Determining the colon ideal in a polynomial ring

Question

I am a grad student having background in a Commutative Algebra. I need help with the following.

Let $$ \begin{aligned} I = \langle &x_1^4,\, x_2^4,\, x_3^4,\, x_4^4, \\ &x_1^3 x_2^3,\, x_1^3 x_3^3,\, x_1^3 x_4^3,\, x_2^3 x_3^3,\, x_2^3 x_4^3,\, x_3^3 x_4^3, \\ &x_1^2 x_2^2 x_3^2,\, x_1^2 x_3^2 x_4^2,\, x_2^2 x_3^2 x_4^2,\, x_1^2 x_2^2 x_4^2, \\ &x_1 x_2 x_3 x_4 \rangle \end{aligned} $$ be an ideal in a polynomial ring $R=\mathbb{R}[x_1,x_2,x_3,x_4]$.

Question: Prove that there does not exist any $f\in R$ with $\mathrm{deg}(f)\leq 5$ such that $(I:f)=\langle x_1,x_2,x_3,x_4 \rangle$. Here $(I:f)=\{r \in R: rf \subseteq I\}$.

Any help would be appreciated. Thank you.

Answer 1

We have $(I:f)\supsetneq\langle x_1,x_2,x_3,x_4 \rangle$ if and only if $1\in (I:f)$, which happens if and only if every monomial in $f$ is divisible by one of the listed monomial generators of $I$.

Thus if $(I:f)= \langle x_1,x_2,x_3,x_4 \rangle$, one of the monomials $m$ of $f$ must not be divisible by any of the listed generators of $I$. On the other hand, increasing any of the exponents of $x_1,x_2,x_3,x_4$ in $m$ would result in a monomial divisible by one of the listed generators of $I$.

Let $e_1, e_2, e_3,e_4$ denote these exponents respectively. We know that $e_{i_1}=0$ for some $i_1\in1,2,3,4$ (or $m$ would be divisible by $x_1x_2x_3x_4$).

Now $mx_{i_1}$ must be divisible by one of the listed generators of $I$. Any such generator dividing $mx_{i_1}$ must be divisible by $x_{i_1}$, or else it would divide $m$. Thus the generator must have exponent exactly $1$ on $x_{i_1}$ (or it would not divide $mx_{i_1}$). We conclude the generator must be $x_1x_2x_3x_4$ so $e_j\geq1$ for $j\neq i_1$, or else

Now we must have $e_{i_2}=1$ for some $i_2\neq i_1$, or $m$ would be divisible by a listed generator of $I$. Thus $mx_{i_2}$ must be divisible by a listed generator of $I$ with exponent exactly $2$ on $x_{i_2}$. Thus $e_j\geq 2$ for $e_j\neq i_1,i_2$, or else $mx_{i_2}$ would not be divisible by any of the listed generators of $I$.

Continuing in this way, we see that $e_{i_3}=2$ for some $i_3\neq i_1,i_2$, and $e_{i_4}=3$ for $i_4\neq i_1,i_2,i_3$. Thus $m=x_{i_2}x_{i_3}^2x_{i_4}^3$, which has degree $6$.