Singular points and tangent lines Find the singular points and the corresponding tangent lines of $X^4+Y^4-X^2Y^2$.  
I get that the singular points are those of the form $(a,a)$ and $(a,-a)$. The corresponding tangent lines are $X=-a, Y=-a$ and $X=-a, Y=a$ respectively. Is this correct? Thanks.
 A: No, what you write is not correct.
Let $\omega $ be  is a primitive third root of $1$. 
We then have  $$X^4-X^2Y^2+Y^4=(Y-i\omega X)(Y+i\omega X)(Y-i\omega^2 X)(Y+i\omega^2 X)$$ so that your affine curve is a cone, the union of the 4 lines $$y= i\omega x,\: y=-i\omega x,\: y=i\omega^2x, \:y=-i\omega^2x$$ The only singularity is   at the origin $(0,0)$ and the tangent lines at the origin are the very same four lines described above, i.e. the irreducible components of the curve. 
I have assumed  up to now that the base field is algebraically closed and of  characteristic $\neq 3, 2$.
 If the characteristic is $3$, the curve has equation $x^4+y^4-x^2y^2=(x^2+y^2)^2=0$ and is singular everywhere.
The tangent lines  at the origin are the two double lines $(y-ix)^2=0, (y+ix)^2=0$ (where, of course,  $i$ is a root in the base field of $z^2+1=0$).
At the other points of the curve the tangent lines consist of the line on which the point lies counted with multiplicity $2$.
Similarly, if the characteristic is $2$,  the curve has equation $x^4+y^4-x^2y^2=(x^2+y^2+xy)^2=0$ and is singular everywhere.
