Answer to Question #306748 in Abstract Algebra for Chinmoy kumar Bera

Question

let i=<x,2> and=<x,3> be ideal in z[x] .prove that IJ=<x,6>

Answer 1

Let us prove it by double inclusion.

First of all, $3x\in IJ, 2x\in IJ \text{ so } 3x-2x=x\in IJ$. Also, $3\cdot 2 = 6\in IJ$, so we have $<x, 6>\subseteq IJ$.

Now, let us prove the other inclusion. An element of $I$ is of a form $x\cdot P(x) + 2a_0$ and an element of $J$ is of a form $x \cdot Q(x) + 3b_0$. Therefore, an element of $IJ$ is of a form $x\cdot (xP(x)Q(x)+2a_0 Q(x)+3b_0P(x)) + 6a_0b_0$. We see, that it is an element of $<x,6>$, so we conclude that

$IJ = <x,6>$