This is a question from an (old) exam at my university on Rings and Modules. It's not part of any ongoing assignment.
We are given a homomorphism $\phi: \mathbb{Z}[X,Y] \rightarrow \mathbb{Z}[T]$, defined by $\phi(X) = T^2, \phi(Y) = T^5$.
The first part of the question is whether $\phi$ is surjective, and it's easy to show it's not: since any monic monomial in $\mathbb{Z}[X,Y]$ is of the form $X^iY^j$, and $\phi (X^iY^j) = T^{2i+5j}$, so for example $T^3 \in \mathbb{Z}[T]$ cannot be an element of the Im$(\phi)$.
The second part of the question is where I'm having doubts: It says to "Find a simple description of the ideal of the kernel of $\phi$."
My approach:I try to find an iff condition for an element of $\mathbb{Z}[X,Y]$ being an element of the kernel. First I note that any $f \in \mathbb{Z}[X,Y]$ may be written as $\sum_{i,j = 0}^n a_{i,j}X^iY^j$, where $a_{i,j} \in \mathbb{Z}$, and $i,j$ range independently over $n \in \mathbb{N}$. Since $\phi$ is a homomorphism, we have $$\phi(f) = \sum_{i,j = 0}^n a_{i,j}\phi(X^iY^j) = \sum_{i,j = 0}^n a_{i,j}T^{2i + 5j}$$
At this point, we have $f\in \text{Ker}(\phi) \iff \sum_{i,j = 0}^n a_{i,j}T^{2i + 5j} = 0 \in \mathbb{Z}[T]$. After applying the homomorphism, the coefficients of different monic monomials in $\mathbb{Z}[X,Y]$ may become the same monic monomial in $\mathbb{Z}[T]$. For example, $\phi (X^8Y^3) = \phi(X^3Y^5) = T^{31}$.
Now I try to identify monic monomials in $\mathbb{Z}[X,Y]$ that reduce to the same monomial in $\mathbb{Z}[T]$ under $\phi$. We note that $\phi (X^iY^j) = \phi (X^aY^b) \iff 2i+5j = 2a+5b$. For some fixed $i,j$, consider the solution set in $a,b$, that is the solution set $\{(a,b): 2i+5j = 2a+5b\} = \{(a,b): 2(i-a)+5(j-b) = 0\}$. Now this is just a linear diophantine equation of the form $2x+5y = 0$, so we get that $a_n = i-5n$ and $b_n = j+2n$ for some $n\in \mathbb{Z}$.
Now the iff condition becomes $$f\in \text{Ker}(\phi) \iff \sum_{n \in \mathbb{Z}} a_{i-5n,j+2n} = 0$$
for every $i,j$ ranging from $0 \text{ to } n$, and such that we interpret $a_{s,t} = 0$ whenever $s$ or $t$ is negative, or whenever $a_{s,t}$ is not a coefficient of some term of $f$.
As an example, if $f = a_{3,5}X^3Y^5 + a_{8,3}X^8Y^3$, then $\phi(f) = 0 \iff a_{3,5}+a_{8,3} = 0$.
I believe the above does indeed characterise the elements of the kernel of $\phi$, but it feels very clunky and I think there must be a better way of approaching the problem. Particularly, I didn't use part (a) where $\phi$ is shown to not be surjective, and my approach doesn't seem very 'algebraic'. Is there a better answer/approach?
Best Answer
Here is what I would consider the standard "more algebraic" approach:
$X^5-Y^2$ is clearly contained in the kernel of $\phi$. That means that we can "insert" $\Bbb Z[X,Y]/\langle X^5-Y^2\rangle$ between $\Bbb Z[X,Y]$ and $\Bbb Z[T]$ like so: $$ \Bbb Z[X,Y]\to\Bbb Z[X,Y]/\langle X^5-Y^2\rangle\to \Bbb Z[T] $$ in such a way that the first homomorphism here is the canonical quotient map, and the second homomorphism (say $\phi'$) is "still" $\phi'(X+\langle X^5-Y^2\rangle)= T^2$ and $\phi'(Y+\langle X^5-Y^2\rangle)= T^5$ (I will skip the coset notation from here on out). Finally, the composition of these two maps is $\phi$. Algebraists summarize all this by saying that $\phi$ "factors through" $\Bbb Z[X,Y]/\langle X^5-Y^2\rangle$.
We want to know whether $\phi'$ is injective. Because if it is, we have found the entire kernel. To do this, we take an element in the kernel of $\phi'$, and show that it must be $0$.
How is this different from working in $\Bbb Z[X,Y]$? Note that we can make all elements of $\Bbb Z[X,Y]/\langle X^5-Y^2\rangle$ have a particular form, namely where there is no $Y$ term with higher exponent than $1$ (remember that $Y^2=X^5$ in the quotient ring). This makes things much easier.
Let $f\in\ker \phi'$ be written on that form. So we have $$ f(X,Y)=f_0(X)+Yf_1(X) $$ for some single-variable integer-coefficient polynomials $f_0, f_1$. Now we apply $\phi'$. We get $$ 0=\phi'(f)(T)=f(T^2,T^5)=f_1(T^2)+T^5f_2(T^2) $$ It isn't difficult to see that we must have $f_1=f_2=0$.