Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$

Problems and solutions of ring theory in abstract algebra

Problem 517

Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.

Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.

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Define the map $\psi: R[x,y] \to R[y]$ by sending $f(x,y)\in R[x,y]$ to $f(0,y)$.
Namely, the map $\psi$ is the substitution $x=0$.
It is straightforward to check that $\psi$ is a ring homomorphism.

For any polynomial $g(y)\in R[y]$, let $G(x, y)=g(y)\in R[x,y]$.
Then we have $\psi(G(x,y))=G(0,y)=g(y)$.
This proves that $\psi$ is surjective.

We claim that the kernel of $\psi$ is the ideal $(x)$.

Suppose that $f(x,y) \in \ker(\psi)$.
We write
\[f(x,y)=f_0(y)+f_1(y)x+\cdots +f_n(y)x^n,\] where $f_i\in R[y]$ for $i=1, \dots, n$.

Since $f(x,y)\in \ker(\psi)$, it yields that
\[0=\psi(f(x,y))=f(0,y)=f_0(y).\] Hence
f(x,y)&=f_1(y)x+\cdots +f_n(y)x^n\\
&=x\left(f_1(y)+\cdots +f_n(y)x^{n-1}\right) \in (x).
Thus, $\ker(\psi) \subset (x)$.

On the other hand, suppose $f(x,y)\in (x)$.
Then there exists $g(x,y) \in R[x,y]$ such that
\[f(x,y)=xg(x,y).\] It follows that
\psi\left(\, f(x,y) \,\right)&=\psi\left(\, xg(x,y) \,\right)=0g(0,y)=0.
It implies that $f(x,y) \in \ker(\psi)$, hence $\ker(\psi) \subset (x)$.

Putting two inclusions together gives $(x)=\ker(\psi)$.

In summary, the map $\psi:R[x,y] \to R[y]$ is a surjective ring homomorphism with kernel $(x)$.

Hence by the isomorphism theorem, we obtain the isomorphism
\[R[x,y]/(x)\cong R[y].\]

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1 Response

  1. 07/26/2017

    […] The third example is the ring of polynomials in two variables $R=Q[x, y]$ over $Q$ and the principal ideal $I=(x)$ generated by $x$. The quotient ring $Q[x,y]/(x)$ is isomorphic to $Q[y]$. (The proof of this isomorphism is given in the post Prove the Ring Isomorphism $R[x,y]/(x) cong R]y]$.) […]

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