# A question regarding Hilbert's Nullstellensatz.

Let $k$ be an algebraically closed field, and $a$ an ideal of the polynomial ring $k[x_1,x_2,\dots,x_n]$. The strong form of Hilbert's Nullstellensatz says that $I(Z(a))=\sqrt{a}$.

Note:- Initially, the question was regarding $I(\langle (x-1)+(y-2)+(z-3)\rangle$. I then replaced $(x-1)+(y-2)+(z-3)$ with $(x-1)^2+(y-2)^2+(z-3)^2$ so as to ensure that the zero locus consists of only the point $(1,2,3)$.

I have a question regarding this statement. Take the polynomial ring $\Bbb{C}(x,y,z)$, where $\Bbb{C}$ is obviously an algebraically closed field. Now take the point $(1,2,3)\in\Bbb{C^3}$. Clearly $\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle$ has $(1,2,3)$ as its only zero. In other words, $Z(\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle)=(1,2,3)$. Now let us consider $I(Z(\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle))$. This will include the polynomial $(x-1)+(y-2)$, among others like $(x-1)$ and $(y-2)$. However, $(x-1)+(y-2)\notin \sqrt{\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle}$. Isn't this a contradiction?

• In algebraic geometry, if you want to define a point $(x_1, \ldots, x_n)$ as a vanishing locus of a set of polynomials in $k^n$ (where $k$ is an algebraically closed field), you must take the ideal $((X_1 -x_1), \ldots, (X_n-x_n))$. In this case, it would be $((x-1),(y-2),(z-3))$. Each equation defines a (hyper)plane, and the intersection of $n$ of them is a point. You can use the Nullstellensatz to see that these ideals are exactly the maximal ideals in $k[X_1, \ldots, X_n]$ (since points are minimal varieties) – Dorebell Jul 26 '14 at 3:10
• Please don't make a substantial change to your post without leaving some reference to it. My answer now appears to have very little to do with your question. – Michael Albanese Jul 26 '14 at 3:18
• I am sorry I will make the due changes again. – fierydemon Jul 26 '14 at 3:19
• @MichaelAlbanese- I have made a note of the fact that initially the question was about $(x-1)+(y-2)+(z-3)$, and not $(x-1)^2+(y-2)^2+(z-3)^2$ – fierydemon Jul 26 '14 at 3:22

Note that the equation $x + y + z = 6$ defines a plane in $\mathbb{C}^3$; in particular, $(1, 2, 3)$ is not the only point which satisfies this equation. For example, $(6, 0, 0)$ lies in the plane, so $$(6, 0, 0) \in Z(\langle x + y + z - 6\rangle).$$ Note that the polynomial $(x - 1) + (y - 2)$ is not zero at $(6, 0, 0)$, so $$(x - 1) + (y -2) \not\in I(Z(\langle x + y + z - 6\rangle)).$$ In particular, there is no contradiction.
• But $(x-1)+(y-2)$ is equal to $0$ at $(1,2,3)$. That is what I meant. Hence, it is a part of $I((1,2,3))$. – fierydemon Jul 26 '14 at 3:13
• Note that $f \in I(V)$ if $f(v) = 0$ for every $v \in V$. As there is at least one point in $V = Z(\langle x + y + z - 6\rangle)$ such that $(x - 1) + (y - 2)$ is not zero (for example, the point $(6, 0, 0)$), $(x - 1) + (y - 2) \not\in I(Z(\langle x + y + z - 6\rangle))$. – Michael Albanese Jul 26 '14 at 3:16
• Your claim is still wrong as there are points in $\mathbb{C}^3$ other than $(1, 2, 3)$ where the polynomial $(x - 1)^2 + (y - 2)^2 + (z + 3)^2$ is zero. For example, $(x, y, z) = (2, 2, -3 + i)$. – Michael Albanese Jul 26 '14 at 3:21