# 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. Add to solve later

## 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. Add to solve later

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