$f \in\mathbb Q[X]$, $ [f] = [f \operatorname{div} X^5] $ in $ \mathbb Q[X]/\equiv_{X^5} $

Question

I have to proof if the following statement is true or false:

$f \in\mathbb Q[X]$, $ [f] = [f \operatorname{div} X^5] $ in $ \mathbb Q[X]/\equiv_{X^5} $

Since I am new to that I have tried to think about the solution for $\mathbb Z$ and not for polynoms to simplify the task:

So I have the values for the class $[f \operatorname{div} 5]$: $$1:5 = 0,2$$ $$2:5 = 0,4$$ $$3:5 = 0,6$$ $$4:5 = 0,8$$ $$6:5 = 1,2$$

and so on...

On the other hand the values for the class $[f]/\equiv_{5}$ should be: $$\{1,6,11,16\}$$ $$\{2,7,12,17\}$$ $$\{3,8,13,18\}$$ and so on...

So I would think that the statement is false.

Question: Is it possible to do the task the way I do? If not, how can I do it correctly? And if someone would be able to show me the way with polynomials I would really appreciate that.

Answer 1

What you write is only vaguely related to the questions. At the end you somewhat correctly gave the classes of integers modulo $5$, this is not quite what you need though. The first part makes no sense, at least not to me, sorry.

So now what are you supposed to do:

First the objects here are polynomials with rational coefficients; congruences modulo integers are not relevant.

The congruence is modulo $X^5$, that is two polynomial $f,g$ are in the same class if their difference is divisible by $X^5$.

And a polynomial is divisible by $X^5$ if it contains neither a constant term, a linear, a quadratic, a third degree, a fourth degree term.

The notation $f \operatorname{div} X^5$ means the remainder of the Euclidean division of $f$ by $X^n$. That is the unique degree $\le 4$ polynomial $r$ such that $f = q X^5 + r$ for some polynomial $q$.

Now what you need to show is: a polynomial $g$ is congruent to $f$ modulo $X^5$ if and only if $g$ is congruent to this $r$ modulo $X^5$.

That is $f-g$ is divisible by $X^5$ if and only if $g-r$ is divisible by $X^5$.

This is not very hard to do, and I'll let you try it. Let me know if you need further assistance.

Answer 2

To prove the desired statement it is a very good idea to first look at the case of integers. Let's see what happens there. Let $f$ be an integer and instead of $X^5$ we have the number $5$. In this case the equivalence classes, as also rightly constructed by you, are the following:

1. $\{\ldots,0,5,10,15,20, \ldots \}$
2. $\{\ldots, 1,6,11,16,21, \ldots \}$
3. $\{\ldots, 2,7,12,17,22, \ldots \}$
4. $\{\ldots, 3,8,13,18,23, \ldots \}$
5. $\{\ldots, 4,9,14,19,24, \ldots \}$

Notice how the difference between integers in the same equivalence class is some multiple of $5$. And conversely that any two numbers whose difference is a multiple of 5 are in the same equivalence class.

Now to check if our claim is actually correct, let us see what happens if $f= 13$. Here $[f]$ is the fourth equivalence class. What about $[f div 5]$? Well first of all $f div 5$ is the remainder we get when we divide $f=13$ by $5$, which is $3$. Clearly $3$ is also in the fourth equivalence class. So in this case $[f] = [f div 5]$.

Now let us take a little broader view to analyze why this happened. The key here is to remember how division works. When we divide $f$ by 5, we get the following by the euclidean algorithm: $$ f= q.5+r$$ where $q$ is the quotient and $r$ is the remainder. Notice that the difference between $f$ and $r$ has to be a multiple of $5$. Hence they have to be in the same equivalence class.

Now think about how you can generalize this argument to the case of polynomials. I am sure you can do it, if not you can ask for more details.