Theorem 4.33. (Nullstellensatz - strong version) Suppose that \(k\) is algebraically closed and let \(I \subset k[X_1,\ldots , X_n]\) be an ideal. Then \[ \mathcal {I}(\mathcal {V}(I)) = \sqrt {I}. \]
Proof. The inclusion \(\sqrt {I} \subset \mathcal {I}(\mathcal {V}(I))\) is easy: if \(f \in \sqrt {I}\), then \(f^\ell \in I\) for some \(\ell \geq 1\), so \(f^\ell \), and hence \(f\), vanishes at every point of \(\mathcal {V}(I)\).
Conversely, write \(I = (f_1,\ldots , f_s)\), using the Hilbert basis theorem, and let \(f \in \mathcal {I}(\mathcal {V}(I))\). Explicitly, this means that if \((a_1,\ldots ,a_n) \in k^n\) satisfies \(f_i(a_1,\ldots ,a_n)=0\) for all \(i\), then \(f(a_1,\ldots ,a_n)=0\). We may assume \(f \neq 0\), and we need to show that there exist \(\ell >0\) and \(h_1,\ldots , h_s\in k[X_1,\ldots , X_n]\) such that \[ f^\ell = \sum _{i=1}^s h_i f_i. \]For this, we use the Rabinowitsch trick: consider the polynomial ring \(k[X_1,\ldots , X_n, Y]\) in one more variable and the ideal \(\tilde {I}= (f_1,\ldots , f_s, 1-Yf) \subset k[X_1,\ldots , X_n, Y]\). We claim that \(\tilde {I}= k[X_1,\ldots , X_{n},Y]\). Indeed, if \(\tilde {I}\) were a proper ideal, then by the weak Nullstellensatz (Theorem 4.29) it would have a zero \((a_1,\ldots ,a_n, a_{n+1}) \in k^{n+1}\). Then \(f_i(a_1,\ldots ,a_n)=0\) for all \(i\), hence \(f(a_1,\ldots , a_n)=0\), but also \[ 1 - f(a_1,\ldots , a_n) a_{n+1} = 0, \] a contradiction. Hence \(1 \in \tilde {I}\), and we can write \[ 1 = \sum _{i=1}^s g_i f_i + g_0 (1-Yf) \text {\ for some \ } g_0, g_1, \ldots , g_s \in k[X_1,\ldots , X_n,Y]. \] Now apply the ring homomorphism \(k[X_1,\ldots , X_n,Y] \to k(X_1,\ldots ,X_n)\) which sends \(X_i \mapsto X_i\) for \(1 \leq i \leq n\) and \(Y \mapsto \frac {1}{f}\). The last term vanishes, and we get \[ 1 = \sum _{i=1}^s g_i (X_1,\ldots , X_n, \tfrac {1}{f}) f_i \] in \(k(X_1,\ldots ,X_n)\). Multiplying by \(f^\ell \), where \(\ell \) is the largest power of \(Y\) appearing in \(g_1,\ldots , g_s\), clears the denominators: \(h_i \coloneqq f^\ell g_i(X_1,\ldots , X_n, \tfrac {1}{f})\) lies in \(k[X_1,\ldots , X_n]\), and \[ f^\ell = \sum _{i=1}^s h_i f_i. \] □
\(\blacktriangleright \) The theorem says that the only way a polynomial can vanish on \(\mathcal {V}(I)\) without belonging to \(I\) is that one of its powers belongs to \(I\). For instance, let \(I = (Y - X^2, Y) \subset k[X,Y]\), which describes the intersection of the parabola \(Y = X^2\) with the line \(Y = 0\), see Figure 5. Then \(\mathcal {V}(I) = \{(0,0)\}\) and \(\mathcal {I}(\mathcal {V}(I)) = (X,Y)\), but \(X \notin I = (X^2, Y)\): the ideal remembers that the line is tangent to the parabola, as \(k[X,Y]/I\) has dimension \(2\) rather than \(1\). By contrast, the line \(Y = X\) meets the parabola transversally at \((0,0)\) and \((1,1)\), and the ideal \((Y - X^2, Y - X) = (X^2 - X, Y - X)\) is radical. The hypothesis that \(k\) is algebraically closed cannot be dropped: over \(\mathbb {R}\), the ideal \((X^2 + Y^2)\) is prime, hence radical, but its variety is the single point \((0,0)\), whose ideal is \((X, Y)\).
