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A really neat well known example is as follows:

Choose a conic $C_1$ and a tangential line $C_2$ in $\mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $\phi:\mathbb{P}^2\rightarrow{X}$ with $C:=\phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $\mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $\mathbb{P}^2$. Note that the degrees of $L'|C_1$ and $L'|C_2$ both coincide with the degree of $L|C$ and are therefore equal. But $L'\cong\mathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'\cong\mathcal{O}$, which is non-effective! A contradiction!

In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme.

In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1\amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism above), and $Z=\mathbb{P}^2$, but the argument is more general provided any finite set of closed points in $Z$ is contained in an affine open set. $X$ will denote the quotient.

Claim: Suppose $j:Y\rightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Y\rightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram (which I don't know how to draw here, but one visualize it easily): $Y\xrightarrow{j}{Z}$, $Y\xrightarrow{g}Y'$, $Y'\rightarrow{X}$, $Z\xrightarrow{h}X$ where $X$ is a scheme, $h$ is finite and induces monomorphisms on coordinate rings and $Y'\rightarrow{X}$ is a closed immersion.

Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $B\subset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $A\rightarrow{A/I}$. Put $A'=j^{-1}(B)$ and $Spec(A')=X$. The claim is clear for $X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$, $X$ would be replaced by $U'=h(U)$ which is an open subset.

Now this guarantees the existence of $X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $X$ that it is actually reduced and irreducible. QED.

I hope this is satisfactory.

4 added 18 characters in body

A really neat well known example is as follows:

Choose a conic $C_1$ and a tangential line $C_2$ in $\mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $\phi:\mathbb{P}^2\rightarrow{X}$ with $C:=\phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $\mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $\mathbb{P}^2$. Note that the degrees of $L'|C_1$ and $L'|C_2$ both coincide with the degree of $L|C$ and are therefore equal. But $L'\cong\mathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'\cong\mathcal{O}$, which is non-effective! A contradiction!

In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme.

In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1\amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism aboveabove), and $Z=\mathbb{P}^2$, but the argument is more general. $X$ will denote the quotient.

Claim: Suppose $j:Y\rightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Y\rightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram(which diagram (which I don't know how to draw here, but you can make a pic in your mind)one visualize it easily): $Y\xrightarrow{j}{Z}$, $Y\xrightarrow{g}Y'$, $Y'\rightarrow{Z'}$, Y'\rightarrow{X}$,$Z\xrightarrow{h}Z'$with Z\xrightarrow{h}X$ where $Z'$ X$is a scheme,$h$is finite and inducing induces monomorphisms on coordinate rings and$Y'\rightarrow{Z'}$Y'\rightarrow{X}$ is a closed immersion.

Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $B\subset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $A\rightarrow{A/I}$. Put $A'=j^{-1}(B)$ and $Spec(A')=Z'$. Spec(A')=X$. The claim is clear for$Z'$. X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$, $Z'$ X$would be replaced by$U'=h(U)$which is an open subset. Now this guarantees the existence of$Z'$X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $Z'$ X$that it is actually reduced and irreducible. QED. I hope this is satisfactory. 3 added 108 characters in body A really neat well known example is as follows: Choose a conic$C_1$and a tangential line$C_2$in$\mathbb{P}^2$and asssociate to a point$P$on$C_1$the point of intersection$Q$of$C_2$and the tangent line to$C_1$at$P$. This gives a birational isomorphism from$C_1$to$C_2$. Identify the curves by this map to get the quotient variety$\phi:\mathbb{P}^2\rightarrow{X}$with$C:=\phi(C_1)$. Now if there was an embedding of$X$in a smooth scheme then, there would surely exist an effective line bundle on$X$, say$L$whose pull back to$\mathbb{P}^2$will obviously be effective. Let us see why this is a contradiction. Let$L'$be the pullback of$L$to$\mathbb{P}^2$. Note that the degrees of$L'|C_1$and$L'|C_2$both coincide with the degree of$L|C$and are therefore equal. But$L'\cong\mathcal{O}(k)$and therefore the degrees in question are$2k$and$k$respectively for$C_1$and$C_2$. Therefore$k=0$and$L'\cong\mathcal{O}$, which is non-effective! A contradiction! In view of VA's comment, I give a complete proof here for constructing$X$as a scheme. In our special case it is a trivial pushout construction: Here I am thinking of$Y$as$C_1\amalg{C_2}$,$Y'=C$(the quotient by the birational isomorphism above, and$Z=\mathbb{P}^2$, but the argument is more general. Claim: Suppose$j:Y\rightarrow{Z}$is a closed subscheme of a scheme$Z$, and$g:Y\rightarrow{Y'}$is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram(which I don't know how to draw here, but you can make a pic in your mind):$Y\xrightarrow{j}{Z}$,$Y\xrightarrow{g}Y'$,$Y'\rightarrow{Z'}$,$Z\xrightarrow{h}Z'$with$Z'$a scheme,$h$is finite and inducing monomorphisms on coordinate rings and$Y'\rightarrow{Z'}$is a closed immersion. Proof: First assume that$Z$is affine, in which case$Y,Y'$are both affine too. Let$A,A/I,B$be their respective coordinate rings. Then$B\subset{A/I}$in a natural way. Let's use$j$again to denote the natural map$A\rightarrow{A/I}$. Put$A'=j^{-1}(B)$and$Spec(A')=Z'$. The claim is clear for$Z'$. Also if$Z$is replaced by an open subset$U$such that$g^{-1}g(U\cap{Y})=U\cap{Y}$,$Z'$would be replaced by$U'=h(U)$which is an open subset. Now this guarantees the existence of$Z'$once it has been shown that$Z$can be covered by affine open subsets$U$such that$g^{-1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of$Z'\$ that it is actually reduced and irreducible. QED.

I hope this is satisfactory.

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