Factor ring

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April 13th 2009, 02:05 PM
Coda202

Factor ring

Let I = <x^2 + 5x +6> = {(x^2 + 5x+6) *f| f in Q[x]} Does the factor ring Q[x]/I have any zero divisors?
Clearly, x^2 +5x+6 is not irreducible as it equals (x+2)(x+3), thus Q[x]/I is not a field, however I am having difficulty showing whether or not it is an integral domain.
April 13th 2009, 02:12 PM
NonCommAlg

Quote:

Originally Posted by Coda202

Let I = <x^2 + 5x +6> = {(x^2 + 5x+6) *f| f in Q[x]} Does the factor ring Q[x]/I have any zero divisors?
Clearly, x^2 +5x+6 is not irreducible as it equals (x+2)(x+3), thus Q[x]/I is not a field, however I am having difficulty showing whether or not it is an integral domain.

$(x+2 +I)(x+3 + I)=0$ and $x+2 +I \neq 0, \ x+3 + I \neq 0.$ so $x+2+I$ and $x+3 +I$ are zero divisors.
April 14th 2009, 11:16 AM
ThePerfectHacker

Quote:

Originally Posted by Coda202

Let I = <x^2 + 5x +6> = {(x^2 + 5x+6) *f| f in Q[x]} Does the factor ring Q[x]/I have any zero divisors?
Clearly, x^2 +5x+6 is not irreducible as it equals (x+2)(x+3), thus Q[x]/I is not a field, however I am having difficulty showing whether or not it is an integral domain.

If $F_1,F_2$ are fields then $F_1\times F_2$ has zero divisors. In fact, all zero divisors are given by: $(a,0),(0,b)$ where $a\in F_1^{\times}, b\in F_2^{\times}$ .

Notice that $x^2 + 5x+6 = (x+2)(x+3)$ . Thus, by Chinese remainder theorem:
$\mathbb{Q}[x]/(x^2+5x+6) \simeq \mathbb{Q}[x]/(x+2) \times \mathbb{Q}[x]/(x+3)\simeq \mathbb{Q}\times \mathbb{Q}$
As you can see the RHS has many many zero divisors.