Mathematics builds on itself. If your foundation is brittle, you cannot grow tall. The following exercise, suitable for a first-year graduate student in algebraic geometry, illustrates this idea. It sits atop a tall hierarchy of prerequisites. Intuitively, it says that the graph of any polynomial is algebraically indistinguishable from the graph of a line.
Claim. The affine coordinate ring of the zeros of the polynomial $f(x,y)=y-g(x)$ is isomorphic to the polynomial ring in one variable over an algebraically closed field $k$.
Proof. Since $f$ is monic of degree one in $k[x][y]$, $f$ is irreducible. Since $k[x][y]$ is a unique factorization domain, the ideal $(f)$ generated by $f$ is prime, hence radical. By Hilbert's Nullstellensatz, the ideal $I(Z(f))$ of the zero set $Z(f)=Z((f))$ of $(f)$ is $\sqrt{(f)}=(f)$. By the universal property of polynomial rings, there exists a ring homomorphism $\phi:k[x][y]\to k[t]$ defined by $x\mapsto t$ and $y\mapsto g(t)$. Since $\phi(h(x))=h$ for any $h\in k[t]$, $\phi$ is surjective. If $h\in (f)$, then $$\begin{align}\phi(h)&=\phi(qf)\\&=\phi(q(y-g(x)))\\&=\phi(q)(\phi(y)-\phi(g(x)))\\&=\phi(q)(g(t)-g(t))\\&=0,\end{align}$$ so $h$ is an element of the kernel $K$ of $\phi$. Conversely, by the division algorithm, if $h\in K$, then $$\begin{align}0&=\phi(h)\\&=\phi(qf+r(x))\\&=\phi(q)\phi(f)+\phi(r(x))\\&=r(t),\end{align}$$ so $h=qf\in (f)$; hence $(f)=K$. By the first isomorphism theorem, $$\begin{align}k[x,y]/I(Z(f))&=k[x,y]/(f)\\&=k[x,y]/K\\&=k[x][y]/K\end{align}$$ is isomorphic to $k[t]$. $$\tag*{$\blacksquare$}$$
On the other hand, not every algebraic set has this property, as the following exercise illustrates.Claim. The affine coordinate ring of the zeros of the polynomial $f(x,y)=xy-1$ is not isomorphic to the polynomial ring in one variable over an algebraically closed field $k$.
Proof. Suppose that $f$ is reducible. Then there exist non-units $g$ and $h$ such that $f=gh$. Since $$2=\deg(f)=\deg(gh)=\deg(g)+\deg(h),$$ we have $0<\deg(g)=\deg(h)=1$. So $g=ax+by+c$, $h=a'x+b'y+c'$, and $$gh=aa'x^2+a'bxy+a'xc+ab'xy+bb'y^2+b'cy+ac'x+bc'y+cc'.$$ Suppose, without loss of generality, that $a=0$. This forces $b\neq0$, $c=0$, $a'\neq0$, $b'=0$, and $c'=0$, but $cc'=0\neq-1$, which is a contradiction. Therefore, $f$ is irreducible.
By the same argument as in the previous claim, $k[x,y]/I(Z(f))=k[x,y]/(xy-1)$. Since $2=\deg(xy-1)>\deg(x-\lambda)=1$ for any $\lambda\in k$, $xy-1$ does not divide $x-\lambda$. So $x-\lambda\notin(xy-1)$; hence $\overline x\in k[x,y]/(xy-1)$ is non-constant, and $\overline x\cdot\overline y=\overline{x\cdot y}=\overline1$. So $\overline x$ and $\overline y$ are units. Therefore, since $k[t]$ has no non-constant units, $k[t]$ is not isomorphic to $k[x,y]/(xy-1)$. $$\tag*{$\blacksquare$}$$