0
$\begingroup$

My question is pretty easy, however I got stuck to get the answer.

Assume the polynomial ring $ R=\mathbb Z[x,y]$ and $I=(xy)$ be an ideal of $R$. I want to compute the radical of $I$.

I am trying to show that $I$ is prime ideal since I know tha radical for prime ideal will be the prime ideal itself. Anyway, what is the radical of $I$.

Any help will be appreciated.

  • 0
    Are you certain that $I$ is prime? It seems like $\bar{x}$ and $\bar{y}$ are nonzero elements in the quotient ring $R/I$, but $\bar{x}\bar{y}=\bar{0}$. – 2017-01-29
  • 0
    I do not assume $I$ Prime ideal. I want to find $ rad(I)$ – 2017-01-29

1 Answers 1

1

If $P\in Rad(I)$ then there exists $k\in\mathbb{N}$ and $Q\in\mathbb{Z}[x,y]$ such that $P^k=xyQ$. And obviously, $k>0$.

Consequently, $x$ and $y$ divide $P^k$.

But $x$ and $y$ are irreducible. Hence, by Euclid's lemma, each of them must divide $P$.

Finally, since $x$ and $y$ are coprime, their lcm (namely $xy$) must divide $P$. So $P\in I$.

Hoping it's correct ...

  • 0
    Please, what is the Euclid's Lemma – 2017-01-29
  • 0
    @Team - Euclid's lemma says that if an irreducible element (of any GCD domain) divides a product, then it divides at least one factor. – 2017-01-29
  • 0
    You mean lcm and not gcd. – 2017-01-29
  • 0
    @Mohan - of course, sorry for the mistake ... Editing – 2017-01-29
  • 0
    No, [Euclid's Lemma](https://en.wikipedia.org/wiki/Euclid's_lemma) doesn't state what you say. That one is a well known property of GCD domains, namely irreducible = prime. – 2017-01-29
  • 0
    Btw, you don't even need this. The elements $x,y$ are prime. – 2017-01-29