4f= (4,2,6)+(8,4,2)=(2,6,8)\)
\(\displaystyle 5f= (2,6,8)+(8,4,2)=(0,0,0)\)

which is what we expect so generally we have the order of $f$ under $\mathbb{Z}_n$ as $|f|=n/ \text{gcd}(a_1,a_2,\cdots ,a_n)$ ,right ?
 

Deveno

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Feb 15, 2012
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Right, you want to ensure the gcd and 10 have no common factors. This is why finite fields (of characteristic $p$) are so nice to work over (and produce vector spaces, instead of mere modules), because we no longer have to worry about that.

Prime factorization (in its basic integer form, and in its extensions to more complicated systems) is just *that* useful.

Your last statement is "almost correct". To see why I say "almost" consider the order of:

$4 + 4x + 8x^2$.

It is clear that gcd(4,4,8) = 4, but the order is not 10/4 = 5/2, it is 5.

In other words the order is actually:

$\dfrac{n}{\text{gcd}(n,\text{gcd}(a_1,a_2,\dots,a_k))}$

for a polynomial in $\Bbb Z_n[x]$ (under addition) of degree $k$

(here it is implicitly assumed the integers $a_1,\dots,a_k$ are already reduced mod $n$).

Given that 10 (our example) only has 2 divisors d with 1 < d < 10, it is unlikely that a polynomial chosen at random has order < 10, but this becomes much more of a concern if working over a highly divisible number like 60.

See if you can guess the order of $30x + 12x^2$ in $\Bbb Z_{60}[x]$ under addition, and then prove it (your proof should consist of just three calculations).
 

Klaas van Aarsen

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Mar 5, 2012
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I think if we have \(\displaystyle f= \sum_{k=0}^n a_k x^k \) then if we restrict that $\text{gcd}(a_1,a_2, \cdots , a_n)=1$ we have the order as $10$.
How about the order of $5x+2$?
 

Deveno

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Feb 15, 2012
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How about the order of $5x+2$?
Not a very good example, we have:

$2(5x + 2) = 4 \neq 0$
$5(5x + 2) = 5x \neq 0$

which shows the order is indeed 10.
 

ZaidAlyafey

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Jan 17, 2013
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Your last statement is "almost correct". To see why I say "almost" consider the order of:

$4 + 4x + 8x^2$.

It is clear that gcd(4,4,8) = 4, but the order is not 10/4 = 5/2, it is 5.

In other words the order is actually:

$\dfrac{n}{\text{gcd}(n,\text{gcd}(a_1,a_2,\dots,a_k))}$
Yup, I was concerned about that !

See if you can guess the order of $30x + 12x^2$ in $\Bbb Z_{60}[x]$ under addition, and then prove it (your proof should consist of just three calculations).
The order of $30$ is $2$ and the order of $12$ is $5$ and they will coincide at the first even integer namely $10$.
I don't understand what you mean by three calculations ?
 

Klaas van Aarsen

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I like $\text{lcm}(|a_1|, ..., |a_k|)$.
 

Deveno

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Feb 15, 2012
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Yup, I was concerned about that !



The order of $30$ is $2$ and the order of $12$ is $5$ and they will coincide at the first even integer namely $10$.
I don't understand what you mean by three calculations ?
Yes, it is clear that the order is at most 10. What two other possibilities need to be ruled out?