\(\blacktriangleright \) The most important consequence of the Nullstellensatz is that, for \(k\) algebraically closed, \(\mathcal {I}\) and \(\mathcal {V}\) are mutually inverse, inclusion-reversing bijections \[ \{\text {affine varieties in } k^n\} \underset {\mathcal {V}}{\overset {\mathcal {I}}\rightleftarrows } \{\text {radical ideals of } k[X_1,\ldots , X_n]\}. \] Indeed, \(\mathcal {V}(\mathcal {I}(V)) = V\) for every affine variety \(V\) (an exercise in Lecture 5), \(\mathcal {I}(U)\) is a radical ideal for any \(U \subset k^n\), and \(\mathcal {I}(\mathcal {V}(J)) = J\) for a radical ideal \(J\) by the theorem. Points of \(k^n\) correspond to maximal ideals, as we saw in Lecture 12, and we will see in the next lecture that irreducible varieties correspond to prime ideals.
Corollary 4.34. Let \(k\) be a field (not necessarily algebraically closed) and \(I = (f_1,\ldots , f_s) \subset k[X_1,\ldots ,X_n]\). Then \(f \in \sqrt {I}\) if and only if \(1\) belongs to the ideal \(\tilde {I} = (f_1,\ldots , f_s, 1-Yf) \subset k[X_1,\ldots ,X_n,Y]\).
Proof. If \(f = 0\) there is nothing to prove. Otherwise, the second half of the proof of Theorem 4.33, from \(1 \in \tilde {I}\) on, works over any field and shows that \(1 \in \tilde {I}\) implies \(f \in \sqrt {I}\). Conversely, suppose \(f \in \sqrt {I}\), so that \(f^\ell \in I\) for some \(\ell >0\). Since also \(1-Yf \in \tilde {I}\), we can write \[ 1 = Y^\ell f^\ell + (1-Y^\ell f^\ell ) = Y^\ell f^\ell + (1-Yf) \cdot (1+ Yf + \ldots + Y^{\ell -1} f^{\ell -1}) \in \tilde {I}. \] □
This corollary gives an algorithm for deciding whether \(f \in \sqrt {I}\): compute a Gröbner basis of \((f_1,\ldots , f_s, 1-Yf)\) and check whether it contains a non-zero constant.
\(\sphericalangle \) Consider the \(2\)-by-\(2\) matrix \(M\) with entries \(X,Y,Z,W\). The equation \(M^2=0\) gives four polynomial equations, which generate the ideal \(I = (X^2+YZ,XY+YW, XZ+WZ,W^2+YZ) \subset k[X,Y,Z,W]\). Is \(I\) a radical ideal? Show that \(\sqrt {I}=(X+W,XW-YZ)\). (Hint: a nilpotent matrix has trace and determinant zero, and the converse holds for \(2\)-by-\(2\) matrices by the Cayley–Hamilton theorem.)
You can also do this in Macaulay2:
R = QQ[x,y,z,w]; I = ideal (x^2+y*z,x*y+y*w,x*z+w*z,w^2+y*z); I == radical I radical I
\(\sphericalangle \) (Hard) Consider an \(n\)-by-\(n\) matrix \(X\) with entries \(X_{i,j}\) for \(1 \leq i,j \leq n\) and an \(n\)-by-\(n\) matrix \(Y\) with entries \(Y_{i,j}\) for \(1 \leq i,j \leq n\). Form the ideal \(I \subset \mathbb {C}[X_{i,j}, Y_{i,j}]_{1 \leq i,j \leq n}\) generated by the entries of the matrix \(XY - YX\). Is \(I\) a radical ideal? It has been known since the 1950s that \(\sqrt {I}\) is prime for all \(n\), and whether \(I\) itself is prime, that is, radical, was a well-known open question going back to Artin and Hochster (1982). For \(n \leq 4\) the answer is yes, by Gröbner basis computations for \(n=3,4\), and a recent preprint8 claims a proof for all \(n\). We can check the case \(n = 3\) in Macaulay2 as follows:
n=3 R = QQ[x_(1,1)..x_(n,n),y_(1,1)..y_(n,n)] M = matrix table (n, n, (i,j) -> x_(i+1,j+1)) N = matrix table (n, n, (i,j) -> y_(i+1,j+1)) I = minors(1, M*N-N*M) isPrime I
If you want to break your computer, you can set \(n\) to a higher value and run the same code.
