Kernel of the following Substitution Homomorphism over $\mathbb{C}[x]$ I'm hoping someone could review my proof for accuracy, thanks! There is another proof of this on stack exchange, but it uses quotient rings, which we haven't learned yet.  
Problem:
Let $\phi: \mathbb{C}[x,y]$ -> $\mathbb{C}[t]$, a the homomorphism that sends x -> $t + 1$ and $ y$ ->  $t^3 - 1 $. Determine the kernel $K$ of $\phi$, and prove that every ideal of $\mathbb{C}[x,y]$ that contains $K$ can be generated by two elements. 
Proof: 
Note that $f = (x-1)^3 - 1 - y \in K$. Suppose there is some $g \in K$ such that $f$ does not divide $g$ in $\mathbb{C}[x,y]$. Then we have that $f$ is monic of degree $1$ when considered as a polynomial in the variable $y$.
Then we have $g = fq + r$, where deg(r) $\lt$ 1 (with respect to the variable $y$). 
This implies that $r(x)$ is either a polynomial in $\mathbb{C}[x]$ that is non constant or is the $0$ polynomial. If it is nonconstant in $x$ then since r$\in$ K we must have that $x+1$ is a root of r. 
Part I think is wrong: But this implies that the polynomial $r$ has infinite roots, as $x$ is a variable that can range over all of $\mathbb{C}$ for example. 
Part I think may work better: Alternatively, plugging in x -> $x + 1$ will not send a non zero polynomial to zero as we are working in an integral domain and the substitution will maintain the degree of the polynomial as distributing $x+1$ will just produce extra terms of lower degree. 
Hence we have r(x) must be the zero polynomial and so f generates all of K and so K is principle and $K$ = $ ( (x-1)^3 - 1 - y )$. 
Then Let $I$ be an ideal that contains K. If I = (1), then I contains K and is generated by 1 element. Suppose $I$ is a proper ideal then.
New attempt: 
Then for any $g$ $\in$ $I$ we have $g = fq + r$ when dividing over $y$. Then deg(r) $\lt$ $1$ and so it is either a constant or a polynomial over $x$, call it $r(x)$. Since $r(x)$ $\in$ $\mathbb{C}[x]$ then the set of all remainders is an ideal and is principle as it is polynomials in one variable over a field.  
Though I’m not sure if we can claim the set of remainders is an ideal. It is certainly a subset of $I$, but I’m not sure if it is an ideal itself. 
______________below this  is wrong__________
Then let $g$ $\in$ I and suppose $f$ does not divide $g$. Then we can similarly divide $g$ by $f$ and conclude the remainder, $r(x)$, must be degree 0 in the variable y which implies it may be of non zero degree of $x$. Indeed it may be as considering $f$ as a function in the variable $x$ now we have that it is monic of degree 3. Then $g = fq + r$ is valid as an equation in the variable $y$, and the variable $x$. So we are dealing with the same $r(x)$ value as previously, and it must satisfy the additional constraint that the degree of $r(x)$ is less than $3$. 
Hence any g $\notin$ K for g$\in$ I is of the form $g = fq + r$, where $r(x)$ $\in$ $\mathbb{C}[x]$. Then $r(x)$ $\in$ I also, as $I$ is closed under addition. Then the the lowest degree of $r$ is $0$, $1$ or $2$. $0$ implies $I$ is the whole ring. $1$ or $2$ implies we can generate $I$ with either $x$ or $x$$^2$ as $\mathbb{C}[x]$ is a ring with only ideals that are principle that are generated by lowest degree monic polynomials, and so $x$ or x$^2$ would suffice in that case. 
In any case we have I =  or , or <1>. 
Hence the claim is shown.  
 A: My review:
Overall, this is a good proof. The idea is mostly correct, though there are a couple problems.
Part you think is wrong vs. Part you think is better:
The part you think is wrong is actually correct, but could be phrased better.
You have that $r(x)\in \newcommand\CC{\mathbb{C}}\CC[x]$ has the property that $r(t+1)=0\in\CC[t]$. Therefore, for all $\alpha\in\CC$, $r(\alpha+1)=0$, so $r$ has infinitely many roots in $\CC$, and must be the zero polynomial.
The part you think is better is also correct though, except you should say $x\mapsto t+1$. Don't mess up the variables, that will confuse your readers.
Your proof that $K=(f)$ is good.
Proving that if $K\subseteq I$, then $I$ is generated by at most two elements.
The idea for this part starts well, but quickly becomes confusing.
The first problem is how you conclude $r$ has degree less than $3$ in $x$. You appear to be dividing by $f$ twice. However the second division by $f$ might cause the degree of $y$ to go up. Thus you can't conclude $\deg r < 3$.
In addition, the degree of $r$ being $1$ or $2$ would not imply that $I = (f,x)$ or $(f,x^2)$. The ideals $(f,x)$ and $(f,x+1)$ are different!
Also it's not clear how you're concluding that there cannot be a third generator. You don't explicitly say that this is not possible. You implicitly conclude this from your attempted characterization of the ideals containing $K$, but you haven't proved that these are the only possible ideals.
How to fix this part
Let $I$ be an ideal containing $K$. $I$ is finitely generated. 
Let $g_1,g_2,\ldots,g_n$ be the fewest possible elements of $I$ such that 
$I$ is generated by $f$, $g_1$, $\ldots$, $g_n$. We will prove that $n=1$.
If $n\ge 2$, then it suffices to prove that there exists a polynomial 
$g_1'$ such that $(f,g_1,g_2)=(f,g_1')$. Then $I=(f,g_1,\ldots,g_n)=(f,g_1',g_3,g_4,\ldots,g_n)$, contradicting the minimality of the collection of generators.
Now by division, $g_1=fq_1+r_1(x)$ and $g_2=fq_2+r_2(x)$. Therefore 
$(f,g_1,g_2)=(f,r_1,r_2)$. Now $r_1$ and $r_2$ are single variable polynomials, 
so if $d=\gcd(r_1,r_2)$, we know $(r_1,r_2)=(d)$. Thus
$(f,g_1,g_2)=(f,d)$. Taking $g_1'=d$, this completes the proof.
Indeed, from the proof, we can see that every ideal 
$I$ containing $K$ is either $(f)=K$, or of the form $(f,g)$, where $g$ has zero $y$-degree.
