I know this is a bit of an odd request, but I am looking to construct the natural map $E^n\to E^{(n)}$ as schemes without knowing how to take a quotient by a finite group action, and do it in a relatively understandable way. My goal in doing this is to show the construction to some fellow graduate students who know some basics of algebraic geometry, but not enough to know how to take a quotient yet. (I'm presenting in a seminar which has some interesting rules about what you must be able to show and what you have to be able to cover.)
It would be enough to do this in the case $n=2$ (and maybe $n=3$, but let's stick with $n=2$ to start). Here I know that $E^{(2)}$ is a ruled surface over $E$ (in fact, $E^{(n)}$ is a $\Bbb P^{n-1}$-bundle over $E$ by some general theory), so I should be able to give a morphism $E^2\to E^{(2)}$ via a line bundle $\mathcal{L}$ on $E^2$ and a surjection from the locally free sheaf $f^*\mathcal{F}\to \mathcal{L}$ where $E^{(2)}=\Bbb P(\mathcal{F})$ and $f:E^2\to E^{(2)}\to E$ is the composite map. From this question, $E^{(2)}$ is the unique ruled surface of invariant $-1$, $\mathcal{F}$ is a nontrivial extension of $\mathcal{O}_E$ and $\mathcal{O}_E(P)$, and the composite map $E^2\to E$ is multiplication.
If I knew about the quotient map $E^2\to E^{(2)}$, I could say that $f^*\mathcal{F}\to\mathcal{L}$ should be the pullback of $\pi^*\mathcal{F}\to\mathcal{O}_{E^{(2)}}(1)\cong\mathcal{O}_{E^{(2)}}(\sigma_P(E))$ where $\sigma_P:E\to E^{(2)}$ is the natural section given from the surjection $\mathcal{F}\to\mathcal{O}_E(P)$, but going the other way is giving me trouble - it's hard for me to come up with a good way to write down what this map is. I'm looking for ways to specify this map, or alternate ways to construct the map $E^2\to E^{(2)}$.
How about the following. Let $\alpha:E\times E\to E$ be the addition map. Let $D\subseteq E\times E$ be the divisor $E\times 0+0\times E$. Then I claim that the adjunction map $\phi:\alpha^*\alpha_*({\cal O}(D))\to{\cal O}(D)$ is a surjection and that $\alpha_*({\cal O}(D))$ is locally free of rank $2$. I also claim that the corresponding map $q:E\times E\to {\bf P}(\alpha_*({\cal O}(D)))=P$ identifies $q$ with the quotient map under the involution swapping the factors (so that $P=E^{(2)}$).
To establish the first claim, let $I:E\times E\to E\times E$ be the map $(a,b)\mapsto (a,a+b)$. This is obviously an isomorphism. We then have $I(D)=\Delta+0\times E$, where $\Delta$ is the diagonal of $E\times E$. If we let $\pi:E\times E\to E$ be the second projection, we then have $\pi\circ I=\alpha.$ Now for any $b\in E$, the pull-back of $I(D)$ to ${E\times b}$ is $(b,b)+(0,b)$. This is an effective divisor of degree $2$ on $E\times b\simeq E$ and so its linear system is of dimension $2$ (by Riemann-Roch) and ${\cal O}((b,b)+(0,b))$ has no cohomology in non zero degree. Also, the linear system of $(b,b)+(0,b)$ is base-point free (see Hartshorne Cor. IV.3.2). So, by the semicontinuity theorem, we see that the adjunction map $\pi^*\pi_*({\cal O}(I(D)))\to{\cal O}(I(D))$ is surjective and that $\pi_*({\cal O}(I(D)))$ is of rank $2$. Applying $I^{*}$ to this we get the first claim.
Now let $\sigma:E\times E\to E\times E$ be the map swapping the factors. We have $\sigma(D)=D$, so the map $\phi$ is $\sigma$-equivariant.
Note that the action of $\sigma$ on $\alpha_*({\cal O}(D))$ is trivial. To show this, apply the relative Lefschetz trace formula
(one could also do this directly by applying $I$ and computing on one fibre of $\pi$). The fixed point set of $\sigma$ is $\Delta\simeq E$, and one obtains the formula $$ {\rm Trace}(\sigma,\alpha_*({\cal O}(D)))= {1\over 2}{\rm Trace}([2]_*({\cal O}(2\cdot 0)))={1\over 2}4=2 $$ and so the action of $\sigma$ on $\alpha_*({\cal O}(D))$ has to be trivial, since its eigenvalues are all $\pm 1$ (here the $0$ in $2\cdot 0$ is the origin of $E$).
So we conclude that the fibres of the map $q:E\times E\to {\bf P}(\alpha_*({\cal O}(D)))=P$ induced by $\phi$ are $\sigma$-invariant. Note that the map $q$ is not constant on any fibre of $\alpha$, since the restriction of $D$ to any fibre of $\alpha$ is effective of degree $2$. Also, since the degree of the restriction of ${\cal O}_{P}(1)$ to any fibre of $P$ over $E$ is $1$ and the degree of the restriction of $D$ to any fibre of $\alpha$ is $2$, we conclude that $q$ is surjective and of degree $2$. So by the universal property of the quotient, we get a map $E^{(2)}\to P$ over $E$, which is birational and finite, and thus by Zariski's main theorem, it must be an isomorphism, establishing the second claim.