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Consider the quotient ring $F[x]/(x^n)$, where $F$ is a field. Every element of this ring can be represented by a $n-1$ degree polynomial, i.e., written as $$f = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + (x^n)$$ (some coset of $(x^n)$).

I'm trying to prove that $f$ is a unit $\iff$ $a_0 \neq 0$.

Now, the unit in this ring is the element represented by $1 + (x^n)$. So for $\Rightarrow$, we can prove by contraposition $a_0 = 0 \Rightarrow f$ is not a unit. So let $a_0 = 0$ and assume $f$ is a unit, then there is a $g = b_0 + b_1x + ... + b_{n-1}x^{n-1} + (x^n)$ such that $fg = 1 + (x^n)$. This means that $fg$ must have a constant term $1$. But then $1 = a_0b_0 = 0\cdot b_0=0$ which is a contradiction (since in a field, $1 \neq 0$).

For the reverse implication, for small $n$ I was able to write down the inverses explicitly, ie; for $n = 2$, $f = a_0 + a_1x + (x^2)$ in $F[x]/(x^2)$, if $a_0 \neq 0$ we can take $g = a_0^{-1} - a_0^{-1}a_1a_0^{-1}x + (x^2)$ which is nonzero since $a_0^{-1}$ is nonzero. then $fg$ = $1 - a_1a_0^{-1}a_1a_0^{-1}x^2 + (x^2) = 1 + (x^2)$

now for $n=3$ we can find something similar, and the only inverse that shows up in the coefficients of $g$ is the inverse of $a_0$ which by assumption exists, the rest of the inverses do not have to be taken and so we won't run into trouble computing the inverse given any $n$ (besides the trouble that for large $n$, writing this out would be a lot of work).

However, I am stuck trying to give an actual proof for the reverse implication. I'm sure I'm missing something here, but I don't know what. Do you guys have any hints to help me solve this problem?

Thanks in advance!

user26857
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Brushy
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4 Answers4

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Suppose $\;f(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}\;,\;\;a_0\neq0\;$ , so $\gcd(x^n,\,f(x))=1$ , and thus there exist $\;p(x),\,q(x)\in\Bbb F[x]\;$ with $\;p(x) x^n+q(x) f(x)=1\;$ , and then

$$(f(x)+(x^n))(q(x)+(x^n))=1+(x^n)$$

Michael Hardy
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DonAntonio
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Why don't try to figure out that $F[X]/(X^n)$ is a local ring whose maximal ideal is $(X)/(X^n)$. Then the invertible elements are those outside the maximal ideal, that is, the images of polynomials which are not in $(X)$ hence having $f(0)\ne 0$.

With respect to your try: if $a_0\ne0$, then $f\notin(X)$, so $\gcd(X^n,f)=1$. Then there are $u,v\in F[X]$ such that $X^nu(X)+f(X)v(X)=1$. Now we get $f(X)v(X)\equiv 1\bmod (X^n)$.

user26857
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There's a simple trick here that also gives a formula for the inverse

$$ (1 + x g(x)) (1 - x g(x)) (1 + x^2 g(x)^2) (1 + x^4 g(x)^4) \cdots = (1 - x^2 g(x)^2) (1 + x^2 g(x)^2) (1 + x^4 g(x)^4) \cdots \\= (1 - x^4 g(x)^4) (1 + x^4 g(x)^4) \cdots \\\vdots \\ = 1 $$

Multiply and divide in a copy of $a$ as needed.

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You can make your approach work inductively. Suppose you know an inverse exists modulo $x^2$. Then, you have a $g$ such that

$$ fg \in b x^2 + (x^3) $$

If you add in a correction term to $g$, you get

$$ f \cdot (g + c x^2) \in (b + a_0 c) x^2 + (x^3) $$

and you can solve for $c$ so as to get an inverse modulo $x^3$. Induct.