# Tight closure/x^4,y^4,xy^3/On P^1 and other curves/Example

We consider the ideal $I={\left(x^{4},y^{4},xy^{3}\right)}$ in $S=K[x,y]$ and in finite graded extensions $S\subseteq R$ (e.g. $R=K[x,y,z]/(F)$, where $F$ is a homogeneous integral equation for $z$ over $K[x,y]$) and describe an algorithm to compute the tight closure $I^{*}$. The graded resolution of the ideal is
    $0\longrightarrow S(-5)\oplus S(-7)\longrightarrow S(-4)\oplus S(-4)\oplus S(-4){\stackrel {x^{4},y^{4},xy^{3}}{\longrightarrow }}S\longrightarrow S/I\longrightarrow 0,$
where the map on the left is given by sending the generators to
    $(0,x,-y){\text{ and }}(y^{3},0,-x^{3}).$
On the projective line ${\mathbb {P} }_{K}^{1}$ this corresponds to
    $0\longrightarrow {\mathcal {O}}(-5)\oplus {\mathcal {O}}(-7)=\operatorname {Syz} {\left(x^{4},y^{4},xy^{3}\right)}\longrightarrow {\mathcal {O}}(-4)\oplus {\mathcal {O}}(-4)\oplus {\mathcal {O}}(-4){\stackrel {x^{4},y^{4},xy^{3}}{\longrightarrow }}{\mathcal {O}}\longrightarrow 0.$
We may pull back this sequence along the finite morphism
    $C=\operatorname {Proj} {\left(R\right)}\longrightarrow {\mathbb {P} }_{K}^{1}$
to obtain the corresponding exact sequence over the curve $C$ which can be used to compute the tight closure of the ideal in $R$. A homogeneous element $h\in R$ of degree $m$ yields a cohomology class in
    $H^{1}(C,\operatorname {Syz} {\left(x^{4},y^{4},xy^{3}\right)}(m))\cong H^{1}(C,{\mathcal {O}}_{C}(m-5))\oplus H^{1}(C,{\mathcal {O}}_{C}(m-7))\,,$
which can be easily computed using Čech cohomology. On $D_{+}(x)$, $h$ comes from $\left({\frac {h}{x^{4}}},0,0\right)$ and on $D_{+}(y)$ it comes from $\left(0,{\frac {h}{y^{4}}},0\right)$. Their difference, the syzygy
    $\left({\frac {h}{x^{4}}},-{\frac {h}{y^{4}}},0\right)$
equals
    $-{\frac {h}{xy^{4}}}\left(0,x,-y\right)+{\frac {h}{x^{4}y^{3}}}\left(y^{3},0,-x^{3}\right).$
Hence the components of this cohomology class are
    $-{\frac {h}{xy^{4}}}\in H^{1}(C,{\mathcal {O}}_{C}(m-5)){\text{ and }}{\frac {h}{x^{4}y^{3}}}\in H^{1}(C,{\mathcal {O}}_{C}(m-7)).$
Therefore the issue whether $h$ belongs to the tight closure of $I$ depends on these two components, which both correspond to a parameter situation.
First of all, if $m\geq 7$, then both degrees are non-negative and therefore these classes are tightly $0$ by (the proof of) fact. If $m=6$, we only have to look at the second component inside $H^{1}(C,{\mathcal {O}}_{C}(-1))$. For the monomial $y^{3}z^{3}$ the second component is $0$ (independent of $F$), hence it belongs to the tight closure, though the first component need not be $0$. The monomial $x^{2}y^{2}z^{2}$ yields ${\frac {z^{2}}{x^{2}y}}$, which is not $0$ unless the equation has low degree. This class is (with some exceptions in small characteristics) not tightly $0$. Hence $x^{2}y^{2}z^{2}$ does not belong to the tight closure. For $m=5$, still only the second component is interesting, therefore $y^{3}z^{2}$ belongs to the tight closure, but $xy^{2}z^{2}$ not (under the same restrictions). For $m\leq 4$ both components lie in negative degree, so an element will belong to the tight closure only if it belongs to the ideal itself. For the element $y^{3}z$ the second component is $0$, but not the first component.