Sachchidanand Prasad

29-11-2023

Problem: Let $I=(x,2)$ and $J=(x,3)$ be ideal in the ring $\mathbb{Z} [x]$.
  • Prove that $IJ = (x,6)$.
  • Prove that the element $x \in IJ$ cannot be written as $x = f(x) g(x)$, where $f(x) \in I$ and $g(x) \in J$.
Solution: Recall that if $I=(a_1,\dots, a_m)$ and $J=(b_1, \dots, b_n)$ are ideals in a commutative ring, then we have \[ IJ=(a_ib_j), \] where $1\leq i \leq m$ and $1\leq j \leq n$.
  • Using the above fact, we have \[ IJ = \left( x^2, 2x, 3x, 6 \right). \] Since $2x, 3x \in IJ$, so $3x - 2x = x \in IJ$. Thus, the first three generators can be generated by $x$ and hence, \[ IJ = (x, 6). \]
  • Since $\deg (fg) = \deg (f) + \deg (g)$, so if $x = f(x) g(x)$, then either of the function has to be constant. If $f(x) $ is constant, then $f \equiv 2$. Thus, $x = 2g(x)$ which implies $g(x) = 2^{-1} x$. But $2$ is not invertible in $\mathbb{Z} $ and hence $2^{-1} x \notin \mathbb{Z} [x]$, therefore $f \equiv 2$ is not possible. Similarly, if $g(x)$ is constant then $g\equiv 3$ which is also not possible as $3$ is not invertible in $\mathbb{Z} $. Thus $x = f(x) g(x)$ is not possible for any $f(x) \in I$ and $g(x) \in J$.

    We can also solve this by using that $\mathbb{Z} [x]$ is UFD. So, $x = (\pm 1)(\pm x) $. Thus, $f(x) = \pm 1$ and $g(x) = \pm x$ or $f(x) = \pm x$ and $g(x) \pm 1$. In any case, either $1 \in I$ or $1 \in J$ which implies either $I = \mathbb{Z} [x]$ or $J = \mathbb{Z} [x]$ which is not true. Thus, we can not write $x$ as a product of elements from $I$ and $J$.