Recall that if $k$ is an algebraically closed field, then any degree $d$ plane curve $X$ will have arithmetic genus $g=(d-1)(d-2)/2$, by a simple calculation of $H^1(X,\mathscr{O}_X)$. This tells us that no smooth projective curve of genus $5$ can be embedded in $\mathbb{P}^2_k$.

This is my thought process so far: if $i:X\hookrightarrow\mathbb{P}^n$ is a closed immersion (with $n\ge2$), then the short exact sequence $$0\to\mathscr{I}\to\mathscr{O}_{\mathbb{P}^n}\to i_*\mathscr{O}_X\to 0$$ gives and the fact that $H^1(\mathbb{P}^n,\mathscr{O}_{\mathbb{P}^n}) = H^2(\mathbb{P}^n,\mathscr{O}_{\mathbb{P}^n})=0$ gives us that $$H^1(X,\mathscr{O}_X)\cong H^1(\mathbb{P}^n,i_*\mathscr{O}_X)\cong H^2(\mathbb{P}^n,\mathscr{I})$$ and so, the problem of finding a curve with arithmetic genus $5$ reduces to finding an ideal sheaf $\mathscr{I}\subset\mathscr{O}_{\mathbb{P}^n}$ cutting out a curve, such that $\dim_k H^2(\mathbb{P}^n,\mathscr{O}_{\mathbb{P}^n})=5$.

Is this the correct approach? If so, where should I go from here?

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    $\begingroup$ May be you should go one more step and consider $\mathbb{P}^1\times\mathbb{P}^1$ (which as you know is a smooth quadric hypersurface in 3-space). If you take a type $(a,b)$ curve then its genus can easily seen to be $(a-1)(b-1)$. So, you only need to find a smooth curve of type $(2,6)$ which can be done. $\endgroup$ – Mohan Apr 18 '17 at 21:40
  • $\begingroup$ I think I might not have enough background knowledge to fully piece together a proof. If we let $j:\mathbb{P}^1\times\mathbb{P}^1\hookrightarrow\mathbb{P}^3$ and let $\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(1) = j^*\mathscr{O}_{\mathbb{P}^3}(1)$, then is there a simple expression for the cohomology $H^i(\mathbb{P}^1\times\mathbb{P}^1,\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(d))$? $\endgroup$ – Monstrous Moonshine Apr 18 '17 at 21:45
  • $\begingroup$ For any hypersurface in projective spaces, these are easy to write down. In this particular case, it is even simpler, since it is a degree 2 hypersurface. $\endgroup$ – Mohan Apr 18 '17 at 22:08
  • $\begingroup$ Ok, I've followed your advice, and worked out the painstaking details, and now I've just reduced to finding some $(6,2)$-bihomogeneous $f\in k[x_0,x_1;y_0,y_1]$ satisfying the usual irreducibility and nonsingularity criteria when we move to the four standard affine opens. Is there any obvious candidate for such an $f$? I'm kind of loath to try to find such an $f$ simply by trial and error at this point. $\endgroup$ – Monstrous Moonshine Apr 19 '17 at 1:48
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    $\begingroup$ @MonstrousMoonshine Consider the closed immersion $i : \mathbb P^1 \times \mathbb P^1 \hookrightarrow \mathbb P^{(a+1)(b+1)-1}$ defined using global sections of $\mathcal O(a,b)$. Notice that a global section of $\mathcal O(a,b)$ is a hyperplane section from the perspective of the $\mathbb P^{(a+1)(b+1)-1}$! That's why you can use Bertini. $\endgroup$ – Kenny Wong Apr 19 '17 at 9:29

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