Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 3

For an ideal $I\subseteq k[T_1,\dots ,T_n]$, defineFor $X\subseteq k^n$, defineand recall that $\sqrt{I}=\{f:f^r\in I\text{ for some }r\ge 1\}$ is the radical of an ideal.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[T_1,\dots ,T_n]$ isfor a unique $a\in k^n$, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

To prove the weak form, let $\mathfrak{m}$ be maximal. The residue fieldis a field finitely generated as a $k$-algebra. By the Zariski lemma, $K/k$ is finite algebraic; algebraic closedness gives $K=k$. If $a_i$ is the image of $T_i$, the quotient map is evaluation at $a=(a_1,\dots ,a_n)$ and its kernel is $(T_1-a_1,\dots ,T_n-a_n)$. This proves the assertion.

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Let $\mathfrak{m}$ be maximal in $\mathbb{Q}[T_1,\dots ,T_n]$ and putThe Zariski lemma makes $K$ a finite extension of $\mathbb{Q}$. Base change givesThis ring is nonzero because the field extension makes $\mathbb{C}$ a faithfully flat module over $\mathbb{Q}$. Choose a maximal ideal $\mathfrak{n}$ of the quotient, or equivalently a maximal ideal of $\mathbb{C}[T_1,\dots ,T_n]$ containing $\mathfrak{m}\mathbb{C}[T_1,\dots ,T_n]$. Its contraction to the rational polynomial ring contains $\mathfrak{m}$ and is proper, so maximality of $\mathfrak{m}$ forces

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LetCertainly $\mathfrak{a}\subseteq (X)$, so $\sqrt{\mathfrak{a}}\subseteq (X)$. Conversely, cubing the second generator shows that $\mathfrak{a}$ contains $X^6(Y-3{)}^3$, while multiplying the first generator by $X^5$ shows that it contains $X^6{Y}^3$. The ideals generated by $Y^3$ and $(Y-3{)}^3$ are comaximal ideals, so a polynomial combination of the two polynomials is $1$. Multiplication by $X^6$ gives $X^6\in \mathfrak{a}$, hence $X\in \sqrt{\mathfrak{a}}$. Thereforegenerated by the single polynomial $X$.

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Write the finitely generated algebra asBy the Weak Hilbert Nullstellensatz, $\mathbb{C}$-algebra homomorphisms $A\to \mathbb{C}$ correspond exactly to the points of the affine algebraic set $V(I)$.

If $V(I)$ is finite, its cardinality is finite. If it is infinite, the complex affine algebraic set cardinality dichotomy giveswhich is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore

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