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Write $A=k[x,y]$ and $\mathfrak{m}=(x,y)$. Since $f(0,0)\ne 0$, the principal open subscheme $D(f)=\mathrm{Spec}{A}_f$ contains $\mathfrak{m}$, andCover $V$ by the two affine opens $D(xf)$ and $D(yf)$, whose intersection is $D(xyf)$. The degree-zero part of the resulting Čech cohomology complex givesinside the fraction field of $A$. The last equality follows because $A_f$ is a unique factorization domain and a rational function regular after localizing at both $x$ and $y$ has no possible prime factor left in its denominator.

The same affine cover is acyclic, so its degree-one Čech group computes sheaf cohomology and givesBefore localizing at $f$, the quotienthas the $k$-basisWriting $f=c+h$ with $c=f(0,0)\in k^{\times }$ and $h\in (x,y)$, multiplication by $h$ is locally nilpotent on $Q$: for each negative monomial, a sufficiently high power of $(x,y)$ moves every term into $A_x+A_y$. Hence $c+h$ acts invertibly on $Q$ by a finite geometric series on each element. Localizing at $f$ therefore leaves $Q$ unchanged, andThe displayed infinite basis proves that this vector space is infinite-dimensional.

Solved by gpt-5.6-sol high.

Write $X=\mathrm{Spec}B$, $Y=\mathrm{Spec}A$, and let the morphism correspond to a homomorphism $\phi :A\to B$. Put $I=\ker \phi$ andThe injection $A/I\hookrightarrow B$ gives $X\to Z$, and the quotient $A\to A/I$ gives a closed immersion $Z\hookrightarrow Y$, so $f$ factors through $Z$.

If $f$ factors through another closed subscheme $Z^{\prime }=\mathrm{Spec}(A/J)$, then $J\subseteq \ker \phi =I$. The resulting quotient $A/J\to A/I$ induces the unique factorization $Z\to Z^{\prime }\to Y$. Thus $Z$ is the scheme-theoretic image.

It remains to identify its underlying set. A principal open subscheme $D(a)\subseteq \mathrm{Spec}A$ misses $f(X)$ exactly when $D(\phi (a))$ is empty, equivalently when $\phi (a)$ is a nilpotent element. Hence the ideal of functions vanishing set-theoretically on $f(X)$ has radical $\sqrt{\ker \phi }$. The closure is thereforeas required.

Solved by gpt-5.6-sol high.