inverse element in a set of polynomials

Question

Let $f\in\mathbb{F}[x]$ be a monic, irreducible polynomial of degree $d$.

We define $\mathbb{K}=\{g\in\mathbb{F}[x] \ | \ deg(g)<deg(f)\}$.

Let $g\in\mathbb{K}$, assume that $g\neq 0_\mathbb{F}$, prove that there exist a polynomial $q\in\mathbb{K}$ such that $g\cdot q=s\cdot f+1_\mathbb{F}$ for some polynomial $s\in\mathbb{F}[x]$.

It is true that $\mathbb{K}$ is not a subfield, since the multiplication is modulo $f$ but it doesn't matter for the problem.@Hurkyl — Don Fanucci, Mar 17 '17 at 15:32

Tsemo Aristide · Accepted Answer · 2017-03-17T15:36:50.283

1

Since $f$ is irreducible $\mathbb{F}[X]/(f)$ is a field. Let $p:\mathbb{F}[X]\rightarrow \mathbb{F}[X]/(f)$ the quotient map, you have $p(g)$ has an inverse $q'$. Thus $p(g)q'=1$. You can write $q'=a_0+a_1p(x)+...+a_np(x^n)$ where the degree of $f$ is $n+1$. Let $q=a_0+...+a_nx^n$. $p(qg)=1$ implies that $qg=sf+1$.

edited Mar 17 '17 at 15:36

answered Mar 17 '17 at 15:27

Tsemo Aristide

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inverse element in a set of polynomials

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