Trefoil knot as an algebraic curve Is the trefoil knot with its usual embedding into affine $3$-space

an algebraic curve (maybe after extending scalars to $\mathbb{C}$)? Is there even some thickening to some algebraic surface? If not, is there at least some similar algebraic curve which describes this type of knot? I hope that this question is not silly, I know almost nothing about this classical stuff on algebraic curves. A google research indicates that there is some connection with the cusp $y^2=x^3$, but I don't really get it.
PS: I am interested in explicit equations. Specifically, is the trefoil cut out by two equations in affine $3$-space?
 A: There is a paper of Stephan Klaus that gives an explicit algebraic surface construction of the solid trefoil.
In addition, I found part of a solution to finding an algebraic curve isotopic to the trefoil by Michael Trott, but unfortunately the final pages are missing from the Google Books preview.
A: Consider two relatively prime integers $p,q \geq 2$ and the complex affine curve $C\subset \mathbb C^2 $given by $x^p+y^q=0$, which has an isolated singularity at the origin $O=(0,0)$.    
If  $C$ is intersected with the real $3$-sphere $S_\epsilon$ of equation $\mid x\mid ^2+\mid y \mid^2=\epsilon ^2$,  Brauner proved in 1928 that the  resulting real algebraic curve $C\cap S_\epsilon$ is a $(p,q)$ knot.
For $p=2, q=3$ , you obtain the trefoil knot, as you correctly conjectured.  
Milnor has written a fantastic booklet (122 pages...) on the subject: Singular points of complex hypersurfaces, published by Princeton in 1968, with   Brauner's theorem
 proved on the second page!   
Edit
The intersection $C\cap S_\epsilon\subset \mathbb C^2=\mathbb R^4$ is a real algebraic curve with completely explicit polynomial equations .
For the trefoil knot with $p=2,q=3 $ for example, we get (writing $x=x_1+ix_2, y=y_1+iy_2$)
$$ x_1^2+x_2^2+y_1^2+y_2^2=\epsilon^2, \quad  x_1^2-x_2^2+y_1^3-3y_1y^2_2=0,\quad 2x_1x_2+3y_1^2y_2-y_2^3=0      $$       
A: I know I am a bit late, but I think it is pretty easy to see that the trefoil can be described in terms of algebraic equations. Consider the parametrization from Wikipedia:
$$
x = \sin t + 2 \sin 2t, \quad y=\cos t - 2 \cos 2t, \quad z=-\sin 3t.
$$
Using the rules for double and triple angle this is equal to
$$
x = \sin t + 4 \sin t\cos t, \quad y=\cos t - 2 +4\sin^2 t, \quad z=-\sin t (4\cos^2 t -1).
$$
Denoting $s=\sin t$ and $c=\cos t$ you have the ideal
$$
I=\langle s + 4 sc-x,c - 2 +4s-y,-s (4c -1)-z,s^2+c^2-1\rangle
$$
describing the trefoil. The equation $s^2+c^2-1=0$ makes the connection between $\sin t$ and $\cos t$ while eliminating the parameter $t$. Now you can eliminate $s$ and $c$ in order to obtain a representation in the variables $x,y$ and $z$. Here is the SINGULAR code:
ring R = 0,(x,y,z,s,c),dp;
ideal I = 4*s*c-x+s, 4*s^2-y+c-2, -4*s*c^2-z+s, s^2+c^2-1;
ideal J = eliminate(I,sc);

Now $J=\langle P_1,P_2,P_3,P_4,P_5,P_6\rangle$ with
$$
\begin{aligned}
P_1&=12x^2y-4y^3-13x^2-13y^2+64z^2+9,\\
P_2&=x^3-3xy^2+4x^2z+4y^2z-9z,\\
P_3&=64y^3z+9x^3+9xy^2+4x^2z+100y^2z-192xz^2\\
&\phantom{=}-256z^3-72xy-180yz+27x-36z,\\
P_4&=256xy^2z-256x^2z^2-256y^2z^2-16y^3-128xyz\\
&\phantom{=}+256yz^2+35x^2-61y^2-240xz+640z^2-36y+9,\\
P_5&=48y^4+112y^3-144xyz-192yz^2-35x^2\\
&\phantom{=}-143y^2-108xz-160z^2-108y+99,\\
P_6&=8xy^3-4x^3+8xy^2-24x^2z-32xz^2-18xy-18yz+9x+27z.
\end{aligned}
$$
Maybe there are also generators with smaller degree. I haven't tried to find them yet.
There is also the famous trick to generate a single equation for the space curve:
$$
T=\sum_{i=1}^6 P_i^2.
$$
Over the real numbers you have $V_\mathbb{R}(J)=V_\mathbb{R}(T)$.
I'd like to post an image of a visualization based on the algebraic equations. But unfortunately, I don't have enough reputation yet.
Edit: Now I have enough reputation to post an image:

A: See the following paper for lots of examples:
G. Freudenburg, "Bivariate analogues of Chebyshev polynomials with application to embeddings of affine spaces", CRM Proceedings and Lecture Notes, vol. 54 (2011), American Math. Society, 39-56. 
GF
A: It all seems rather complicated to me, but no doubt I've missed something.
I just used polar coordinates  $1/r = 1 + 0.5 \cos (3A/2)$.
This gives a lovely plane trefoil.  The cartesian equation has 11 terms & is of the sixth
degree in x & y - can give it if anyone interested
David R Roberts
