How can we prove that the ideals $\,I=(x_1+x_2, x_2^2)\,$ and $\,J=(x_1+x_2, x_1^2)$ are equal?
I was thinking of looking at $ {\rm Mon}(I) = {\rm Mon}(J)\,$. Thank you.
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What is Mon ?... – lhf Oct 26 '20 at 12:46
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monomial ideals – Wuestenfux Oct 26 '20 at 13:20
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1note $I\ni (x_1-x_2)(x_1+x_2)+x_2^2=x_1^2$ – Atticus Stonestrom Nov 13 '20 at 05:57
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In fact you just have to show that $x_1^2 \in I$. Consider $x_1(x_1+x_2) - x_2^2$.
monikernemo
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As in the mod reduction step in the Euclidean algorithm for the gcd, for a polynomial $\,f(t)\in \Bbb Z[t]$
$$\begin{align}\bmod\, &x\!+\!y\!:\,\ x\equiv -y\,\Rightarrow\, \color{#c00}{f(x)\equiv f(-y)}\\[.4em] \Longrightarrow\ \ &\!\!(x\!+\!y,\, \color{#c00}{f(x)}) \,=\, (x\!+\!y,\, \color{#c00}{f(-y)})\\[.4em] {\rm by}\ \ \ \ \ &\!\!(a, \color{#c00}{b}) = (a,\color{#c00}{b'})\ \ \,{\rm if}\,\ \ \color{#c00}{b\equiv b'}\!\!\!\pmod{\!a} \end{align}\qquad $$
OP is special case $\,f(t) = t^2.\,$ Generally for ideals - just as for gcds - the ideal is preserved by reducing a generator modulo another or, more generally, by unimodular transformations.
Bill Dubuque
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