How can I find the kernel of $\phi$? We have the homomorhism $\phi: \mathbb{C}[x,y] \to \mathbb{C}$ with $\phi(z)=z, \forall z \in \mathbb{C}, \phi(x)=1, \phi(y)=0$. I have shown that for $p(x,y)=a_0+\sum_{k,\lambda=1}^m a_{k \lambda} (x-1)^k y^{\lambda}$, we have $\phi(p(x,y))=a_0 \in \mathbb{C}$. How can I find the kernel of $\phi$ ?
 A: As you suspect, the kernel is indeed $\langle x-1, y \rangle$. 
To see this, note that your homomorphism is really just the evaluation homomorphism at $(1, 0)$. Then $\langle x-1, y \rangle \subset \ker \phi$, since $p(x ,y) \in \langle x-1, y \rangle$ implies $p(x,y) = (x-1)p_{1}(x,y) + yp_{2}(x,y)$ for some $p_{1}(x, y), p_{2}(x,y) \in \mathbb{C}[x,y]$, i.e. $p(1,0) = 0$. 
Consider the evaluation map $\rho: \mathbb{C}[x,y]\rightarrow \mathbb{C}[y]$ given by evaluation at $x=1$. Clearly, the kernel of $\rho$ is $\langle x-1 \rangle$, so $\mathbb{C}[x,y]/\langle x-1\rangle \cong \mathbb{C}[y]$. Then, to see that $\ker \phi \subset \langle x-1, y \rangle$, let $p(x,y) \in \ker \phi$. Then $p(1,0) = 0$. As noted by user Krish, we can divide by the monic $x-1$ viewing $x-1$ as a polynomial in $x$ over the ring $\mathbb{C}[y]$. Doing so gives $p(x,y) = (x-1)g(x,y) + k(x,y)$ for some $g(x,y), k(x,y) \in \mathbb{C}[x,y]$. Evaluating at $1$, we find $p(1,y) = k(1,y) = \alpha(y)$ for some $\alpha(y) \in \mathbb{C}[y]$. Then $p(1,0) = \alpha(0) = 0$, so in particular, $y \mid \alpha(y) = k(1,y)$, so $y \mid k(x,y)$. Hence, $p(x,y) \in \langle x-1, y \rangle$. 
A: @AWertheim: $\mathbb{C}[x, y]$ is not an Euclidean domain. It's not even a PID. But your answer is right. We can consider $x - 1$ as a polynomial in $x$ over the ring $\mathbb{C}[y]$, and since it is a monic polynomial (in $x$), we can use division algorithm. The same holds for $y$. (since I don't have enough reputation point, I can't add comments. that's why I'm forced to write this as a separate answer. sorry.)
A: There is no need at all to compute with polynomials.
Since $x-1$ and $y$ lie in the kernel of $\phi$, it lifts to a homomorphism $\psi : \mathbb{C}[x,y]/(x-1,y) \to \mathbb{C}$. It is bijective, i.e. the kernel of $\phi$ is precisely $(x-1,y)$, since we can construct an inverse homomorphism: Every $\mathbb{C}$-algebra $A$ has a unique homomorphism $\mathbb{C} \to A$.  Applying this to $A=\mathbb{C}[x,y]/(x-1,y)$, we obtain a homomorphism $\alpha : \mathbb{C} \to A$ such that $\psi \alpha = \mathrm{id}$ (because of uniqueness). In order to check $\alpha \psi = \mathrm{id}_A$, it suffices to check this on the two algebra generators $[x]$ and $[y]$ of $A$. But $\alpha(\psi([x]))=\alpha(1)=[1]=[x]$ and $\alpha(\psi([y]))=\alpha(0)=[0]=[y]$.
You should think of this as follows: $\mathbb{C}[x,y]$ is the universal $\mathbb{C}$-algebra on two generators $x,y$. Taking the quotient by $(x-1,y)$ means that we enforce $x=1$ and $y=0$ (and nothing more), so that the generators become redundant and the quotient reduces to the free $\mathbb{C}$-algebra on $0$ generators, i.e. $\mathbb{C}$. No computation is needed. Now evaluate $\mathbb{C}[x,y]/(x^2-y)$ or $\mathbb{C}[x,y,z]/(xy-z,y-1)$ without computation.
