Describe the ring structure of

Also, indicate how the ring structure differs from

Would appreciate help with this exercise

Peter

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- April 4th 2013, 03:05 PMBernhardPolynomial Ring Z[x]/(x^2)
Describe the ring structure of

Also, indicate how the ring structure differs from

Would appreciate help with this exercise

Peter - April 4th 2013, 05:15 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
- April 4th 2013, 09:19 PMBernhardRe: Polynomial Ring Z[x]/(x^2)
Thanks ... but I still need some clarification ...

My thinking ... and then some of my issues/problems follow ...

Following an example I found in Gallian (page 257), first consider where is the ring of polynomials with real co-efficients.

Then

But is a Euclidean Domain and hence possesses a Division Algorithm, so we may write:

where r(x) = 0 or r(x) has degree less than 2.

so we can write r(x) = ax + b where a, b

Thus

= since the ideal absorbs the term

=

Thus

Now, by a similar argument we can demonstrate that

which makes the two rings and look to have the same structure?

One of my questions is how exactly are these two ring structures different?

A second worry is that the above demonstration works because is a Euclidean Domain ... so the same argument as above does not apply to

because is not a Euclidean Domain and hence we cannot use the Division algorithm.

How do we rigorously demonstrate that

and

Can someone please help clarify the above problems and issues?

Peter - April 4th 2013, 09:41 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
Write a surjective homomorphism with kernel . The first isomorphism theorem for rings gives an isomorphism between and .

As for how the structures are different: let's look at the and . The first one introduces a non-trivial element which squares to zero. There is no such element in the second one, which introduces a square root for . It is easy to see that this second field is actually . If you try to write an ring isomorphism from the first ring to the second ring, you need to send this to a nilpotent element in , but there is no such element. As such, the rings are completely different. - April 4th 2013, 09:50 PMBernhardRe: Polynomial Ring Z[x]/(x^2)
Thanks Gusbob

But I am still not sure how you rigorously demonstrate that

Can you elaborate?

Peter - April 4th 2013, 09:58 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
- April 4th 2013, 10:08 PMBernhardRe: Polynomial Ring Z[x]/(x^2)
Sorry to be slow but what is exactly?

Further you say that "one introduces a non-trivial element which squares to zero" - but the elements of are cosets.

Do you mean is the coset ? Then you would bbe saying that every element of this like, for example is nilpotent?

Please correct me if necessary.

Peter - April 4th 2013, 10:22 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
Sorry for the confusion. I was referring to my previous notation . You can check that this is a ring by itself. Here, is meant to be a variable used to distinguish it from . When I say we introduce a non-trivial element, what I actually meant was adding an element to . I apologise for the ambiguity. In general, when we want to add an element satisfying some polynomial property to a ring , we add it by forming the associated polynomial ring, then quotient out by the relation it satisfies. For example, to add satisfying , we write

It is true that every element of are cosets, but I was writing a ring homomorphism from that ring to . If you wish, does correspond to , in which case your third sentence is correct.