On the finite algebra $\Bbb F_q[X]_{<n}$ over $\Bbb F_q$

Question

Let us take the vector space of all the polynomials of degree less than $n$ over the finite field $\Bbb F_q$, \begin{eqnarray*} \Bbb F_q[X]_{<n} &:=& \{ f(X)\in \Bbb F_q[X]:\deg f(X) <n\} \leq \Bbb F_q[X]. \end{eqnarray*}

Clearly, this is not a ring under the usual multiplication of polynomials. On the other hand, let us endow the abelian group $(\Bbb F_q[X]_{<n},+)$ with the following multiplication: \begin{align*} \odot: \Bbb F_q[X]_{<n}\times \Bbb F_q[X]_{<n} \longrightarrow \Bbb F_q[X]_{<n},\quad (f(X),g(X))\longmapsto f(X)\odot g(X):=r(X), \end{align*} where $r(X)\in \Bbb F_q[X]$ is the remainder of the polynomial $f(X)g(X)\in \Bbb F_q[X]$, when divided by $X^n-1$. This multiplication is known as multiplication modulo $\boldsymbol{ X^n-1}$. Then, $(\Bbb F_q[X]_{<n},+,\odot)$ forms a commutative ring with $1_{\Bbb F_q[X]_{<n}}\in \Bbb F_q[X]_{<n}$.

But I stack on the rest of the properties in order to show the algebra structure.

So, please let me expose my thoughts. For the associativity of $\odot$, \begin{equation} (f(X)\odot g(X))\odot h(X)=f(X)\odot (g(X)\odot h(X)) \tag{$*$} \end{equation} my thought is to apply the Division Algorigthm. That is, by dividing $f(X)g(X)$ by $X^n-1$, there exists unique $q(X),r(X)\in \Bbb F_q[X]$ such that $f(X)g(X)=(X^n-1)q(X)+r(X)$, where $r(X)=0$ or $\deg r(X)<n$. Thus, $f(X)\odot g(X)=r(X)$. Then, $(f(X)\odot g(X))\odot h(X)=r(X)\odot h(X)$. Thus, in the same spirit, there exists $q'(X),r'(X)\in \Bbb F_q[X]$ such that $r(X)h(X)=q'(X)(X^n-1)+r'(X)$, where $r'(X)=0$ or $\deg r'(X)<n$. Then, \begin{eqnarray*} (f(X)\odot g(X))\odot h(X) &=& r'(X) \\ &=& r(X)h(X)-q'(X)(X^n-1) \\ &=& (f(X)g(X)-(X^n-1)q(X))h(X)-q'(X)(X^n-1)\\ &=& f(X)g(X)h(X)-(X^n-1)q(X)h(X)-q'(X)(X^n-1)\\ &=& f(X)g(X)h(X)-(X^n-1)(q(X)h(X)+q'(X)). \end{eqnarray*} The latter is of degree $<n$ and as we can see it is the remainder of the division of $f(X)g(X)h(X)$ by $X^n-1$, and it is unique. By working likewise in the RHS of $(*)$, we obtain the same result. Hence the equality holds.

Is that argument complete and correct? Do we work likewise for the rest axioms and, furthermore, is there any faster way?

Answer 1

There is a faster way. I think it is worth learning about quotients of rings by ideals. In your case, there is an isomorphism* of $\mathbb{F}_q[X]_{<n}$ and the quotient ring $\mathbb{F}_q[X]/I$ where $I$ is the principal ideal generated by $x^n$. The general theory then tells you immediately (just from coset business) that $\mathbb{F}_q[X]/I$ is a commutative ring with identity because the "mother" $\mathbb{F}_q[X]$ is. This way you can avoid checking all the axioms and put all the work on the abstract quotient ring construction.

*The correspondence is that the polynomial $a_0 + a_1x + \cdots + a_{n-1}x^{n-1} \in \mathbb{F}_q[X]_{<n}$ maps to the coset $a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + I$.