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Prove that there is no surjective homomorphism from $\mathbb{Z}[X]$ to $\mathbb{Q}$.

Attempt: If possible let $\phi(x) :\mathbb{Z}[X] \to \mathbb{Q} $ be an onto homomorphism, then it is easy to prove that $\phi$ fixes $\mathbb{Z}$, if we determine where $x$ goes to then we end up determining the homomophism $\phi$.

Say $x\to \frac{p}{q}$, then any element in the image of the homomorphism can be written as $a_0+a_1\frac{p}{q}+a_2(\frac{p}{q})^2+ \dots+a_n(\frac{p}{q})^n$.

I am unable to proceed further

spaceman_spiff
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1 Answers1

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Continuing from where you left of:

So every element in the image is of the form $$\frac{q^n a_0 + q^{n-1} a_1 p + \dots + a_n p^n}{q^n}.$$ So, elements of ${\mathbb Q}$ that have a prime factor in their denominator that does not divide $q$, are not in the image.

Magdiragdag
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  • thank for the answer.... so consider $\frac{1}{r}$, such that (q,r) = 1.Now we can deduce that $\frac{q^n}{r} = integer$, from the above assumption, which is absurd. – spaceman_spiff Apr 13 '17 at 08:57