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

Question #306748

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


1
Expert's answer
2022-03-07T17:23:02-0500

Let us prove it by double inclusion.

First of all, 3xIJ,2xIJ so 3x2x=xIJ3x\in IJ, 2x\in IJ \text{ so } 3x-2x=x\in IJ. Also, 32=6IJ3\cdot 2 = 6\in IJ, so we have <x,6>IJ<x, 6>\subseteq IJ.

Now, let us prove the other inclusion. An element of II is of a form xP(x)+2a0x\cdot P(x) + 2a_0 and an element of JJ is of a form xQ(x)+3b0x \cdot Q(x) + 3b_0. Therefore, an element of IJIJ is of a form x(xP(x)Q(x)+2a0Q(x)+3b0P(x))+6a0b0x\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><x,6>, so we conclude that

IJ=<x,6>IJ = <x,6>


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment