# Radical ideal of $(x,y^2)$

How does one show that the radical of $(x,y^2)$ is $(x,y)$ over $\mathbb{Q}[x,y]$? I have no idea how to do, please help me.

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Do you know the definition of radical of an ideal? After studying that, try proving first the containment $\sqrt{ (x,y^2) } \supseteq (x,y)$ and then $\subseteq$. – Andrea Aug 21 '11 at 17:53
$(x,y)$ is maximal – yoyo Aug 21 '11 at 21:19
$(x^2,xy,y^2)=(x,y)^2 \subsetneq (x,y^2) \subsetneq (x,y)$, then take radical. – Ehsan M. Kermani Dec 1 '11 at 22:11

Recall that the radical of an ideal $\mathfrak{a}$ is equal to the intersection of the primes containing $\mathfrak{a}$. Here, let $\mathfrak{p}$ be a prime containing $(x, y^2)$. Then $y^2 \in \mathfrak{p}$ implies $y \in \mathfrak{p}$. Can you finish it off from there?

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thanks for the answer! – Jr. Aug 21 '11 at 17:58

This is a longer (somewhat self contained), more general answer concerning the radical of any ideal generated by monomials. The answer to your question is f) of the proposition below. If you don't really care about all the proofs, just remember and use f).

Definitions: $K$ will denote an arbitrary field; ideal generated by $f_1,\ldots,f_k$ will be denoted by $\langle f_1,\ldots,f_k\rangle$. $I\unlhd K[\mathbf{x}]$ is a monomial ideal if it is generated by monomials, i.e. $I=\langle m_1,\ldots,m_k\rangle$ for some monomials $m_i$. A monomial $m$ is squarefree if every variable in $m$ has coefficient $1$ (or $0$), i.e. $m$ is of the form $x_{j_1}\cdots x_{j_l}$. A squarefree monomial ideal is an ideal that is generated by square-free monomials. Let $\sqrt{~}$ be the map that sends a monomial to the product of its variables, i.e. $x_{j_1}^{a_1}\cdots x_{j_l}^{a_l}\mapsto x_{j_1}\cdots x_{j_l}$ when all $a_i\neq 0$. Notice that the $\mathrm{gcd}$ and $\mathrm{lcm}$ of monomials is again a monomial.

Proposition: a) If $\mathfrak{a}$ is a monomial ideal and $f=\sum_ic_in_i\in\mathfrak{a}$ and $c_i\in K\setminus\{0\}$ and $n_i$ are different monomials, then all $n_i\in\mathfrak{a}$.

b) A monomial ideal in $K[\mathbf{x}]$ is prime if and only if it is of the form $\langle x_{i_1},\ldots,x_{i_k}\rangle$, i.e. generated by variables.

c) The only maximal monomial ideal in $K[\mathbf{x}]$ is $\langle x_1,\ldots,x_n\rangle$, i.e. generated by all variables.

d) If $\mathfrak{a}_i,\mathfrak{b}$ are monomial ideals, then $(\sum_i\mathfrak{a}_i)\cap\mathfrak{b}=\sum_i(\mathfrak{a}_i\cap \mathfrak{b})$.

e) For monomial ideals, $\langle m_1,\ldots,m_s\rangle\cap\langle n_1,\ldots,n_t\rangle=\sum_{i,j}\langle m_i\rangle\cap\langle n_j\rangle=\sum_{i,j}\langle \mathrm{lcm}(m_i,n_j)\rangle$.

f) If $m_1,\ldots,m_s$ are monomials, then $$\sqrt{\langle m_1,\ldots,m_s\rangle} = \langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle.$$

Proof: a) If $\mathfrak{a}=\langle m_1,\ldots,m_k\rangle$ and $m_i$ monomials, then $f\in\mathfrak{a}$ is of the form $f=\sum_{j=1}^kh_j(\mathbf{x})m_j$, so every $n_i$ contains some $m_j$.

b) $(\Rightarrow):$ If $\mathfrak{a}=\langle m_1,\ldots,m_k\rangle$ is prime and $m_i$ is not a variable, i.e. $m_i=m_i'm_i''$, then either $m_i'\in\mathfrak{a}$ or $m_i''\in\mathfrak{a}$, so $m_i$ can be replaced as a generator by one of these. $(\Leftarrow):$ if $\mathfrak{a}=\langle x_1,\ldots,x_k\rangle,$ then $K[x_1,\ldots,x_n]/\mathfrak{a} \cong K[x_{k+1},\ldots,x_n]$ is a domain, so $\mathfrak{a}$ is prime.

c) A maximal ideal is always prime, so by b) it must be generated by variables. But $K[x_{k+1},\ldots,x_n]$ is a field only when $k=n$.

