From Ravi Vakil's Foundations of Algebraic Geometry, exercise 7.4.C: Let $R = k[x, y, z]/(xz, yz, z^2)$. Show that $\operatorname{Proj} R \cong \mathbb{P}_k^1$ (as schemes).
My thought was to induce a map of schemes via the ring morphism $\phi: R \rightarrow k[s, t]$, with $x \mapsto s, \; y \mapsto t, \; z \mapsto 0$. But I don't know how to show that this induces an isomorphism of schemes. In particular, in $\operatorname{Proj} R$ the two ideals $(x + y)$ and $(x + y + z)$ (for example) are different, which doesn't seem to have a counterpart in $\mathbb{P}_k^1$.
Edit: It's been suggested that we can check this locally. To check locally, define $\alpha:k[s,t]→R$ by $s,t↦x,y$. This induces $\pi:\operatorname{Proj}R→\mathbb{P}_1^k$ such that $\pi^{-1}(D(s))=D(x)$ and $\pi^{-1}(D(t))=D(y)$. I can see that $\pi$ gives a bijection on each of these open sets, but I can't see that it gives isomorphisms of schemes: for example, taking $\varphi:\mathcal{O}_{\operatorname{Proj}R}→\mathcal{O}_{\mathbb{P}_1^k}$ induced by $\alpha$ , we get $\varphi(D(s)):k[\frac{t}{s}]→k[\frac{y}{x},\frac{z}{x}]$ , which is not an isomorphism. Am I doing this wrong?
You're on the right track with checking locally, but you have made a small error in calculating the sections of $\operatorname{Proj} R$ on the open sets $D(x)$ and $D(y)$. Recall that $\mathcal{O}_{\operatorname{Proj} R}(D(x))\cong R_{(x)}$, the homogeneous localization of $R$ at the element $x$. In other words, this is $(R_x)_0$, the degree-zero portion of the graded ring $R_x$. This is $$k\left[\frac{y}{x},\frac{z}{x}\right]/\left(\frac{z}{x},\frac{yz}{x^2},\frac{z^2}{x^2}\right)\cong k\left[\frac{y}{x},\frac{z}{x}\right]/\left(\frac{z}{x}\right)\cong k\left[\frac{y}{x}\right]$$ and indeed the map of sections here is an isomorphism, identifying $\frac{y}{x}$ and $\frac{t}{s}$.
Alternatively, as suggested in the comments, one may use the fact that if $S=\bigoplus_n S_n$ is a graded ring, the Veronese subring $S^{(d)}=\bigoplus_n S_{nd}$ has the same Proj. In this case, since $R=k[x,y,z]/(xz,yz,z^2)$ has $R_d\cong k[x,y]_d$ for any $d\neq 1$, then we can use this fact twice to obtain the claim.