## ZZ[x] and QQ[x] are not isomorphic

Prove that the rings $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.

We begin with some lemmas.

Lemma 1: Let $R$ and $S$ be rings with 1. If $\varphi : R \rightarrow S$ is a surjective ring homomorphism, then $\varphi(1_R) = 1_S$. Proof: Let $s \in S$. Now $s = \varphi(r)$ for some $r \in R$. Note that $\varphi(1_R) \cdot s = \varphi(1_R \cdot r)$ $= \varphi(r)$ $= s$, and likewise $s \varphi(1_R) = s$. Since the 1 in $S$ is unique, we have $\varphi(1_R) = 1_S$. $\square$

Lemma 2: Let $R$ and $S$ be rings with 1. If $\varphi : R \rightarrow S$ is a ring homomorphism such that $\varphi(1) = 1$, then $\varphi[R^\times] \subseteq S^\times$. Proof: If $u \in R^\times$ is a unit, we have $uv = 1$ for some $v \in R$. Then $\varphi(u) \varphi(v) = 1$, so that $\varphi(u)$ is a unit. $\square$

Lemma 3: Let $R$ and $S$ be rings with 1. If $\varphi : R \rightarrow S$ is a surjective ring homomorphism, then $\varphi[R^\times] = S^\times$. Proof: The $(\subseteq)$ direction follows from the previous lemma. Now let $w \in S^\times$. Since $\varphi$ is surjective, $w = \varphi(u)$ for some $u \in R$. Note that $1_S = wz$ for some $z \in S^\times$. Then $1_R = u \varphi^{-1}(z)$, and thus $u \in R^\times$. $\square$

Lemma 4: Let $R$ and $S$ be rings with 1. If $\varphi : R \rightarrow S$ is a ring isomorphism, then $\varphi|_{R^\times} : R^\times \rightarrow S^\times$ is a multiplicative group isomorphism. Proof: By the previous lemma, $\varphi|_{R^\times}$ indeed maps $R^\times$ surjectively to $S^\times$. Moreover, as the restriction of an injective function, $\varphi|_{R^\times}$ is injective. Finally, $\varphi|_{R^\times}$ is a multiplicative homomorphism since $\varphi$ is. $\square$

Note that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are rings with 1. If $\varphi : \mathbb{Z}[x] \rightarrow \mathbb{Q}[x]$ is a ring isomorphism, then the restriction $\psi$ of $\varphi$ to the units is a group isomorphism $\psi : \mathbb{Z}[x]^\times \rightarrow \mathbb{Q}[x]^\times$. However, note that by Proposition 4, $\mathbb{Z}[x]$ has 2 units while $\mathbb{Q}[x]$ has infinitely many units; thus we have a contradiction.

• Danny  On November 30, 2011 at 5:26 pm

I dont know if Im just confused or not. In the second lemma, you proved that a unit element goes to a unit element under ring homomorphism but you stated the statement of the lemme and the image of R is contained in s , is that wrong? I mean I don`t see any connection between the statement of the lemma and what you proved!! Am I missing something?

• nbloomf  On November 30, 2011 at 6:58 pm

By $R^\times$ I mean the set of units in $R$. So the line $\varphi[R^\times] \subseteq S^\times$ just means that units map to units under $\varphi$.

• Gobi Ree  On January 19, 2012 at 1:56 am

Another proof: In $\mathbb{Q}[x]$, $f(x)+f(x)=g(x)$ has a solution for all $g(x)$, namely $f(x)= \frac{1}{2} g(x)$. But in $\mathbb{Z}[x]$, $f(x)+f(x)=1$ has no solution.

• nbloomf  On January 19, 2012 at 10:02 am

Nice!