Prove that there is a nontrivial ring homomorphism from $\mathbb{Q}[x,y]/(x,y)^2$ to $\mathbb{C}$.

Question

I am currently studying for my qualifying exam in algebra. I am thinking about an old question that asks you to look at $R = \mathbb{Q}[x,y]/I$ where $I=(x,y)^2$. It asks for the following three things:

1. Find all ideals in $R$.
2. Describe the units in $R$.
3. Show that there is a nontrivial ring homomorphism from $R$ to $\mathbb{C}$.

Here is a screenshot of the question in its original form:

Usually I would proceed using the correspondence theorem for part 1. I am thinking that the only ideals of $\mathbb{Q}[x,y]$ containg $I$ are $(x,y)$, $(x^2,y)$ and $(x,y^2)$. However, I don't know how to justify that these are the only ideals containing $I$.

The units in $R$ are $\{ a + bx + cy : a,b,c \in \mathbb{Q} \}$. We should be able to equate coefficients in $$ 1 = (a + bx + cy)(a' + b'x + c'y)$$ We get the system: \begin{align*} 1 &= aa' \\ 0 &= ab' + a'b \\ 0 &= ac' + a'c \end{align*} since $a'$ is completely determined by $a$, so are $b'$ and $c'$ determined by the given $b$ and $c$ and the existence of $a'$. So there elements are all invertible. So every element with a constant term in $R/I$ is invertible. Moreover, any element that doesn't have a constant term is not invertible (this is clear). So we have found all the units.

It is still unclear to me how to recover $\mathbb{C}$ from taking a quotient. Since we have found all the ideals in $R$, one of the ideals listed in part 1 (taking the quotient with $I$) must be precisely the kernel of the map that $R$ to $\mathbb{C}$.

Answer 1

There is only one homomorphism. You need to send $x$ to an element of $\mathbb C$ that squares to zero, but the only such element is $0$, and similarly for $y$, and so all that is left is to determine the image of $\mathbb Q$. However, since $1$ is sent to $1$, the requirements of a homomorphism guarantee that $n$ is sent to $n$, and then $p/q$ is sent to $p/q$. Therefore, $\mathbb Q$ is sent to the copy of $\mathbb Q$ that we normally think of sitting inside the complex numbers, $x$ and $y$ are both sent to zero, and this actually determines a well defined homomorphism.

Answer 2

Regarding the question in the title, which curiously does not appear in the body of the post, it's obvious that no, $\Bbb Q[x,y]/J$ is not isomorphic to $\Bbb C$. For example $\Bbb Q[x,y]$ is countable, hence $\Bbb Q[x,y]/J$ is countable.