Let $I$ be the ideal in $\displaystyle\,k\left[x_1, x_2, x_3, y_1, y_2, y_3\right]$ generated by the $2\times 2$ permanents of the matrix $$ \begin{bmatrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3 \end{bmatrix}. $$
So,
$$ I = \begin{pmatrix}x_1y_2+y_1x_2,&x_2y_3+y_2x_3,&x_3y_1+x_1y_3\end{pmatrix} $$
How do I compute, by hand, a primary decomposition of $I$?
(I can compute it in M2, but I am curious about how I would do this by hand)
If char $k$ is 2, then the permanents are same as determinants, so $I$ just is a prime ideal. So, let's assume that char$\neq 2$ now.
First, notice that $I$ is radical because its initial ideal (pick, say, just graded lex order) is radical. So, there are no embedded primes. Hence, what we should do is finding all the irreducible components of the projective variety defined by $I$.
To do so, take affine charts; i.e. dehomogenize. There are many symmetries here, so let's do dehomogenizing at $x_1$ for an example. Plugging in $x_1 = 1$, we get $(y_2 + y_1x_2, x_2y_3+y_2x_3, x_3y_1+y_3)$ for the ideal, and hence, noting $y_2 = -y_1x_2,\ y_3 = -x_3y_1$ we are really computing the components of $k[x_2, x_3, y_1]/(2x_2x_3y_1) = k[x_2, x_3, y_1]/(x_2x_3y_1)$ (notice we used char$\neq 2$ here). Well, the components are just $y_1=0$ or $x_2=0$ or $x_3 = 0$. Indeed, $y_1=0$ gets us the component $(y_1, y_2, y_3)$, and $x_2 = 0$ gets us $(x_2, y_2, x_3y_1 + x_1y_3)$, and so forth.
Using symmetry, one can thus see that we have five components (and hence, the primary decomposition of $I$): $$(x_1, x_2, x_3), (y_1, y_2, y_3), (x_1,y_1, x_2y_3+x_3y_2), (x_2, y_2, x_3y_1 + x_1y_3), (x_3, y_3, x_1y_2+x_2y_1)$$
and we are done!
N.B. This generalizes to $2\times n$ fairly easily, so no need to limit to $n=3$ case.