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Let $R$ be a ring and $f, g \in R[x], \deg(g) > 0$ and the dominant coefficient of $g$ an inversible element of $R$. There exists unique $q, r \in R[x]$ such that $$f(x) = q(x)g(x) + r(x)$$ with $\deg(r) < \deg(g)$.

This is basically the theorem of euclidean division of polynomials. But what happens if the dominant coefficient is not an inversible element of $R$ ? (If the ring is $\mathbb Z$ for example and the dominant coefficient of $g$ is $\neq ±1$.

Kilkik
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  • The theorem will hold true in $F[x]$ where $F$ is the field of fractions of $R$. I believe it's equivalent to say that there exists an element $s$ of $R$ such that $sf(x)$ can be written as $q(x)g(x)+r(x)$ in that way, with uniqueness up to further scaling by elements of $R$. – Greg Martin Jun 22 '21 at 17:28
  • Would it be fair to say "Euclidean division of polynomials when the leading coefficient of the divisor is not invertible" rather than "a non inversible ring"? I do not think the latter is proper usage. – rschwieb Jun 22 '21 at 18:57
  • This question is there but it simply asks if the theorem fails or not. This one asks "what goes wrong" so I guess it is different. – rschwieb Jun 22 '21 at 18:59

1 Answers1

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Write $g(x)=\sum_0^k g_ix^i$ and $f(x)=\sum_0^n f_nx^i$.

The first step of the algorithm, as I understand the steps, is to find an element $r\in R$ such that $g_kr=f_n$, so that you can subtract $g(x)rx^{n-k}$ from $f(x)$ to get something with lesser degree than $f(x)$.

So as you can see, this all depends on $f_n$ being in the ideal $g_kR$. If $f_n\notin g_kR$ then we simply are at a loss of how to execute the first step. Further along the line, you're going to have other, less predictable coefficients that also have to lie in $g_kR$.

If $g_k$ is a unit, then you have no problem because $g_kR=R$, and the coefficients cannot fail to be multiples of $g_k$.

rschwieb
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