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| \textbf{Definition: }\, A polynomial $p(x)=a_nx^n+\cdots+a_0$ over a UFD |
\textbf{Definition: }\, A polynomial $p(x)=a_nx^n+\cdots+a_0$ over a UFD |
| $R$ is said to be \emph{primitive} if its coefficients are not all divisible |
$R$ is said to be \emph{primitive} if its coefficients are not all divisible |
| by any element of $R$ other than a unit. |
by any element of $R$ other than a unit. |
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| \textbf{Proposition (Gauss): }\, |
\textbf{Proposition (Gauss): }\, |
| Let $R$ be a UFD and $F$ its field of fractions. \, |
Let $R$ be a UFD and $F$ its field of fractions. \, |
| If a polynomial $p\in R[x]$ is reducible in $F[x]$, then it is |
If a polynomial $p\in R[x]$ is reducible in $F[x]$, then it is |
| reducible in $R[x]$. |
reducible in $R[x]$. |
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| \textbf{Remark:}\, |
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| The above statement is often used in its contrapositive form. For an example of this usage, see \PMlinkname{this entry}{AlternativeProofThatSqrt2IsIrrational}. |
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| \textbf{Proof: }\, We may assume that $p$ is primitive. Suppose |
\textbf{Proof: }\, We may assume that $p$ is primitive. Suppose |
| $p=qr$ with $q,r\in F[x]$. \,There are unique elements $a,b\in F$ |
$p=qr$ with $q,r\in F[x]$. \,There are unique elements $a,b\in F$ |
| such that $q/a$ and $r/b$ are in $R[x]$ and are primitive. \,But |
such that $q/a$ and $r/b$ are in $R[x]$ and are primitive. \,But |
| $p/ab=(q/a)(r/b)$. \,Since $p$ is primitive, it follows from |
$p/ab=(q/a)(r/b)$. \,Since $p$ is primitive, it follows from |
| Gauss's lemma I that $ab$ is a unit, and therefore so are |
Gauss's lemma I that $ab$ is a unit, and therefore so are |
| $a$ and $b$. \,This completes the proof. |
$a$ and $b$. \,This completes the proof. |
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| \textbf{Remark: }\, Another result with the same name is |
\textbf{Remark: }\, Another result with the same name is |
| Gauss' lemma on quadratic residues. |
Gauss' lemma on quadratic residues. |
