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Revision difference : ideal decomposition in Dedekind domain |
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Version 4 |
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According to the entry ``\PMlinkname{fractional ideal}{FractionalIdeal}'', we can \PMlinkescapetext{state} that in a Dedekind domain $R$, each non-zero integral ideal $\mathfrak{a}$ may be written as a product of finitely many prime ideals $\mathfrak{p}_i$ of $R$,
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According to the entry ``\PMlinkname{fractional ideal}{FractionalIdeal}'', we can \PMlinkescapetext{state} that in a Dedekind domain $D$, each non-zero integral ideal $\mathfrak{a}$ may be written as a product of finitely many prime ideals $\mathfrak{p}_i$ of $D$,
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| $$\mathfrak{a} = \mathfrak{p}_1\mathfrak{p}_2...\mathfrak{p}_k.$$ |
$$\mathfrak{a} = \mathfrak{p}_1\mathfrak{p}_2...\mathfrak{p}_k.$$ |
| The \PMlinkescapetext{product decomposition is unique up to the order of the factors}. |
The \PMlinkescapetext{product decomposition is unique up to the order of the factors}. |
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\textbf{Corollary.}\, If $\alpha_1$, $\alpha_2$, ..., $\alpha_m$ are elements of a Dedekind domain $R$ and $n$ is a positive integer, then one has
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\textbf{Corollary.}\, If $\alpha_1$, $\alpha_2$, ..., $\alpha_m$ are elements of a Dedekind domain $D$ and $n$ is a positive integer, then one has
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| \begin{align} |
\begin{align} |
| (\alpha_1,\,\alpha_2,\,...,\,\alpha_m)^n = |
(\alpha_1,\,\alpha_2,\,...,\,\alpha_m)^n = |
| (\alpha_1^n,\,\alpha_2^n,\,...,\,\alpha_m^n) |
(\alpha_1^n,\,\alpha_2^n,\,...,\,\alpha_m^n) |
| \end{align} |
\end{align} |
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for the ideals of $R$.
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for the ideals of $D$.
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| This corollary may be proven by induction on the number $m$ of the \PMlinkescapetext{generators (not on the exponent} $n$). |
This corollary may be proven by induction on the number $m$ of the \PMlinkescapetext{generators (not on the exponent} $n$). |
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| It's interesting to notice that an equation like (1), even in the case\, $m = 2$,\, implies the invertibility of the ideal\, $(\alpha_1,\,\alpha_2)$ in any commutative ring $R$ with unity, provided that at least one of $\alpha_1$ and $\alpha_2$ is regular element.\, We prove a bit more: |
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| \begin{thmplain} |
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| Let $R$ be a commutative ring with non-zero unity.\, If an ideal\, $(a,\,b)$\, of $R$, with $a$ or $b$ regular, obeys the multiplication rule |
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| \begin{align} |
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| (a,\,b)(c,\,d) = (ac,\,ad\!+\!bc,\,bd) |
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| \end{align} |
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| with all ideals $(c,\,d)$, then\, $(a,\,b)$ is an invertible ideal. |
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| \end{thmplain} |
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| {\em Proof.}\, The rule gives |
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| $$(a,\,b)^2 = (a,\,-b)(a,\,b) = (a^2,\,ab\!-\!ba,\,b^2) = (a^2,\,b^2).$$ |
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| Thus the product $ab$ may be written in the form |
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| $$ab = ua^2\!+\!vb^2,$$ |
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| where $u$ and $v$ are elements of $R$.\, Let's assume that e.g. $a$ is regular.\, Again applying the rule yields |
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| $$(a,\,b)(va,\,a-vb)(a^{-2}) = (va^2,\,a^2-vab+vab,\,ab-vb^2)(a^{-2}) = |
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| (va^2,\,a^2,\,ua^2)(a^{-2}) = (v,\,1,\,u) = R.$$ |
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| Consequently the ideal\, $(a,\,b)$\, has an inverse ideal, which settles the proof. |
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