Let $A$ be a commutative ring which is not an integral domain. I try to find a polynomial $P$ of $A[X]$ such that $d°P = 1$ and $P$ admits no root in any ring $B$ such that $A$ is a subring of $B$.

  • 2
    $\begingroup$ Why are you interested in this? If this is related to your research, then how did this arise? (If you're interested in this only because it is a homework problem, it would be better to ask on math.SE.) $\endgroup$ Jun 28 '14 at 16:50
  • 1
    $\begingroup$ I am reading a course on Galois Theory in which a note is asserting that such a polynomial does exist. Even if it is considered as being evident, I couldn't find it. $\endgroup$
    – Gaussian
    Jun 28 '14 at 16:58

Take a ring $A$ that is not an integral domain, and let $d\in A$ be a zero-divisor. Consider the polynomial $dx-1$. Since $d$ is a zero-divisor in $A$, it is a zero-divisor in every ring $B$ containing $A$ as a subring. In particular, $d$ is not a unit in any such $B$. Therefore, the polynomial $dx-1$ cannot have a root in any $B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.