# Polynomial without roots in a ring

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 $B$.

• 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.) – Jeremy Rouse Jun 28 '14 at 16:50
• 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. – 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 $B$.