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The following question is about understanding a basic definition. the example involved is simple.

According to an answer to another question on this site, The set generated by $x,y$ is an ideal in $\mathbb{Q}[x,y]$. I can't understand why it is an ideal. In what way does it absorb other elements of $\mathbb{Q}[x,y]$ (by multiplication operation). Maybe the question is what elements does $(x,y)$ contain, since I know what $\mathbb{Q}[x,y]$ means and what polynomial multiplication means.

Thanks.

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    $\begingroup$ $(x,y)$ denotes by definition the ideal generated by $x,y$. Hence, it is an ideal by definition. Its elements are $xf(x,y)+yg(x,y)$. $\endgroup$
    – Crostul
    Commented Mar 22, 2017 at 13:46
  • $\begingroup$ @Crostul Thank you, can you please tell me whether the set generated by $x^m,y^n$ is an ideal in $\mathbb{Q}[x,y]$? By your definition it intuitively looks to me like it is. $\endgroup$
    – Friedman
    Commented Mar 22, 2017 at 13:56

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As Crostul mentions in his comment, the ideal $(x, y)$ by definition consists of the elements of the form $$x f(x, y) + y g(x, y)$$ (but NB this representation need not be unique).

Now, any element of $\Bbb Q[x, y]$ can be uniquely written as $$a_{00} + \sum_{i, j > 0} a_{ij} x^i y^j$$ (for all but finitely many $a_{ij}$ nonzero), so $p(x, y) \in \Bbb Q[x, y]$ is in the ideal $(x, y)$ iff the constant term $a_{00}$ is zero. On the other hand, $a_{00} = p(0, 0)$, so $(x, y)$ consists precisely of the polynomials that vanish at the origin.

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  • $\begingroup$ By vanish at the origin do you mean a discontinuous point or just equals zero? And please answer the question I asked Crostul: Is the set generated by $x^m,y^n$ is an ideal in $\mathbb{Q}[x,y]$? By your definitions it looks to me looks like it is. $\endgroup$
    – Friedman
    Commented Mar 22, 2017 at 14:16
  • $\begingroup$ I don't quite understand the first question, but I mean more precisely that $(x, y) = \{p(x, y) \in \Bbb Q[x, y] : p(0, 0) = 0\}$. And you should ask your other question in a separate post. $\endgroup$
    – Travis Willse
    Commented Mar 22, 2017 at 14:23
  • $\begingroup$ Yes guess I was unclear but never mind I got it thank you. $\endgroup$
    – Friedman
    Commented Mar 22, 2017 at 14:26
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    $\begingroup$ @Friedman Yes, the ideal generated by $x^m$ and $y^n$ is also an ideal, by definition. It consists of polynomials of the form $x^m \cdot f(x,y) + y^n \cdot g(x,y)$ - that is, any monomial in it has at least $m$-th power of $x$ or $n$-th power of $y$. $\endgroup$
    – lisyarus
    Commented Mar 22, 2017 at 14:27

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