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.

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Comments

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

    I don`t know if I`m 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.

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