Which is maximal? Have you factored the polynomials?Mr Davis 97 said:Homework Statement
Sorry for the multiple postings. I actually solved the other problems, so I have this last one:
Are ##\mathbb{Z}_3[x] / \langle x^2 + 2x + 1 \rangle ## and ##\mathbb{Z}_3[x] / \langle x^2 + x + 2 \rangle## isomorphic
Homework Equations
The Attempt at a Solution
It seems like they wouldn't be, because one ideal is maximal while the other isn't, but I am not sure if this is correct.
##x^2 + 2x + 1 = (x+1)^2## but ##x^2 + x + 2## can't be factored because it has no roots in the field, hence it is maximalfresh_42 said:Which is maximal? Have you factored the polynomials?
Also, one more thing. How does one list out the elements of quotient rings such as these? Also, is there a general way to count the number of elements in ##fresh_42 said:Yes, you're right. I thought it was ##x^2+x-2##. So one quotient ring is a field and the other one not even an integral domain.
So in this case, for both of the quotient rings, would we have ##3^2 = 9## elements?fresh_42 said:Yes, the equivalence classes of ##{R}[x]/\langle p(x) \rangle## are all of the form ##[r(x)]_{p(x)} = r(x) + \{s(x)\cdot p(x)\,\vert \,s(x) \in R[x]\}## which means, one can push all monomials of degree ##m := \operatorname{deg}p## and higher into the ideal. However, it gets a little bit more complicated, if the highest coefficient isn't a unit (and I would have to do the math first). Let's therefore assume the highest coefficient is ##1## and ##p(x) = x^m +\ldots + p_1x+p_0##. Then ##[x^m]_{p(x)}= -[p_{m-1}x^{m-1}]_{p(x)} - \ldots - [p_{1}x]_{p(x)} - [p_{0}]_{p(x)}## which means all elements ##x^n## with ##n \geq m## can be reduced to lower powers. Now multiply the possible coefficients with the degree and you have the number of elements. Or better: ## \{[1]_{p(x)}, \, \ldots \,, [x^{m-1}]_{p(x)} \}## forms a basis of the ##R-## module (or vector space in case ##R## is a field). Of course if ##R## isn't finite, you will still get a finite dimension, but infinitely many elements.
I get six.Mr Davis 97 said:So in this case, for both of the quotient rings, would we have ##3^2 = 9## elements?
I'm trying to reconcile your work with comment I found on https://math.stackexchange.com/ques...nts-of-the-quotient-ring-of-a-polynomial-ring. In the comment why does he use exponentiation?fresh_42 said:I get six.
Because (for short without the annoying ##[.]_{p(x)}##) we have a basis ##\{1,x,x^2,x^3,\ldots , x^{m-1}\}##. These are all monomials of lower degree than ##m##. All higher can be reduced with subtractions by ##p(x)## to a lower one. And ##p(x)=0## in this quotient ring, so the subtractions do no harm to the ring element. This means ##m## basis vectors (or elements) and each can carry ##n## elements.Mr Davis 97 said:I'm trying to reconcile your work with comment I found on https://math.stackexchange.com/ques...nts-of-the-quotient-ring-of-a-polynomial-ring. In the comment why does he use exponentiation?
An isomorphic quotient ring is a ring that has the same structure and properties as another ring, but with a different set of elements. In other words, two quotient rings are isomorphic if there exists a ring homomorphism between them that preserves the ring structure.
To show that two quotient rings are isomorphic, you need to find a ring homomorphism between them. This can be done by defining a function between the two rings and proving that it preserves the ring structure, i.e. it preserves addition, multiplication, and the identity element. If such a function exists, then the two quotient rings are isomorphic.
The kernel of a ring homomorphism is the set of elements in the domain ring that map to the zero element in the codomain ring. If two quotient rings have the same kernel, then they are isomorphic. This is because the kernel determines the structure of the ring, and if the kernel is the same for both rings, then the ring structure must also be the same.
Yes, two quotient rings can be isomorphic even if they have different underlying sets. This is because the isomorphism is based on the structure and properties of the rings, not just the elements in them. As long as the two rings have the same operations and properties, they can be considered isomorphic.
No, not all quotient rings are isomorphic to each other. Isomorphism is a specific relationship between two rings, and not all rings will have this relationship. Two quotient rings can be isomorphic if they have the same structure and properties, but if they differ in any way, they will not be isomorphic.