# The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic

## Problem 494

Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.

## Proof.

We give three proofs.
The first two proofs use only the properties of ring homomorphism.

The third proof resort to the units of rings.

If you are familiar with units of $\Z[x]$, then the third proof might be concise and easy to follow.

### The First Proof

Assume on the contrary that the rings $\Z[x]$ and $\Q[x]$ are isomorphic.
Let
$\phi:\Q[x] \to \Z[x]$ be an isomorphism.

The polynomial $x$ in $\Q[x]$ is mapped to the polynomial $\phi(x)\in \Z[x]$.

Note that $\frac{x}{2^n}$ is an element in $\Q[x]$ for any positive integer $n$.
Thus we have
\begin{align*}
\phi(x)&=\phi(2^n\cdot \frac{x}{2^n})\\
&=2^n\phi\left(\frac{x}{2^n}\right)
\end{align*}
since $\phi$ is a homomorphism.

As $\phi$ is injective, the polynomial $\phi(\frac{x}{2^n})\neq 0$.
Since $\phi(\frac{x}{2^n})$ is a nonzero polynomial with integer coefficients, the absolute values of the nonzero coefficients of $2^n\phi(\frac{x}{2^n})$ is at least $2^n$.

However, since this is true for any positive integer, the coefficients of the polynomial $\phi(x)=2^n\phi(\frac{x}{2^n})$ is arbitrarily large, which is impossible.
Thus, there is no isomorphism between $\Q[x]$ and $\Z[x]$.

### The Second Proof

Seeking a contradiction, assume that we have an isomorphism
$\phi:\Q[x] \to \Z[x].$

Since $\phi$ is a ring homomorphism, we have $\phi(1)=1$.
Then we have
\begin{align*}
1&=\phi(1)=\phi \left(2\cdot \frac{1}{2}\right)\\
&=2\phi\left( \frac{1}{2} \right)
\end{align*}
since $\phi$ is a homomorphism.

Since $\phi\left( \frac{1}{2} \right)\in \Z[x]$, we write
$\phi\left( \frac{1}{2} \right)=a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0,$ for some integers $a_0, \dots, a_n$.

Since $2\phi\left( \frac{1}{2} \right)=1$, it follows that
$2a_n=0, \dots, 2a_1=0, 2a_0=1.$ Since $a_0$ is an integer, this is a contradiction.
Thus, such an isomorphism does not exists.
Hence $\Q[x]$ and $\Z[x]$ are not isomorphic.

### The Third Proof

Note that in general the units of the polynomial ring $R[x]$ over an integral domain $R$ is the units $R^{\times}$ of $R$.

Since $\Z$ and $\Q$ are both integral domain, the units are
$\Z[x]^{\times}=\Z^{\times}=\{\pm 1\} \text{ and } \Q[x]^{\times}=\Q^{\times}=\Q\setminus \{0\}.$ Since every ring isomorphism maps units to units, if two rings are isomorphic then the number of units must be the same.

As seen above, $\Z[x]$ contains only two units although $\Q[x]$ contains infinitely many units.
Thus, they cannot be isomorphic.

• The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain Let $R$ be a commutative ring with $1$. Prove that the principal ideal $(x)$ generated by the element $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain. Prove also that the ideal $(x)$ is a maximal ideal if and only if $R$ is a […]
• Rings $2\Z$ and $3\Z$ are Not Isomorphic Prove that the rings $2\Z$ and $3\Z$ are not isomorphic.   Definition of a ring homomorphism. Let $R$ and $S$ be rings. A homomorphism is a map $f:R\to S$ satisfying $f(a+b)=f(a)+f(b)$ for all $a, b \in R$, and $f(ab)=f(a)f(b)$ for all $a, b \in R$. A […]
• Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$ 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.   Proof. Define the map $\psi: R[x,y] \to […] • Nilpotent Element a in a Ring and Unit Element$1-ab$Let$R$be a commutative ring with$1 \neq 0$. An element$a\in R$is called nilpotent if$a^n=0$for some positive integer$n$. Then prove that if$a$is a nilpotent element of$R$, then$1-ab$is a unit for all$b \in R$. We give two proofs. Proof 1. Since$a$[…] • Ring of Gaussian Integers and Determine its Unit Elements Denote by$i$the square root of$-1$. Let $R=\Z[i]=\{a+ib \mid a, b \in \Z \}$ be the ring of Gaussian integers. We define the norm$N:\Z[i] \to \Z$by sending$\alpha=a+ib$to $N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.$ Here$\bar{\alpha}$is the complex conjugate of […] • A ring is Local if and only if the set of Non-Units is an Ideal A ring is called local if it has a unique maximal ideal. (a) Prove that a ring$R$with$1$is local if and only if the set of non-unit elements of$R$is an ideal of$R$. (b) Let$R$be a ring with$1$and suppose that$M$is a maximal ideal of$R$. Prove that if every […] • Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal Let$R$be a commutative ring. Then prove that$R$is a field if and only if$\{0\}$is a maximal ideal of$R$. Proof.$(\implies)$: If$R$is a field, then$\{0\}$is a maximal ideal Suppose that$R$is a field and let$I$be a non zero ideal: $\{0\} […] • 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 […] #### You may also like... ###### More in Ring theory ##### 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...