$\frac{F[x]}{\langle x^2\rangle} \cong \{{\begin{bmatrix} a_0 & a_1 \\ 0 & a_0 \\ \end{bmatrix} \text{such that } a_i \in F \text{ for all i} }\}$

I'm trying to generalize the following isomorphisms:

$$\frac{F[x]}{\langle x^2\rangle} \cong \{{\begin{bmatrix} a_0 & a_1 \\ 0 & a_0 \\ \end{bmatrix} \text{such that } a_i \in F \text{ for all i} }\}$$

$$\frac{F[x]}{\langle x^3\rangle} \cong \{{\begin{bmatrix} a_0 & a_1 & a_2 \\ 0 & a_0 & a_1 \\ 0 & 0 & a_0 \end{bmatrix} \text{such that } a_i \in F \text{ for all i} }\}$$

to any $n$

And i'm trying to do so as cleanly and efficiently as possible....

Since $x^n$ is a monic polynomial, we have that: $$\frac{F[x]}{\langle x^n\rangle} \cong \{a_0+a_1t+...a_{n-1}t^{n-1} | a_i \in F \text{ and } t^n=0\}$$

So, an arbitrary element of $\frac{F[x]}{\langle x^n\rangle}$ can be written as: $$a(x) = a_0+a_1t+...+a_{n-1}t^{n-1}.$$

Now, I will define the proposed ring isomorphism by:

$\gamma(a(x)) = (a_{i,j})$

where $(a_{i,j}) = \begin{cases} a_{i,i}=a_0 & 1 \leq i \leq n \\ a_{i,i+1}=a_1 & 1\leq i\leq n-1 \\ ..... & .....\\ a_{i,i+k}=a_k & 1\leq i\leq n-k \\ ..... & .....\\ a_{i,i+(n-2)}=a_{n-2} & 1\leq i\leq 2 \\ a_{i,i+(n-1)}=a_{n-1} & i=1\\ 0 & i>j \end{cases}$

Since $\gamma(a(x))=I \rightarrow a(x) = 1$, this map is injective. Thus this map is bijective onto it's image, and the image of this map generalizes the cases of $n=2$ and $n=3$.

Also, notice that given $a(x),b(x) \in \frac{F[x]}{\langle x^n\rangle}$, we that that:

$$a(x)b(x) = \Sigma_{k=0}^{n-1} \Sigma_{i+j=k}a_ib_jt^{i+j}$$ Thus, we have that:

$\gamma(a(x)b(x)) = \gamma(\Sigma_{k=0}^{n-1} \Sigma_{i+j=k}a_ib_jt^{i+j})=$

$$\begin{cases} a_{i,i}=\Sigma_{i+j=0}a_ib_j & 1 \leq i \leq n \\ a_{i,i+1}=\Sigma_{i+j=1}a_ib_j & 1\leq i\leq n-1 \\ ..... & .....\\ a_{i,i+k}=\Sigma_{i+j=k}a_ib_j & 1\leq i\leq n-k \\ ..... & .....\\ a_{i,i+(n-2)}=\Sigma_{i+j=n-2}a_ib_j & 1\leq i\leq 2 \\ a_{i,i+(n-1)}=\Sigma_{i+j=n-1}a_ib_j & i=1\\ 0 & i>j \end{cases}$$

Now, I'm trying to show that $\gamma(a(x))\gamma(b(x))$ has the same form and thus show that the proposed map preserves the ring multiplication. I can see why this is true, but I've been having trouble writing it up cleanly and would appreciate some help! Thanks

Solution 1:

For an efficient proof, I would proceed as follows.

Define a ring homomorphism $\phi:F[x] \to M_n(F)$ by $\phi(f) = f(N)$, where $$ N = \pmatrix{0&1\\&0&1\\&&\ddots&\ddots\\&&&0&1\\ &&&&0}. $$ From this definition, it is clear that $\phi$ preserves addition and multiplication. We simply have $$ \phi(f + g) = (f+g)(N) = f(N) + g(N) = \phi(f) + \phi(g),\\ \phi(fg) = (fg)(N) = f(N)g(N) = \phi(f)\phi(g). $$

We can prove inductively that $$ [N^k]_{i,j} = \begin{cases} 1 & j-i = k,\\ 0 & \text{otherwise}. \end{cases} $$ See the proof that I give at the end of my post here, for instance. Thus, we can conclude that if $f(x) = a_0 + a_1 x + \cdots + a_dx^d$ with $d \geq n-1$ (with the leading coefficient possibly zero), then we have $$ f(N) = \pmatrix{a_0 & a_1 & \cdots & a_{n-1}\\&a_0&\ddots&\vdots\\ &&\ddots&a_1\\ &&&a_0}. $$ Thus, letting $R$ denote the subring of $M_n$ consisting of matrices of the form $$ \pmatrix{a_0 & a_1 & \cdots & a_{n-1}\\&a_0&\ddots&\vdots\\ &&\ddots&a_1\\ &&&a_0}, \quad a_0,a_1,\dots,a_{n-1} \in F, $$ it is clear that the image of $\phi$ is equal to $R$. Moreover, if $f(x) = a_0 + a_1 x + \cdots + a_dx^d$ with $d \geq n-1$, then it is clear that $\phi(f) = 0$ iff $a_0 = a_1 = \cdots = a_{n-1} = 0$. Thus, we have $\ker f = \langle x^n \rangle$.

Putting all of that together, the first isomorphism theorem for rings allows us to conclude that $$ F[x]/\ker \phi \cong \operatorname{im}(\phi) \implies F[x]/\langle x^n \rangle \cong R, $$ which was what we wanted.

Another interesting approach is to consider the map $\psi:F[x]/\langle x^n \rangle \to M_n(F)$ as follows. For any $f \in F[x]/\langle x^n \rangle$, let $\mu_f$ denote the associated "multiplication operator". That is, let $\mu_f:F[x]/\langle x^n\rangle \to F[x]/\langle x^n\rangle$ denote the $F$-linear map given by $$ \mu_f(g(x) + \langle x^n \rangle) = f(x)g(x) + \langle x^n \rangle. $$ We define $\psi(f)$ to be the matrix of $\mu_f$ relative to the basis $$ \mathcal B = \{x^{n-1} + \langle x^n \rangle, \dots, x + \langle x^n \rangle,1 + \langle x^n \rangle\}. $$ This kind of map may seem contrived, but it actually is important to the usual proof of the existence of the Frobenius normal form