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Let $A$ be a commutative ring, and $p\in A[X]$ a polynomial of degree $d>0$. If $A$ is an integral domain, we can find a ring $B$ such that $A\subseteq B$ and $p$ has a root in $B$. For example take $B$ an algebraic closure of the field of fractions of $A$.

Now, if $A$ is not an integral domain, can we find a ring $B$ such that $A\subseteq B$ and $p$ has a root in $B$?

H. Jackson
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    The claim holds if $p$ is monic (the usual proof works). – user26857 Jun 04 '15 at 11:33
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    In the usual proof $B=A[X]/(p)$, and the root is the residue class of $X$ modulo the ideal generated by $p$. In this case the canonical ring homomorphism $A\to B$ is injective. – user26857 Jun 04 '15 at 11:43
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    Irreducibility of polynomials is only defined in a domain. – Eoin Jun 04 '15 at 23:28

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I don't think so. Let $A=\mathbb Z_6$, and $p(X)=2X+1$. Suppose there is $B\supset A$ which contains a root of $p$, say $b$. We have $2b+1=0$ in $B$, so $3=0$ in $B$, and therefore $3=0$ in $A$, a contradiction.

user26857
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