Saturation By Elimination

Prove that if $I \subset R$ is an ideal, and $f \in R$ is a polynomial, then $(I:f^\infty) = (I,ft-1)\cap R$ (where $(I,ft-1)\subset R[t]$). How might you find the smallest $\ell$ such that $(I:f^\ell) = (I:f^\infty)$? Use your method to compute $(I:(zw)^\infty)$, where $I = (x^2, y^3, xz+yw)$ "by hand" in Macaulay2. (You need to make a new ring, with appropriate monomial order, and look at an appropriate Groebner basis. Check you work using the "saturate" function.)


Suppose $g \in (I:f^\infty)$ so $gf^n \in I$ for some $n$. Now we show that $f^nt^n -1 \in (I,ft-1)$ by induction. Suppose $f^{k-1}t^{k-1}-1 \in (I,ft-1)$ for some $k$. Then $f^kt^k -1 = ft(f^{k-1}t^{k-1} - 1) + (ft -1)$ is in $(I,ft-1)$. As a result $g = gf^nt^n - g(f^nt^n -1)$ is in $(I,ft-1)$. Therefore $(I:f^\infty) \subset (I,ft-1)\cap R$.

Suppose $g \in (I,ft-1)\cap R$ so it can be expressed as

\begin{align} g = \sum_{i=0}^\infty g_i t^i + (ft-1)\sum_{i=0}^\infty h_i t^i \end{align}

with each $g_i \in I$ and each $h_i \in R$.

\begin{align} g = g_0 - h_0 + \sum_{i=1}^\infty (g_i + fh_{i-1} - h_i)t^i. \end{align}

Since $g \in R$, all terms with $t$-degree > 0 must cancel. Let $n$ be the largest integer for which $h_n \neq 0$. Then $g_{n+1} + fh_n = 0$, which implies $h_n \in (I:f)$. Then $g_n + fh_{n-1} - h_n = 0$ implies $h_{n-1} \in (I:f^2)$. Repeating this argument $n$ times, $h_0 \in (I:f^{n+1})$. Therefore $g = g_0- h_0 \in (I:f^{n+1})$.

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