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My text book states that "if $p$ and $q$ are polynomials, with $q \ne 0$, then there exist polynomials $G$ and $R$ such that $p/q = G + R/q$, and $\deg R < deg q$ or $R=0$"

So if $p(x) = 1$ and $q(x) = 4$, then $p/q only = G + R/q$ where $R/q = 0$ and $p/q = G$.

Is my reasoning correct? Does the quoted rule apply to constant polynomials?

If so, what type of rule is it. You can make a rule stating $widgit = something + x$ is $something$ or $x = 0$. If $something$ is allowed to be $widget$.

vonbrand
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naftalimich
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    If the coefficients are from a field, like the rationals or the reals, it is a theorem. It applies to constant polynomials. In your case $G=\frac{1}{4}$ and $R=0$. – André Nicolas Jun 20 '14 at 20:53
  • You don't state your reasoning. But I explained the essence of the matter in my answer. If that is not clear then please feel welcome to post questions in the comments. – Bill Dubuque Jun 21 '14 at 01:12

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It certainly applies, the constant polynomials are also polynomials.

More generally, definitions are often very carefully tailored to make important results (like this one certainly is) be covered uniformly, even in corner cases (like constant polynomials like here). The case of the zero polynomial ($q$ and $R$ here) has to be handled differently in many cases.

And yes, you are right. Definitions are essentially arbitrary. But you want them to be useful, saving words (and work), that's why important definitions are't random at all.

vonbrand
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Your conclusion is correct, but you don't state your reasoning, so that cannot be judged. Since this extremal case is extremely important (it yields uniqueness of prime factorizations), it is worth examining it in further detail. Let $\,R = F[x]\,$ be a polynomial ring over a field $\,F.$

Eliminating the fractions yields an equivalent statement, the (Euclidean) Division Algorithm for polynomials: if $\,f,\,g\in F[x]\,$ are polynomials over a field and $\,g\ne 0\,$ then there are polynomials $\,q\,$ (= quotient) and $\,r\,$ (= remainder) in $\,F[x]\,$ such that $\, f = q\, g + r\,$ and $\,r = 0\,$ or $\,\deg r < \deg g.\,$ Said equivalently, if $\,g\nmid f\,$ then the remainder $\, r = (f\ {\rm mod}\ g)\,$ has smaller degree than that of $\,g.\ $

Thus if $\,g\,$ has minimal degree, i.e. $\,\deg g = 0,\,$ then $\,r = 0\ $ (else $\,\deg r < \deg g = 0,\,$ contradiction). Hence the minimal degree elements (constants $\,c\ne 0)$ divide every element of $\,R.\,$ In particular, $\,c\mid 1,\,$ i.e. $\,c\,$ is a unit (invertible). Therefore, necessarily, the coefficient ring $\,F\,$ must be a field.

The same reasoning applies to any subset $\,I\subset R\,$ that is closed under the modulo operation, in particular any ideal $\,I\ne 0.\,$ Thus the least degree element $\,0\ne g\in I\,$ divides every element of $\,I,\,$ hence $\, I = (g)\,$ is a principal ideal. From this one immediately deduces that irreducibles are prime, therefore factorizations into irreducibles are unique (up to order and unit factors).

Bill Dubuque
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