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| Title of object: |
Gauss's Lemma I |
| Canonical Name: |
GausssLemmaI |
| Type: |
Definition |
| Created on: |
2002-11-04 05:39:33.97117-05 |
| Modified on: |
2002-11-04 05:39:33.97117-05 |
| Classification: |
msc:12E05 |
Preamble:
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Content:
There are a few different things that are sometimes called "Gauss's Lemma". See also Gauss's Lemma II.
\\\emph{Gauss's Lemma I:} If $f(x)$ and $g(x)$ are both primitive in R[x], so is $f(x) g(x)$.
\\ \emph{Proof:}
Suppose $f(x) g(x)$ not primitive. We will show either $f(x)$ or $g(x)$ isn't as well. $f(x) g(x)$ not primitive means the gcd of the coefficients of $f(x) g(x)$ is not a unit. Let $p$ be a prime factor of that gcd. Look at the image of everthing mod $p$.
R/p is an integral domain (\emph{why?}), so R/p[x] is an integral domain (\emph{why?}). And we have
\begin{align*}
\overline{f(x)} \ \overline{g(x)} = 0
\end{align*}
So $\overline{f(x)} = 0$ or $\overline{g(x)} = 0$. So $f(x)$ or $g(x)$ is divisible by $p$, so one of them is not primitive. |
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