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Let $I subset mathbbC[x_1,x_2]$ be a proper ideal.
How does it follow from Hilbert's Nullstellensatz, that there must exist $z_1,z_2 in mathbbC$ such that $I subset leftlangle x_1-z_1,x_2-z_2 rightrangle$?

A consequence of Nullstellensatz is that, for every proper ideal $I$ of $mathbbC[x_1,dots,x_n]$,
$$
sqrtI=bigcap_(a_1,dots,a_n)in V(I)(x_1-a_1,dots,x_n-a_n)
$$
where $V(I)=(a_1,dots,a_n)inmathbbC^n:f(a_1,dots,a_n)=0text, for all $fin I$$.

In particular, a maximal ideal of $mathbbC[x_1,dots,x_n]$ has the form $(x_1-a_1,dots,x_n-a_n)$, for a unique $(a_1,dots,a_n)inmathbbC^n$.

Hilbert's Nullstellensatz says that if $I$ is a proper ideal, the zero locus of $I$ is non-empty as we are working over an algebraically closed field.
Thus the zero locus $V(I)$ of $I$ contains a point, say $(z_1,z_2)$.

On the other hand, the ideal $J = langle x_1-z_1,x_2-z_2rangle$ is maximal with zero locus $V(J)=(z_1,z_2)$. Thus $V(J)subseteq V(I)$. Taking ideal operator $cal I$ as in the strong version of Hilbert's Nullstellensatz gives $sqrt I = cal I(V(I)) subseteq cal I(V(J)) = sqrt J$. But $Isubseteqsqrt I$ and $Jsubseteq sqrt J$. Since $J$ is maximal, $J=sqrt J$. Hence, $Isubseteq J$ as required.