Prove that the elements $2x$ and $x^2$ have no LCM in the ring of integral polynomials with even coefficient of $x$

Question

Let $A$ be the subring of $\Bbb Z[x]$ consisting of all polynomials with even coefficient of $x$. Prove that the elements $2x$ and $x^2$ have no lowest common multiple.

Hints please!

Answer 1

Going by the definition of LCM in commutative rings, if $M$ is a least common multiple and $N$ is any common multiple, then $M$ must divide $N$. Thus $p(x)=LCM(2x,x^2)$ must divide the product $2x\cdot x^2=2x^3$. But $4x^2$ is also a common multiple of $2x$ and $x^2$ in $A$. So $p(x)$ divides $4x^2$. As $4x^2$ does not divide $2x^3$ in $A$, $p(x)$ must be a proper divisor of $4x^2$ that still divides $2x^3$. Thus either $p(x)=x^2$ or $p(x)=2x$. But as each fails to divide the other, there is no such $p(x)$.