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My argument goes as follows:

Observe that $I = \langle x,y \rangle$ is exactly the set of polynomials with no constant term.

Then for any $p(x) \in \mathbb{Q}[x,y]$, we can write $p(x) = p'(x) + p_0$, for some $p'(x) \in I$ and $p_0 \in \mathbb{Q}$. Then in the quotient space $\mathbb{Q}[x,y]/I$, we have $\overline{p(x)} = \overline{p_0}$. That is, every coset in $\mathbb{Q}[x,y] / I$ can be written as $\overline{q}$ for some $q \in \mathbb{Q}$ ...


So at this point, it's "obvious", but I don't know how to concisely finish the argument without setting up a map $\phi: \mathbb{Q} \rightarrow \mathbb{Q}[x,y]/I$ by $\phi(q) = \overline{q}$ and showing that its bijective, which just feels distracting and long for such a relatively simple point.

Is there a more elegant way to make this argument?

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  • $\begingroup$ The argument seems good to me. We need to add that the constant term of a sum is the sum of the constant terms, ditto for product. One might as well be explicit about the isomorphism. $\endgroup$ – André Nicolas Feb 17 '16 at 0:08
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Define $\phi : Q[x,y] \to \mathbb{Q}$ by $\phi(f(x,y)) = f(0,0)$. We see that $\phi$ is surjective by taking $f(x,y) = c$ for any $c \in \mathbb{Q}$. We also have $\ker\phi = \left<x,y\right>$. Then by the first isomorphism theorem, $$\frac{\mathbb{Q}[x,y]}{\left<x,y\right>} \cong \mathbb{Q}.$$

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  • $\begingroup$ Ah, I really like this. It sort of "cuts out the middleman" by setting up a smart map directly to the target space. Thanks! $\endgroup$ – cemulate Feb 17 '16 at 0:14
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    $\begingroup$ No problem! The first isomorphism theroem is usually very hand any time you want to show what a quotient is isomorphic to. $\endgroup$ – Ethan Alwaise Feb 17 '16 at 0:15

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