Finding the equation for image of a map

Question

Consider the morphism $\mathbb A^1 \to \mathbb A^2$ given by $f(t) = (t^n,(1-t)^n)$. What is the defining equation of the image? Is there a systematic way to do this?

Answer 1

Too long for a comment.

In M2

R=QQ[t,x,y,MonomialOrder=>Eliminate 1]
I1=ideal(x-t,y-(1-t))
toString (gens gb I1)_0_0
-- x+y-1

I2=ideal(x-t^2,y-(1-t)^2)
toString (gens gb I2)_0_0
-- x^2-2*x*y+y^2-2*x-2*y+1

I3=ideal(x-t^3,y-(1-t)^3)
toString (gens gb I3)_0_0
-- x^3+3*x^2*y+3*x*y^2+y^3-3*x^2+21*x*y-3*y^2+3*x+3*y-1

I4=ideal(x-t^4,y-(1-t)^4)
toString (gens gb I4)_0_0
-- x^4-4*x^3*y+6*x^2*y^2-4*x*y^3+y^4-4*x^3-124*x^2*y-124*x*y^2-4*y^3+6*x^2-124*x*y+6*y^2-4*x-4*y+1

The leading terms seem to be $(x+y)^n$ for $n$ odd and $(x-y)^n$ for $n$ even.

In geogebra

Answer 2

The area of math that deals with this sort of thing is called elimination theory. Given the equations $x=t^n,y=(1-t)^n$, we want to eliminate $t$ from these equations. We can determine the unique up to scaling polynomial which vanishes on your curve as $\operatorname{Res}_t(x-t^n,y-(1-t)^n)$ where $\operatorname{Res}$ is the resultant.

The resultant can be computed as the determinant of the Sylvester matrix of the two polynomials (see Wikipedia for the definition). In our case, this matrix is somewhat nice: it's a $2n\times 2n$ matrix, where $A$ the upper left $n\times n$ block is the identity matrix and $C$ the lower left block matrix is the $n\times n$ diagonal matrix with $x$ along the diagonal. The right-hand blocks are somewhat less nice to describe, but they're both triangular matrices: $B$ the upper right $n\times n$ matrix is lower-triangular with $(-1)^n$ on the diagonal, and $D$ the lower right matrix is upper triangular with $1+y$ on the diagonal. Since the two lower blocks commute, we can compute the determinant of the Sylvester matrix as $\det(AD-BC)=\det(D-xB)$. Unfortunately, I don't see any further simplifications to make to this computation in general, and for instance checking what this determinant is over the first few values of $n$ doesn't give me a pattern that I can generalize. It's clear that these equations should be symmetric in $x$ and $y$ (think of the involution $t\mapsto 1-t$), and the highest degree term should be $(x-(-1)^ny)^n$ based on the diagonal, but I don't see much of a pattern after that.

If you want to look at the results of the resultant for various values of $n$, I'd ask a computer algebra system to do it for you. For instance, Wolfram Alpha will do it for you - the linked example is the $n=4$ case, and you can change it around by changing the exponents in the box at the top of the page.