Let \(R\) be a commutative ring. Recall that an element \(r\) of \(R\) is nilpotent if \(r^n=0\) for some positive integer \(n\) and that the nilradical of \(R\) is the set \(N(R)\) of nilpotent elements.
Prove that \[ N(R)=\bigcap_{P\text{ prime}}P. \]
Hint: given a non-nilpotent element \(r\) of \(R\), you may wish to construct a prime ideal that does not contain \(r\) or its powers.
Given a positive integer \(m\), determine the nilradical of \(\mathbb Z/(m)\).
Determine the nilradical of \(\mathbb C[x,y]/(y^2-x^3)\).
Let \(p(x,y)\) be a polynomial in \(\mathbb C[x,y]\) such that for any complex number \(a\), \(p(a,a^{3/2})=0\). Prove that \(p(x,y)\) is divisible by \(y^2-x^3\).
Solution
1. For every commutative ring \(R\),\[ N(R)=\bigcap_{P\text{ prime}}P. \]
If \(r\in N(R)\), then \(r^n=0\) for some \(n\). Every prime ideal contains \(0\), and if \(r^n\in P\) with \(P\) prime, repeated use of primality gives \(r\in P\). Hence \[ N(R)\subseteq\bigcap_P P. \]
Conversely, suppose \(r\) is not nilpotent. Let \[ S=\{1,r,r^2,r^3,\ldots\}. \] Then \(0\notin S\). Consider the set of ideals of \(R\) disjoint from \(S\), ordered by inclusion. Every chain has an upper bound given by its union, still disjoint from \(S\), so Zorn’s lemma gives a maximal such ideal \(P\).
We claim \(P\) is prime. Suppose \(ab\in P\) but \(a,b\notin P\). By maximality, both \(P+(a)\) and \(P+(b)\) meet \(S\). Thus for some \(m,n\ge0\), \[ r^m=p_1+xa,\qquad r^n=p_2+yb \] with \(p_1,p_2\in P\) and \(x,y\in R\). Multiplying, \[ r^{m+n}=p_1p_2+p_1yb+p_2xa+xyab\in P, \] contradicting \(P\cap S=\varnothing\). Thus \(P\) is prime and \(r\notin P\). Hence any element lying in every prime ideal must be nilpotent.
2. If\[ m=\prod_{i=1}^s p_i^{e_i} \]is the prime factorization of \(m\), then\[ N(\mathbb Z/(m)) =\bigl(\overline{p_1p_2\cdots p_s}\bigr). \]
Write \[ \operatorname{rad}(m)=p_1\cdots p_s. \] If \(\bar a\in\mathbb Z/(m)\) is nilpotent, then \(m\mid a^N\) for some \(N\). Hence every prime divisor \(p_i\) of \(m\) divides \(a^N\), and therefore divides \(a\). Thus \(\operatorname{rad}(m)\mid a\).
Conversely, if every \(p_i\) divides \(a\), choose \(N\ge\max_i e_i\). Then \[ p_i^{e_i}\mid a^N \] for every \(i\), so \(m\mid a^N\). Hence \(\bar a\) is nilpotent. Therefore the nilpotent classes are exactly the multiples of \(\operatorname{rad}(m)\) modulo \(m\).
3. One has\[ N\bigl(\mathbb C[x,y]/(y^2-x^3)\bigr)=0. \]
The polynomial \(y^2-x^3\) is irreducible in \(\mathbb C[x,y]\). Indeed, regarding it as a monic quadratic in \(y\) over the UFD \(\mathbb C[x]\), reducibility would force \(x^3\) to be a square in \(\mathbb C[x]\), which it is not. Since \(\mathbb C[x,y]\) is a UFD, irreducible elements are prime. Hence \((y^2-x^3)\) is a prime ideal, so the quotient is an integral domain. An integral domain has no nonzero nilpotent elements.
4. Let\[ \phi:\mathbb C[x,y]\longrightarrow\mathbb C[t], \qquad \phi(x)=t^2,\quad \phi(y)=t^3. \]Then\[ \ker\phi=(y^2-x^3). \]
Certainly \(y^2-x^3\in\ker\phi\). Conversely, divide any \(p(x,y)\in\mathbb C[x,y]\) by the monic polynomial \(y^2-x^3\) as a polynomial in \(y\) over \(\mathbb C[x]\). There are unique \(q\in\mathbb C[x,y]\) and \(a,b\in\mathbb C[x]\) such that \[ p=q(y^2-x^3)+a(x)+y b(x). \] If \(p\in\ker\phi\), then \[ 0=a(t^2)+t^3 b(t^2). \] The first summand contains only even powers of \(t\), while the second contains only odd powers \(t^{2j+3}\). Hence both summands vanish separately. Thus \(a=0\) and \(b=0\), so \(p\in(y^2-x^3)\).
5. If \(p(a,a^{3/2})=0\) for every complex \(a\) in the sense of the cusp parametrization, then \(y^2-x^3\) divides \(p(x,y)\).
Take an arbitrary \(t\in\mathbb C\) and set \(a=t^2\), choosing the square root \(t\), so that \(a^{3/2}=t^3\). The hypothesis gives \[ p(t^2,t^3)=0 \] for every \(t\in\mathbb C\). Hence the polynomial \(\phi(p)\in\mathbb C[t]\) has infinitely many roots, so \(\phi(p)=0\). By 4, \[ p\in\ker\phi=(y^2-x^3). \] Therefore \(y^2-x^3\) divides \(p\).