Corollary 4.35. Let \(k\) be algebraically closed and let \(I \subset k[X_1,\ldots , X_n]\) be a proper ideal, so that \(V =\mathcal {V}(I)\) is non-empty by the weak Nullstellensatz. Then \[ \mathrm {dim}(V) = \mathrm {deg} HP_{I} (t), \] where \(\mathrm {dim}(V) \coloneqq \mathrm {deg} HP_{\mathcal {I}(V)}(t)\). In other words, the dimension of \(V\) can be computed from any ideal \(I\) with \(\mathcal {V}(I) = V\).
Proof. Since \(I \subset \mathcal {I}(V)\), we have \(I_{\leq s} \subset \mathcal {I}(V)_{\leq s}\), hence \(HF_{\mathcal {I}(V)}(s) \leq HF_I(s)\) for all \(s\) and \(\mathrm {deg} HP_{\mathcal {I}(V)}(t) \leq \mathrm {deg} HP_{I} (t)\). For the other inequality, let \(d = \mathrm {deg}HP_{I}(t)\). By Corollary 4.18, there exist \(1 \leq i_1 < i_2 < \ldots < i_d \leq n\) such that \(I \cap k[X_{i_1}, X_{i_2}, \ldots , X_{i_d}] = \{0\}\), and by the same corollary it suffices to show that \[ \mathcal {I}(V) \cap k[X_{i_1}, X_{i_2}, \ldots , X_{i_d}] = \{0\}. \] Let \(f \in \mathcal {I}(V) \cap k[X_{i_1}, X_{i_2}, \ldots , X_{i_d}]\). Since \(\mathcal {I}(V) = \sqrt {I}\) by Theorem 4.33, there exists \(\ell \geq 1\) such that \(f^\ell \in I\). But \(f^\ell \) also lies in \(k[X_{i_1},X_{i_2},\ldots , X_{i_d}]\), so \(f^\ell \in I \cap k[X_{i_1},X_{i_2}, \ldots , X_{i_d}] = \{0\}\). Thus \(f=0\). □
Corollary 4.36. Let \(k\) be algebraically closed and let \(I \subset k[X_1,X_2,\ldots , X_n]\) be a proper ideal. Then \(\sqrt {I}\) is the intersection of the maximal ideals containing \(I\): \[ \sqrt {I} = \bigcap _{\mathfrak {m} \supset I} \mathfrak {m}. \]
Proof. The inclusion \(\sqrt {I} \subset \bigcap _{\mathfrak {m} \supset I} \mathfrak {m}\) holds in any ring, since a maximal ideal is prime, so \(I \subset \mathfrak {m}\) implies \(\sqrt {I} \subset \mathfrak {m}\). Conversely, let \(f \in \bigcap _{\mathfrak {m} \supset I} \mathfrak {m}\). By the weak Nullstellensatz (Theorem 4.29), the maximal ideals containing \(I\) are the ideals \(\mathfrak {m}_a = (X_1- a_1,\ldots , X_n-a_n)\) with \(a = (a_1,\ldots , a_n) \in \mathcal {V}(I)\), and \(f \in \mathfrak {m}_a\) means \(f(a) = 0\). Therefore \(f\) vanishes on \(\mathcal {V}(I)\), and by the strong Nullstellensatz \(f \in \mathcal {I}(\mathcal {V}(I)) = \sqrt {I}\). □
The last corollary should be contrasted with Proposition 2.7, which says that \(\sqrt {I}\) is the intersection of the prime ideals containing \(I\). Firstly, in \(k[X_1,\ldots , X_n]\) it suffices to take the maximal ideals containing \(I\). Secondly, we have an explicit description of these maximal ideals, or equivalently of the maximal ideals of the ring \(k[X_1,\ldots ,X_n]/I\): they are in bijection with the points of \(\mathcal {V}(I)\).
A ring \(R\) is called a Jacobson ring or Hilbert ring if every prime ideal of \(R\) is an intersection of maximal ideals. Equivalently, for every ideal \(I\) of \(R\), the nilradical and the Jacobson radical of \(R/I\) coincide. The last corollary says that \(k[X_1,\ldots , X_n]\) is a Jacobson ring when \(k\) is algebraically closed. A field is obviously a Jacobson ring, and so is \(\mathbb {Z}\), but the local ring \(k[[X]]\) is not: \((0)\) is a prime ideal, but the only maximal ideal is \((X)\). A generalisation of Hilbert’s Nullstellensatz states that any finitely generated algebra over a Jacobson ring is Jacobson; in particular, \(k[X_1,\ldots , X_n]\) is a Jacobson ring for every field \(k\).
8J. Sempliner, X. Yang, On the normality of the commuting scheme, arXiv:2609.27360 (2026).