$\operatorname{Proj}k[x,y,z]/(xz,yz,z^2)$ isomorphic to $\mathbb{P}^{1}_{k}$ While dealing with the Proj construction, I encountered with this seemingly-simple question, but somehow I can't get the point at this moment.

Is the scheme $$\operatorname{Proj}k[x,y,z]/(xz,yz,z^2)$$ isomorphic to the projective line $\mathbb{P}^{1}_{k}$, where $k$ is a field? 

I considered the morphism 
$$k[x,y,z]/(xz,yz,z^2)\to k[x_{0},x_{1}]$$ given by the evaluation $$(x,y,z)\mapsto(x_{0},x_{1},0)$$ and its induced morphism $$\mathbb{P}^{1}_{k}\to \operatorname{Proj}k[x,y,z]/(xz,yz,z^2)$$ which should be well-defined in this case. But I don't know what to do to check this is indeed an isomorphism. 
Any help or hint will be appreciated!
 A: Hint: The surjective homomorphism $f^\# : k[x,y,z]/(xz,yz,z^2) \to k[x,y]$ (with $z \mapsto 0$) of graded algebras is an isomorphism when localized at $x$, $y$ or $z$ resp. the images.
More details: Let $X=\mathrm{Proj} k[x,y,z]/(xz,yz,z^2)$ and let $f : \mathbb{P}^1 \to X$ be the closed immersion induced by $f^\#$. Since the graded algebra is generated by $x,y,z$, we have $X=D_+(x) \cup D_+(y) \cup D_+(z)$. Therefore, it suffices to prove that $f^{-1}(D_+(a)) \to D_+(a)$ is an isomorphism for $a \in \{x,y,z\}$. But this precisely means that $$(f^\#)_{(a)} : k[x,y,z]/(xz,yz,z^2)_{(a)} \to k[x_0,x_1]_{(f^\#(a))}$$
is an isomorphism of rings, where $(a)$ denotes homogeneous localization at the element $a$. In fact, we have an isomorphism on the whole localization  $(f^\#)_a : k[x,y,z]/(xz,yz,z^2)_{a} \to k[x_0,x_1]_{f^\#(a)}$: For $a=x$ we have $$k[x^{\pm 1},y,z]/(xz,yz,z^2)=k[x^{\pm 1},y,z]/(z,yz,z^2) = k[x^{\pm 1},y].$$
For $a=y$ we have
$$k[x,y^{\pm 1},z]/(xz,yz,z^2)=k[x,y^{\pm 1},z]/(xz,z,z^2) = k[x,y^{\pm 1}].$$
For $a=z$ we have
$$k[x,y,z^{\pm 1}]/(xz,yz,z^2)=0=k[x,y]_{f^\#(z)}.$$
Global solution. Projective schemes have a beautiful description via their functors of points. $\mathbb{P}^1$ represents the functor which takes a $k$-scheme $T$ to the set of (iso-classes of) tuples $(\mathcal{L},x,y)$, where $\mathcal{L}$ is an invertible sheaf on $T$ and $x,y$ are two global sections of $\mathcal{L}$ which generate $\mathcal{L}$. Also, $X$ represents the functor which takes a $k$-scheme $T$ to the set of tuples $(\mathcal{L},x,y,z)$, where $\mathcal{L}$ is an invertible sheaf on $T$ and $x,y,z$ are two global sections of $\mathcal{L}$ which generate $\mathcal{L}$ and which satisfy the relations $x \otimes z = 0$, $y \otimes z = 0$, $z \otimes z = 0$ (these are global sections of $\mathcal{L}^{\otimes 2}$). Now our task is simply to prove $z=0$! Since $\mathcal{L}$ is invertible, it suffices to prove that $\mathrm{id}_{\mathcal{L}} \otimes z : \mathcal{L} \to \mathcal{L} \otimes \mathcal{L}$ is zero. Since $(x,y,z) : \mathcal{O}_T^3 \to \mathcal{L}$ is an epimorphism, it suffices to prove that the composition $\mathcal{O}_T^3 \to \mathcal{L} \otimes \mathcal{L}$ is zero. But this is precisely what the relations $x \otimes z = 0$, $y \otimes z = 0$, $z \otimes z = 0$ tell us!
