Primary ideals and Radicals

hbghlyj

A proper ideal $I$ is primary if whenever $a b \in I$ and $a \notin I$, then $b^n \in I$ for some $n>0$. The radical of an ideal $I$, denoted $\sqrt{I}$, is the set $\sqrt{I}=\{f: f^n \in I\text{ for some }n>0\}$.

• Prove that if $I$ is primary, then $\sqrt{I}$ is prime.
• Is the ideal $J=(x^2, x y) \subset \mathbb{Q}[x, y]$ primary?

hbghlyj

• Assume $I$ is a primary ideal in a commutative ring with unit. To show $\sqrt{I}$ is prime, suppose $ab \in \sqrt{I}$. Then $(ab)^m = a^m b^m \in I$ for some $m > 0$. If $a \notin \sqrt{I}$, then $a^m \notin I$. Since $I$ is primary and $a^m b^m \in I$ with $a^m \notin I$, it follows that $(b^m)^k = b^{mk} \in I$ for some $k > 0$. Thus $b \in \sqrt{I}$. Therefore, $\sqrt{I}$ is prime.
• The ideal $J = (x^2, xy) \subset \mathbb{Q}[x, y]$ is not primary. Consider $a = x$ and $b = y$. Then $ab = xy \in J$. However, $x \notin J$ because elements of $J$ are of the form $f x^2 + g xy$ for $f, g \in \mathbb{Q}[x, y]$, which have total degree at least 2, while $x$ has degree 1. Moreover, $y^n \notin J$ for any $n > 0$ because monomials in $J$ must be divisible by $x^2$ or $xy$ and thus contain a factor of $x$, but $y^n$ does not.

hbghlyj

Let $I=(r)$ be an ideal in a PID $R$.

• Prove that $\sqrt{I}$ is generated by the largest squarefree factor of $r$.
• What is the radical of the ideal $(x^3-x^2)$ in $\mathbb{R}[x]$?

hbghlyj

• Let $R$ be a PID and $I = (r)$ a principal ideal. Since $R$ is a UFD, write $r = u p_1^{a_1} \cdots p_k^{a_k}$ where $u$ is a unit, the $p_i$ are distinct primes, and each $a_i \geq 1$. The largest squarefree factor of $r$ is $s = p_1 \cdots p_k$ (up to units). To show $\sqrt{I} = (s)$, first suppose $f \in \sqrt{I}$, so $f^n \in I$ for some $n > 0$, meaning $r$ divides $f^n$. For each prime $p_i$ dividing $r$, the valuation satisfies $n v_{p_i}(f) \geq a_i$. Since $n$ can be arbitrarily large, this requires $v_{p_i}(f) \geq 1$ (otherwise $n \cdot 0 < a_i$). For primes not dividing $r$, there is no restriction. Thus $f$ is divisible by each $p_i$, so $f \in (s)$.
  
  Conversely, suppose $f \in (s)$, so $f = s \cdot t$ for some $t \in R$. Then $f^n = s^n t^n = (p_1 \cdots p_k)^n t^n = p_1^n \cdots p_k^n t^n$. Now $r$ divides $f^n$ if $n \geq a_i$ for each $i$, which holds for $n \geq \max a_i$. Thus $f \in \sqrt{I}$.
• In $\mathbb{R}[x]$, which is a PID, $I = (x^3 - x^2) = (x^2(x-1))$. The distinct irreducible factors are $x$ and $x-1$, so the largest squarefree factor is $x(x-1) = x^2 - x$. By part (a), $\sqrt{I} = (x^2 - x)$.