Just started my algebraic geometry course so my understanding of ‘curves’ is shaky at best. Question reads:
Show that there exists a cubic curve $f(x,y)$ such that every point on the parametrized curve $$x(t) = 1 + t^2,\qquad y(t) = t + t^3$$ satisfies the equation $$f(x(t),~y(t)) = 0$$ Conversely, show that for any point $(x,y)$ with $f(x,y) = 0$, with one exception, there exists a real number t with $x = x(t),~ y = y(t)$.
So I just made $$f(x,y) = tx - y = 0$$ which is trivially true (might be misusing trivially here, but whatever the equation is always 0). This seems sneaky but acceptable to me, and I feel that I made a function of $x(t)$ and $y(t)$ that satisfies $f(x,y) = 0$ for all t. I answered the question without breaking any rules, or so I think. So everything seems kosher but I’m still suspicious of my answer, am wondering what another answer to the question might be, and I still fail to understand how to go about answering the second half of the question. Please help, I’m just a boy.
Sorry about the long winded and multifaceted questions, I’m a stranger to algebraic geometry that also happened to suck at most elementary algebra. Thanks.
You have a first step, but you must completely eliminate $t$.
Multiplying the first equation by $x^2$,
$$x^3=x^2+x^2t^2$$
and using your relation,
$$x^3=x^2+y^2.$$
Conversely,
$$y=\pm x\sqrt{x-1}$$ or
$$\begin{cases}x=|t|,\\y=t\sqrt{|t|-1}\end{cases}$$ will work for all $t\notin(-1,1)$. Of course, the initially given parameterization works as well (without exception !).