Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F$ be a field, and $F[x,y]$ the ring of polynomials in $x,y$. Let $J$ be the subset of all polynomials $P(x,y)$ in $F[x,y]$ such that $P(0,0)=0$. Then $J$ is an ideal. Is $J$ a principal ideal?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The ideal you describe is called the irrelevant ideal and is given by $J=(x,y)$. It is evident that $J$ is not a principal ideal, since it is generated by two coprime irreducible polynomials $x,y$. Note though that $J$ is a maximal ideal, since $F[x,y]/(x,y) \cong F$.

share|improve this answer
    
I agree it is generated by two coprime irreducible polynomials. Why does it follow that it is not a principal ideal? –  Mika H. Jan 16 at 15:56
2  
@MikaH.: If it were a principal ideal, then it would be of the form $J=(p)$ for some polynomial $p$. Then $p$ would divide both $x$ and $y$. But $x$ is divisible only by $x$ and $1$. Similarly $y$ is divisible only by $y$ and $1$. This shows that $p$ must be $1$, contradiction. –  Manos Jan 16 at 16:33

Hint $\ $ Notice $\,x\,$ is prime since $\,F[x,y]/(x) = F[y]\,$ is a domain. Similarly $\,y\,$ is prime. Being prime, $\,x,y\,$ are irreducible. So if $\,J = (x,y) = (f)\,$ then either $\,f\,$ is a unit, or $\,f\,$ is associate to $\,x\,$ and $\,y,\,$ so $\,x\,$ and $\,y\,$ are associate: $\,x=uy,\,$ so evaluating at $\,x,y = 1,0\,\Rightarrow\, 1=0,\,$ contra $\,F\,$ is a field. Hence $\,f\,$ is a unit, $\,(f)= (1) = (x,y)\,\Rightarrow 1 = x f(x,y) + y\, g(x,y)\,\Rightarrow\, 1=0\,$ by evaluating at $\,x,y = 0,0.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.