Radicals Of Zero Dimensional Ideals

Problem
Let $I \subset k[x_1,\ldots, x_n]$ be a zero dimensional ideal over a field of characteristic zero. For each $1 \leq i \leq n$ let $f_i$ be the unique monic generator for $I\bigcap k[x_i]$.

(a) Show that if $f\in k[z]$ (polynomial ring in the single variable $z$) and $f'$ denotes the derivative with respect to $z$, then $\sqrt{f}:=f/gcd(f,f')$ is the product of all the irreducible factors of $f$.

(b) Show that $\sqrt{I}=(I,\sqrt{f_1},\sqrt{f_2},\ldots, \sqrt{f_n})$.


Solution

(a) Since $k[z]$ is a UFD, factor $f$ as $f=g_1^{a_1}\ldots g_k^{a_k}$ where $g_1,\ldots,g_k$ are the irreducible factors appearing in $f$. Now

(1)
\begin{align} f' &= \sum_{i=1}^k a_ig_1\cdots g_i^{a_i-1} \cdots g_k^{a_k} g'_i\\ &= (g_1^{a_1-1}\cdots g_k^{a_k-1})(\sum_{i=1}^k a_ig_1\cdots g'_i \cdots g_k) \end{align}

Write $h$ for the sum appearing on the right hand side in the second equality. For $i=1,\ldots,k$, $g_i$ divides every term of $h$ except the term involving $g'_i$ (since over a field of characteristic 0, an irreducible polynomial $p$ and its derivative $p'$ are always relatively prime). Hence it follows that $g_i$ does not divide $h$ for any $i$, so $gcd(f,f')=g_1^{a_1-1}\ldots g_k^{a_k-1}$. From this it is clear that $\sqrt{f}=g_1\cdots g_k=f/gcd(f,f')$.

(b) Write $\sqrt{f_i}=g_i$ and let $g_{i,1},\ldots, g_{i,m_i}$ be the $m_i$ distinct irreducible factors appearing in $f_i$. In this notation we have $g_i=g_{i,1}\ldots g_{i,m_i}$.

We have:

(2)
\begin{align} I \subset (I,g_1,\ldots,g_n) \subset \sqrt{I} \end{align}

The first inclusion is trivial and the second follows since some power of $g_i$ is divisible by $f_i$ (hence lies in $I$) for each $i$. We argue that $(I,g_1,\ldots,g_n)$ is itself a radical ideal, which will complete the proof.

Note that for any fixed $i,g_{i,j}$ and $g_{i,k}$ are relatively prime for all $1\leq j < k \leq m_i$. This is because distinct irreducible polynomials are always relatively prime. This implies that the ideal generated by $g_{i,j}$ and $g_{i,k}$ for $j\neq k$ is the whole polynomial ring. We are in the position to apply a previous exercise, which stated that if $f$ and $g$ generate the unit ideal and $I$ is any ideal, then $(I,fg)=(I,f)\bigcap (I,g)$. Successive applications of this exercise yield

(3)
\begin{align} (I,g_1,\ldots,g_n)=\bigcap_{1\leq i_1 \leq m_i, \ldots, 1\leq i_k \leq m_k} (I,g_{1,i_1},g_{2,i_2},\ldots,g_{k,i_k}) \end{align}

Since any finite intersection of radical ideals is radical, it suffices to prove that the ideals appearing in the intersection on the right hand side are radical. We finish by proving the following claim:

Claim: Let $R=k[x_1,\ldots,x_n]$ be a polynomial ring over a field of characteristic 0 and let $p_i \in k[x_i]$ for $i=1,\ldots,n$ be irreducible polynomials. If $I\subset R$ is any ideal, then $(I,p_1,\ldots,p_n)$ is a radical ideal.

Proof of Claim. We prove that the quotient $S=R/(I,p_1,\ldots,p_n)$ is reduced, i.e. it does not contain any nilpotent elements. Let $R_i=R/(p_1,\ldots,p_i)$.

I claim we are done if we prove that $R_n=R/(p_1,\ldots,p_n)$ is a product of fields. In this case the image $\bar{I}$ of $I$ in $R_n$ under the quotient map is an ideal in a product of fields, hence is itself a product of some of these fields. It follows that $R_n/\bar{I}=S$ is again a product of fields, and hence reduced.

We now show that $R_n=R/(p_1,\ldots,p_n)$ is a product of fields. Since $p_1$ is an irreducible polynomial over $k[x_1],R_1=R/(p_1)\cong k(\alpha)[x_2,\ldots,x_n]$, where $\alpha$ is a root of $p_1$ in some extension field. Now suppose $p_2$ factors into $z_2$ irreducibles $p_2=p_{2,1}\ldots p_{2,z_2}$ over the field $k(\alpha)$. There are no repeated factors in this decomposition because the polynomial $p_2$ has distinct roots (it is an irreducible polynomial over $k$, which is a separable field since it has characteristic 0). It follows that

(4)
\begin{align} R_2\cong & k(\alpha)[x_2,\ldots,x_n]/(p_2)\\ \cong & k(\alpha,\alpha_{2,1})[x_3,\ldots,x_n]\times k(\alpha_1,\alpha_{2,2})[x_3,\ldots,x_n] \times \cdots \times k(\alpha_1,\alpha_{2,z_2})[x_3,\ldots,x_n], \end{align}

where $\alpha_{2,1},\ldots,\alpha_{2,z_2}$ are roots of the polynomials $p_{2,1},\ldots,p_{2,z_2}$ in some extention field of $k(\alpha)$. Let $F_{2,i}$ denote the field $k(\alpha,\alpha_{2,i})$ for $i=1,\ldots,z_2$. Now the ideal generated by the image of $p_3$ in $R_2$ is the same as the ideal generated by the tuples $(p_3,\ldots,0),(0,p_3,\ldots,0),\ldots,(0,\ldots,p_3)$ in $R_2 \cong F_{2,1}[x_3,\ldots,x_n]\times \ldots \times F_{2,z_2}[x_3,\ldots,x_n]$. It follows that we can repeat the above argument for $p_3$ in each of the factors, obtaining that $R_3$ is a product of fields. Then successively applying the same arguments yields that $R_n$ is a product of fields.

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