Difficulty constructing an isomorphism form quotient ring of multivariate polynomials I am trying to construct an isomorphism
$$
\frac{\mathbb{C}[x,y,z]}{(x^5-z,x^3-y)}\simeq \mathbb{C}[x]
$$
using the map
$$
\phi : \frac{\mathbb{C}[x,y,z]}{(x^5-z,x^3-y)}\rightarrow \mathbb{C}[x]\\
1\mapsto 1,x\mapsto x, y\mapsto x^3,z\mapsto x^5
$$
Surjectivity seems clear given that we are projecting, in some sense. 
For injectivity, that $(x^5-z,x^3-y)\subset \ker(\phi)$ is clear. However I am struggling with the other direction of inclusion. I have gotten as far as concluding that for some
$$
r(x,y,z)\in \frac{\mathbb{C}[x,y,z]}{(x^5-z,x^3-y)},\; r(x,y,z)=\bar{r}(x,y,z)+I
$$
Where $I$ is the ideal. Since we have in the quotient ring $x^5=z$ and $x^3=y$ we can rewrite this polynomial as
$$
\bar{r}(x,x^3,x^5)
$$
Then if $\phi$ kills it we have
$$
\phi(\bar{r}(x,x^3,x^5))=0
$$
Is it fair then to conclude that $\bar{r}(x,y,z)\equiv 0$ since it vanishes whenever $x=x$ or $y=x^3$ or $z=x^5$, or always in this quotient ring? 
 A: Although a solution was given in the comments by Camilo Arosemena, allow me to complete the solution proposed by the OP.
The key point is this: euclidean division by a polynomial is possible if its leading term is invertible.
I will view $\phi$ as a morphism from $\mathbb{C}[x,y,z]$ to $\mathbb{C}[x]$, defined as the $\phi$ in the question.
Let $P\in \ker(\phi)$.  View it as an element in $\mathbb{C}[x,y][z]$, that is, as a polynomial in $z$ with coefficients in $\mathbb{C}[x,y]$.  Since the leading coefficient of $z-x^5$ is invertible in that ring, we can perform euclidean division:
$$P = (z-x^5)\cdot Q + R,$$
where $\deg_z(R)=0$.  In other words, $z$ "does not appear" in $R$, and we can see $R$ as being an element of $\mathbb{C}[x,y]$.  
Now, apply the same reasoning to divide $R$ by $y-x^3$:
$$ R = (y-x^3)\cdot S + T, $$
where $\deg_y(T)=0$.  Hence $y$ and $z$ don't appear in $T$.  We can view it as an element of $\mathbb{C}[x]$.
Now, we have
$$P=(z-x^5)\cdot Q +(y-x^3)\cdot S + T.$$
Applying $\phi$, we get
$$0=\phi(T).$$
But since $T$ is an element of $\mathbb{C}[x]$, this is only possible if $T=0$.  Hence $P\in (x^5-z, x^3-y)$.
