Ideal $I=({ X }^{ 2 },2X)$ of $\mathbb Z[X]$ generated by ${ X }^{ 2}$ and $2X$ is not primary.

Question

Let $I=({ X }^{ 2 },2X)$ ideal of $\mathbb Z[X]$ generated by ${ X }^{ 2}$ and $2X$. Show that $I$ is not primary.

I tried to find $$\sqrt { I } =\sqrt { ({ X }^{ 2 })+(2X) } =\sqrt { \sqrt { { (X }^{ 2 }) } +\sqrt { (2X) } } =\sqrt { \sqrt { { (X) }^{ 2 } } +\sqrt { (2X) } } =\sqrt { (X)+(2X) } =\sqrt { (X) } =(X).$$ It is wrong?

Answer 1

There is no need to find $\sqrt I$ in order to show that $I$ is not primary. We have $2X\in I$, but $X\notin I$ and $2\notin\sqrt I$.

Answer 2

This is basically the same answer as above, but it sometimes help to think of primary ideals this way:

An ideal $I$ of $\mathbb{Z}[x]$ is primary if all zero divisors in $R=\mathbb{Z}[x]/I$ are nilpotent. Let $\overline{2},\overline{x}$ be the images of $2$ and $x$, respectively, in $R$. Then we have that $\overline{2}\overline{x}=0$, but $\overline{2}^n\neq{}0$ for any $n\geq{}0$.

Therefore we have a zero-divisor that is not nilpotent, which means $I$ is not primary.