d) $(\supseteq)$ is trivially true for any ideals in any ring. $(\subseteq):$ By induction we can reduce the problem to two ideals. Let $f_1+f_2\in\mathfrak{b}$ with $f_i\in\mathfrak{a}_i$, $i=1,2$. By a), every monomial of $f_i$ is in $\mathfrak{a}_i$. If the same monomial occurs in both $f_1$ and $f_2$, say $c_1m$ is a term of $f_1$ and $c_2m$ is a term of $f_2$, then $m\in\mathfrak{a}_1,\mathfrak{a}_1$ so we can write $f_1+f_2=(f_1-c_1m)+(f_2+c_1m)$ with $f_1-c_1m\in\mathfrak{a}_1$ and $f_2+c_1m\in\mathfrak{a}_2$. Hence we may assume that the monomials in $f_1$ are different from those in $f_2$. By a) we conclude that each monomial in $f_1$ or $f_2$ belongs to $\mathfrak{b}$. Hence $f_1,f_2\in\mathfrak{b}$.

e) is a direct consequence of d), because in any ring we have $\langle m_1,\ldots,m_s\rangle=\langle m_1\rangle+\ldots+\langle m_s\rangle.$

f) Put $k_i:=(\text{greatest exponent of any variable in }m_i)\in\mathbb{N}$, i.e. if $m_i=x_{j_1}^{a_1}\cdots x_{j_l}^{a_l}$ then $k_i=\max\{a_1,\ldots,a_l\}$. Now put $k:=k_1+\cdots+k_s-s+1$. We have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq \langle m_1,\ldots,m_s\rangle$, because every term of $\prod_{j=1}^k(f_{1,j}\sqrt{m_1}+\cdots+f_{s,j}\sqrt{m_s})$ has the form $f\sqrt{m_1}^{\beta_1}\cdots\sqrt{m_s}^{\beta_s}$ where $b_j\in\mathbb{N}_0$ and $\beta_1+\cdots+\beta_s=k$, which means at least one $\beta_j\geq k_j$. Therefore $$\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq\langle m_1,\ldots,m_s\rangle\subseteq\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle/\sqrt{~},$$ $$\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}\subseteq\sqrt{\langle m_1,\ldots,m_s\rangle}\subseteq\sqrt{\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle}.$$ Thus it remains to show that $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}=\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle$, i.e. that squarefree monomial ideals are radical.

If $\sqrt{m_1}=x_{j_1}\cdots x_{j_l}$, we have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_{r=1}^l\langle x_{j_r},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle$, because by e), $$\langle x_{j_1},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle\cap\langle x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle= \sum\sum\langle\mathrm{lcm}(\ast,\ast)\rangle= \langle x_{j_1}x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s} \rangle.$$ Next, $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_r\bigcap_{r'}\langle x_{j_r},x_{j_{r'}},\sqrt{m_3},\ldots,\sqrt{m_s}\rangle$, and so on. Therefore $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle = \bigcap_\lambda\mathfrak{p}_\lambda$ for some ideals $\mathfrak{p}_\lambda$, generated by variables. But $\mathfrak{p}_\lambda$ are prime by b), hence $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle} = \sqrt{\cap_\lambda\mathfrak{p}_\lambda}=\cap_\lambda\sqrt{\mathfrak{p}_\lambda}=\cap_\lambda\mathfrak{p}_\lambda$, since prime ideals are radical. $\blacksquare$

Example: we have $\sqrt{\langle x,y^2\rangle}=\langle \sqrt{x},\sqrt{y^2}\rangle=\langle x,y\rangle$ and $\sqrt{\langle xy,xz,yz\rangle}=\langle xy,xz,yz\rangle$ and $\sqrt{\langle xy,(x-y)x\rangle}=\sqrt{\langle xy,x^2-xy\rangle}=\sqrt{\langle xy,x^2\rangle}=\langle xy,x\rangle=\langle x\rangle$. $\blacklozenge$

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HINT $\:$ Applying the basic laws $\ \rm r({\frak a}+{\frak b})\ =\ r(r({\frak a}) + r({\frak b}))\:,\$ and $\rm \ \ r({\frak p}^n)\ =\ {\frak p}\:$ for prime $\frak p$

$\quad {\rm r}(x,y^2)\: =\: {\rm r}({\rm r}(x),{\rm r}(y^2))\: =\: {\rm r}(x,y)\: =\: (x,y)\$ since $\:x,y,(x,y)$ are all prime in $\:\mathbb Q[x,y]\:.$

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I'm interested in basic laws on the radical in any (commutative unital) ring. If $\mathfrak{a},\mathfrak{b}$ are any ideals, do the following hold: (1) $\sqrt{\mathfrak{a}\mathfrak{b}}\!=\!\sqrt{\mathfrak{a}}\sqrt{\mathfrak{b}}\!=\! \sqrt{\mathfrak{a}}\!\cap\!\sqrt{\mathfrak{b}}\!=\! \sqrt{\mathfrak{a}\!\cap\!\mathfrak{b}}$ (2) doesn't $\sqrt{\mathfrak{a}^k}\!=\!\sqrt{\mathfrak{a}}$ hold for any ideal? – Leon Dec 2 '11 at 0:00
@Leon There is a list of identities somewhere in Atiyah-Macdonald, although it probably isn't exhaustive. – Dylan Moreland Dec 2 '11 at 1:53

Clearly $(x,y)\subset r(x,y^2)$ and $1\notin r(x,y^2)$. Since $(x,y)$ is maximal, it is the radical of $(x,y^2)$.

[This is probably Andrea's argument; see his comment to the question.]

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