Equivalence class $r$ in $R$ is represented by $r=a_0 + a_1 X + \cdots+ a_{n-1} X^{n-1},$ with $a_i \in K.$ Show $r \in R$ is a unit iff $a_0 \neq 0$.

Question

Given that $K$ is a field and $R= K [X] / (X^n)$ where $n \in \mathbb{Z}_{\geq 1},$ $x$ is defined by $x:= X + (X^n) \in R$.

I want to show that the equivalence class $r$ in $R$, represented by $r=a_0 + a_1 X + \cdots+ a_{n-1} X^{n-1}$ with $ a_i \in K$ is a unit if and only if $a_0 \neq 0$.

The fact that it is a unit if $a_0 \neq 0$ then shows that an element $r$ with $a_0 \neq 0$ has an inverse, and I am trying to find this inverse.

Any help is appreciated.

Answer 1

If $a_0 = 0$, we can show that $r$ is a $0$ -divisor. Take $s \in R$ to be the element $s = X^{n-1} \neq 0$. Then $rs = a_1X^n +a_2X^{n+1} + \dots + a_{n-1}X^{2n-1} = 0$ in $R$. Of course, $r$ being a $0$-divisor prevents it from being a unit.

Now, if $a_0 \neq 0$, then we let $s = b_0 + b_1X +b_2X^2 ... +b_{n-1}X^{n-1}$. We pick $b_0 = a_0^{-1}$. Having picked $b_i$, we pick $b_{i+1}$ such that the sum $$b_0a_{i+1} + b_1 a_i \dots + b_{i+1}a_0 = 0$$ for $1\leq i \leq n-2$.

We find that the product $rs$ is $a_0b_0 + (a_0 b_1 +a_1 b_0)X \dots + \big(b_0a_{n-1} + b_1 a_{n-2} \dots + b_{n-1}a_0)X^n + kX^{n} = a_0b_0 = 1$ (where $k$ is some polynomial)

Answer 2

Try brute force.

We want to find $f\in K[X]$ so that $fr-1=g(X^n)$ for some $g\in K[X]$. So write

$$ \left(\sum_{k=0}^p c_kX^k\right)\cdot \left(\sum_{i=0}^m a_iX^i\right) = (X^n)\cdot \left(\sum_{j=0}^\ell b_jX^j\right) +1 = \sum_{j=0}^\ell b_jX^{j+n}+1, $$ which gives $$ \sum_{k+i=q} c_ka_iX^q = \sum_{j=0}^\ell b_jX^{j+n}+1. $$ Then solve for what the $c_k$ would have to be.

Answer 3

I think it might help clarify the situation here if we start off with the observation that the cosets $1, X, X^2, \ldots, X^{n - 1} \in R = K[X]/(X^n)$ are linearly independent over $K$; for in not, there are $k_i \in K$, $0 \le i \le n - 1$, not all zero, with

$\displaystyle \sum_0^{n - 1} k_i X^i = 0 \in R = K[X]/(X^n), \tag 1$

that is, the polynomial $k(X) \in K[X]$, where

$k(X) = \displaystyle \sum_0^{n - 1} k_i X^i \in (X^n) \subset K[X], \tag 2$

which implies there is some

$0 \ne g(X) = \displaystyle \sum_0^m g_i X^i \in K[X] \tag 3$

with

$0 \ne \displaystyle \sum_0^{n - 1} k_i X^i = X^n g(X) = \sum_0^m g_i X^{i +n} \in K[X]; \tag 4$

then, comparing degrees of the terms on either side, we see that the least possible degree of any term occurring on the right is $n$, whereas the highest possible degree of any term on the left is $n - 1$; such a situation is clearly impossible, so we must in fact have all $k_i = 0$ and thus $\{1, X, X^2, \ldots, X^{n - 1} \}$ is a linearly independent set over the field $K$.

In the above, we have of course allowed ourselves the slight abuse of notation whereby we employ $X^i$ to denote either the polynomial $X^i \in K[X]$ or the coset $X^i + (X^n) \in R = K[X]/(X^n)$; hopefully the context serves to distinguish one usage from the other.

Now suppose

$r = a_0 + a_1 X + a_2 X^2 + \ldots a_{n - 1} X^{n - 1} \in R = K[X]/(X^n) \tag 5$

is a unit in $K[X]/(X^n)$ and we let

$s = b_0 + b_1 X + b_2 X^2 + \ldots + b_{n - 1} X^{n - 1} \in R = K[X]/(X^n) \tag 6$

be the representative of a coset such that

$sr = rs = 1_R = 1_K + (X^n) \in R = K[X]/(X^n); \tag 7$

then

$sr = ( b_0 + b_1 X + b_2 X^2 + \ldots + b_{n - 1} X^{n - 1})(a_0 + a_1 X + a_2 X^2 + \ldots a_{n - 1} X^{n - 1})$ $= b_0 a_0 + (b_0 a_1 + b_1 a_0) X + (b_0 a_2 + b_1 a_1 + b_2 a_0) X^2 + \ldots$ $+ (b_0 a_{n - 1} + b_1 a_{n - 2} + \ldots + b_{n - 1} a_0) X^{n - 1} = 1_R = 1_K + (X^n) \in R; \tag 8$

the terms containing higher powers of $X$, that is $X^m$ where $m \ge n$, vanish since $X^n = 0$ in $R$; since as we have seen the $X^i$, $0 \le i \le n - 1$, are linearly independent over $K$ we find

$b_0 a_0 = 1_K \in K, \tag{9}$

$b_0 a_1 + b_1 a_0 = b_0 a_2 + b_1 a_1 + b_2 a_0 = \ldots = b_0 a_{n - 1} + b_1 a_{n - 2} + \ldots + b_{n - 1} a_0 = 0; \tag{10}$

it follows from (9) that $0 \ne a_0 \in K$ (since $a_0 b_0 = 1$) is invertible with

$a_0^{-1} = b_0 \in K; \tag{11}$

furthermore, the remaining coefficients $b_1, b_2, \ldots, b_{n - 1}$ of $s$ are inductively determined from (10), viz.

$b_1 = -\dfrac{a_1 b_0}{a_0} = -a_0^{-1}(a_1 b_0), \tag{12}$

$b_2 = -\dfrac{a_2 b_0 + a_1 b_1}{a_0} = -a_0^{-1}(a_2 b_0 + a_1 b_1), \tag{13}$

and so forth, successively determing the $b_i$ until we reach $i = n - 1$:

$b_{n - 1} = -a_0^{-1}(b_0 a_{n - 1} + b_1 a_{n - 2} + \ldots + b_{n - 2} a_1); \tag{14}$

these considerations show that, not only is $a_0 \ne 0$ invertible but also that the entire polynomial $s$ is uniquely determined by the requirement (7) that $r$ be a unit in $R = K[X]/(X^n)$.

We can in fact follow the above argument in the reverse direction and thus see that if $0 \ne a_0 \in K$ we may successively define the $b_i$ via (12)-(14) etc. and thus arrive at $s$ as in (6) which satisfies (7), showing $r \in R = K[X]/(X^n)$ is indeed a unit.