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A polynomial in a polynomial ring in one variable over a field generates a radical ideal iff it has no multiple roots. Is there a sufficient condition for a polynomial in several variables to generate a radical ideal? Like an ideal generated by a polynomial is prime if and only if it is irreducible.

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Look up "square free" –  Bill Cook Dec 18 '11 at 1:12
    
@Bill Cook: I know square free monomial ideals are radical and radical monomial ideals are square free. I was looking for conditions on polynomials that are not monomial. –  Gene Simmons Dec 18 '11 at 1:36
    
@GeneSimmons, you should really look up square free! :D –  Mariano Suárez-Alvarez Dec 18 '11 at 2:29
    
@MarianoSuárez-Alvarez: I tried various searches and looked up over a 100 articles but could not find anythIng close to the answer to my question. Perhaps I don't really understand the hint. I found some criteria for zero dimensional ideals in terms of square free polynomials, but not much else. If anyone has a pointed reference I would prefer that to random google searches. –  Gene Simmons Dec 18 '11 at 4:20
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@Gene: Prove it! If $f$ is squarefree and $f|g^n$, then for every irreducible factor $p$ of $f$, $p|g^n$, hence $p|g$. Therefore (since distinct irreducible factors are relatively prime), the product of all distinct irreducible factors of $f$ divides $g$; but this product is (an associate of) $f$, because $f$ is squarefree. So if $f$ is squarefree, $g^n\in (f)\Rightarrow g\in (f)$, so $(f)$ is radical. Conversely, if $f$ is not squarefree, then the squarefree root of $f$ has a power that lies in $(f)$ but does not itself lie in $f$. –  Arturo Magidin Dec 18 '11 at 5:09

2 Answers 2

up vote 2 down vote accepted

Let $k$ be a field and consider the polynomial ring $A = k[x_1,...,x_n]$.

Claim: Given $f \in A - \{0\}$, (f) is radical if and only if $f$ factors into a product of irreducibles of multiplicity $1$.

Proof:

$\Leftarrow$: We know $A$ is a UFD. So, let $f = f_1...f_m$ be a product of $f$ into irreducible factors such that for all $i \neq j$, $(f_i) \neq (f_j)$. Then $(f) = (f_1...f_m) = (f_1) \cap ... \cap (f_m)$ (I am using unique factorization for the second equality). Thus, $(f)$ is an intersection of prime ideals of $A$ and hence radical.

$\Rightarrow$: Suppose $(f)$ is radical. Again, let $f = {f_1}^{e_1}...{f_m}^{e_m}$ be a product of $f$ into irreducibles where $i \neq j$ $\Rightarrow$ $(f_i) \neq (f_j)$.

Our goal is to show that each $e_i = 1$. Well, suppose not. Then there exists $e_i$ such that $e_i > 1$. Then $({f_1}^{e_1}...{f_i}^1...{f_m}^{e_m}) \subset (f) \subset ({f_1}^{e_1}...{f_i}^1...{f_m}^{e_m})$. The first inclusion is because ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m} \in Rad((f)) = (f)$, and the second inclusion follows from the fact that ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m}|f$.

But, this means that there is some $u \in A^*$ such that ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m} = u{f_1}^{e_1}...{f_i}^{e_i}...{f_m}^{e_m}$, which contradicts unique factorization.

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Thanks. One question, why is the ideal generated by the product of the irreducible factors equal to the intersection of the ideals generated by the individual factors? –  Gene Simmons Dec 18 '11 at 5:20
    
$\subset$ is elementary, and $\supset$ follows from unique factorization. –  Rankeya Dec 18 '11 at 5:22
    
Thanks. One follow up question. This doesn't extend to non-principal ideals right? –  Gene Simmons Dec 18 '11 at 5:23
    
What is the precise statement you are trying to make for non-principal ideals? –  Rankeya Dec 18 '11 at 5:27
    
Are ideals generated by several square free polynomials radical? –  Gene Simmons Dec 18 '11 at 5:33

Let $(p_1),\dots,(p_n)$ be distinct prime ideals of a unique factorization domain, and let $k_1,\dots,k_n$ be positive integers. Then the radical of $$ (p_1^{k_1}\cdots p_n^{k_n}) $$ is clearly $$ (p_1\cdots p_n). $$

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