Show that in the ring $R = \mathbb{Q} [x, y]$ there are ideals that require at least two generators (for example, the ideal $I =\{f\in R: f (0,0) = 0\}$)
What would be the generators for the example I put above? I think it's $\{0,x,y\}$, is this correct?
In what other cases is there more than one generator?
On
Yes indeed, $I=(x,y)$. Given a polynomial $f(x,y)$, we have $f(0,0)=0$ iff the constant term of $f$ is $0$ iff every monomial of $f$ contains at least one variable iff $f\in (x,y)$.
On
How about $\mathscr I=(x^2,y)$, for instance, as another example... or $\mathscr I=(x^n,y^m)$ for $n,m\gt1$ etc...
On
"In what other case is there more than one generator?" is not really the right question. If $I$ is any non-zero ideal you can give a set of generators with more than one element. For example if $I$ is generated by $x$ then $I$ is also generated by $x$ and $2x$.
You meant to ask when it happens that the ideal is not generated by just one element. The answer to that is that happens when it happens.
Btw you still have to show that the ideal in your question is not generated by one polynomial...
Write a polynomial $f(x,y)\in\mathbb{Q}[x,y]$ as $$ f(x,y)=g_0(x)+g_1(x)y+\dots+g_n(x)y^n $$ where $g_i(x)\in\mathbb{Q}[x]$. The condition $f(0,0)=0$ is equivalent to $g_0(0)=0$, that is, $g_0(x)=a_1x+\dots+a_mx^m$ (zero constant term).
Therefore, if $f(0,0)=0$, we have $$ f(x,y)=x(a_1+\dots+a_mx^{m-1})+y(g_1(x)+\dots+g_n(x)y^{n-1} $$ belongs to the ideal $(x,y)$.
Conversely, if $f(x,y)=xf_1(x,y)+yf_2(x,y)$, then $f(0,0)=0$.
Can $(x,y)$ be generated by a single polynomial?
Suppose it is, so $(x,y)=(p(x,y))$. In particular, $x=p(x,y)q(x,y)$ and $y=p(x,y)r(x,y)$. Can you derive a contradiction?