# Curve in $\mathbb{A}^3$ that cannot be defined by 2 equations

According to a problem in Shafarevich 1.6, every curve in $\mathbb{A}^3$ can be cut out by 3 equations.

Can someone give me an example of a curve in $\mathbb{A}^3$ that is not cut out by 2 equations?

• Hartshorne gives (parametrically) the example of $(t^3, t^4, t^5)$ for an ideal of height $2$ which cannot be generated using just two elements. But I'm blanking on whether this is the same as what you're asking -- the difference between being a "complete intersection" or a "set-theoretic complete intersection" seems real. – Dylan Moreland Feb 1 '12 at 4:26
• The ideal of polynomials vanishing on Dylan's curve is generated by $\{x z-y^2,y z-x^3,z^2-x^2 y\}$. Can you generate it with two polynomials? – Mariano Suárez-Álvarez Feb 1 '12 at 5:18

This is mostly a long comment.

Consider the curve $C$ parametrized by $t\mapsto(t^3,t^4,t^5)$, as in Dylan's comment.

Let $A=k[x,y,z]$ be the polynomial ring in three variables, and let us consider it as graded ring with $x$, $y$ and $z$ of degrees $3$, $4$ and $5$, respectively.

Suppose $p=\sum_{i,j,k\geq0}\alpha_{i,j,k}x^iy^jz^k\in A$ is a polynomial such that $p(t^3,t^4,t^5)=0$, so that $p$ vanishes on the curve $C$. Then $$\sum_{\ell\geq0}\Bigl(\sum_{\substack{i,j,k\geq0\\3i+4j+5k=\ell}}\alpha_{i,j,k}\Bigr)t^\ell=0$$ so we see that for each $\ell\geq0$ we have $$\sum_{\substack{i,j,k\geq0\\3i+4j+5k=\ell}}\alpha_{i,j,k}=0.$$ Now, if for each $\ell\geq0$ we define the polynomial $$p_\ell=\sum_{\substack{i,j,k\geq0\\3i+4j+5k=\ell}}\alpha_{i,j,k}x^iy^jz^k,$$ we have first that $$p_\ell(t^3,t^4,t^5)=0,$$ so that $p_\ell$ vanishes on $C$, and $p_\ell$ is homogeneous of degree $\ell$ in our graded ring $A$. Our polynomial $p=\sum_{\ell\geq0}p_\ell$ is therefore a sum of homogeneous elements of $I$.

This means that the ideal $I\subseteq A$ of the polynomials which vanish on $C$ is homogeneous: it is generated by the homogeneous elements it contains.

Now, it is easy to see that the subspaces of $I$ of homogeneous elements of degrees $8$, $9$, and $10$ are $1$-dimensional, spanned respectively by the three polynomials $x z-y^2$, $y z-x^3$ and $z^2-x^2 y$. Moreover, it is also easy to see that all the homogeneous components of $I$ of degree less than $8$ are zero: in fact, the monomials $1$, $x$, $y$, $z$, $x^3$ and $xy$ span the direct sum of the homogeneous components of $A$ of degree less than $8$, and no non-zero linear combination of them vanishes on $C$, as a little computation will show.

Thinking a bit about what all this means, we see that the three polynomials $x z-y^2$, $y z-x^3$ and $z^2-x^2 y$ must belong to any generating set of $I$. This shows that $C$ is not a complete intersection. This may or may not be what Shafarevich has in mind, as "cut" may also mean set-theoretically, as Dylan notes.

If he means the latter, then this example does not work: Ernst Kunz shows, in his Introduction to commutative algebra and algebraic geometry, that all monomial curves in affine $3$-space are set-theoretically complete intersections. 