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.

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
\begin{align*}
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).
\end{align*}
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
\begin{align*}
\psi\left(\, f(x,y) \,\right)&=\psi\left(\, xg(x,y) \,\right)=0g(0,y)=0.
\end{align*}
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
\[R[x,y]/(x)\cong R[y].\]

A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]

The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$.
Let
\[R=\left\{\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\]
Then the usual matrix addition and multiplication make $R$ an ring.
Let
\[J=\left\{\, \begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix} […]

$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]

Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]

Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$
Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\]
be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]
Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?
Solution.
We […]

Generators of the Augmentation Ideal in a Group Ring
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]
where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […]

Ring Homomorphisms and Radical Ideals
Let $R$ and $R'$ be commutative rings and let $f:R\to R'$ be a ring homomorphism.
Let $I$ and $I'$ be ideals of $R$ and $R'$, respectively.
(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.
(b) Prove that $\sqrt{f^{-1}(I')}=f^{-1}(\sqrt{I'})$
(c) Suppose that $f$ is […]

[…] 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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[…] 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 $R[x,y]/(x) cong R]y]$.) […]