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How can I show that $\mathbb{Q}[x,y]/(y)$ is isomorphic to $\mathbb{Q}[x]$ as rings?


I know that first isomorphism theorem is required here but I could not find the proper homomorphism.

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$\mathbb Q[x,y]\to\mathbb Q[x]$, $f(x,y)\mapsto f(x,0)$ is a homomorphism that is onto and has precisely $(y)$ as kernel.

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And to see what this map does in a slightly different way: note that $(y)$ consists of all polynomials with the shape $y p(x,y)$. When we reduce an element $f(x,y) \in \mathbb{Q}[x,y]$ modulo $(y)$, we're free to throw away any part of it that's in $(y)$. So any term in $f$ with a $y$ in front of it is redundant in the quotient. Thus we end up with a unique representative that only has $x$ terms, and moreover there is essentially no other restriction. The most succinct way to do this is as rendered in the other answer.

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