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$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).

Note. Please don't try to read this. I've let it here with the hope that someone can tell me what is going on with my LaTeX and help me to fix it.

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$. As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).

Note. Please don't try to read this. I've let it here with the hope that someone can tell me what is going on with my LaTeX and help me to fix it.

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$. As localization commutes with local cohomology $H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]_P}{(5x+4y)}\right).$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$. As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).