Describe the ring structure of $\displaystyle \mathbb{Z} [x]/(x^2) $

Also, indicate how the ring structure differs from $\displaystyle \mathbb{Z} [x]/(x^2 + 1) $

Would appreciate help with this exercise

Peter

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- Apr 4th 2013, 02:05 PMBernhardPolynomial Ring Z[x]/(x^2)
Describe the ring structure of $\displaystyle \mathbb{Z} [x]/(x^2) $

Also, indicate how the ring structure differs from $\displaystyle \mathbb{Z} [x]/(x^2 + 1) $

Would appreciate help with this exercise

Peter - Apr 4th 2013, 04:15 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
- Apr 4th 2013, 08:19 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 $\displaystyle \mathbb{R}[x]/<x^2> $ where $\displaystyle \mathbb{R}[x] $ is the ring of polynomials with real co-efficients.

Then $\displaystyle \mathbb{R}[x]/<x^2> = \{ g(x) + <x^2> | g(x) \in \mathbb{R}[x] \} $

But $\displaystyle \mathbb{R}[x] $ is a Euclidean Domain and hence possesses a Division Algorithm, so we may write:

$\displaystyle g(x) = q(x)(x^2) + r(x) $ where r(x) = 0 or r(x) has degree less than 2.

so we can write r(x) = ax + b where a, b $\displaystyle \in \mathbb{R} $

Thus $\displaystyle g(x) + <x^2> = q(x)(x^2) + r(x) + <x^2> $

= $\displaystyle r(x) + <x^2> $ since the ideal $\displaystyle <x^2> $ absorbs the term $\displaystyle q(x)(x^2) + r(x) $

= $\displaystyle ax + b + <x^2> $

Thus $\displaystyle \mathbb{R}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{R} \}$

Now, by a similar argument we can demonstrate that

$\displaystyle \mathbb{R}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{R} \} $

which makes the two rings $\displaystyle \mathbb{R}[x]/<x^2> $ and $\displaystyle \mathbb{R}[x]/<x^2 +1> $ 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 $\displaystyle \mathbb{R}[x] $ is a Euclidean Domain ... so the same argument as above does not apply to

$\displaystyle \mathbb{Z}[x] $ because $\displaystyle \mathbb{Z}[x] $ is not a Euclidean Domain and hence we cannot use the Division algorithm.

How do we rigorously demonstrate that

$\displaystyle \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} $ and

$\displaystyle \mathbb{Z}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z} \} $

Can someone please help clarify the above problems and issues?

Peter - Apr 4th 2013, 08:41 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
Write a surjective homomorphism $\displaystyle \mathbb{Z}[x] \to \mathbb{Z}[i]$ with kernel $\displaystyle \langle x^2 +1 \rangle$. The first isomorphism theorem for rings gives an isomorphism between $\displaystyle \mathbb{Z}[x]/\langle x^2+1\rangle$ and $\displaystyle \mathbb{Z}[i]$.

As for how the structures are different: let's look at the $\displaystyle \mathbb{R}/\langle x^2 \rangle$ and $\displaystyle \mathbb{R}/\langle x^2+1 \rangle$. The first one introduces a non-trivial element $\displaystyle \gamma$ which squares to zero. There is no such element in the second one, which introduces a square root for $\displaystyle -1$. It is easy to see that this second field is actually $\displaystyle \mathbb{C}$. If you try to write an ring isomorphism from the first ring to the second ring, you need to send this $\displaystyle \gamma$ to a nilpotent element in $\displaystyle \mathbb{C}$, but there is no such element. As such, the rings are completely different. - Apr 4th 2013, 08:50 PMBernhardRe: Polynomial Ring Z[x]/(x^2)
Thanks Gusbob

But I am still not sure how you rigorously demonstrate that

$\displaystyle \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} $

Can you elaborate?

Peter - Apr 4th 2013, 08:58 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
Basically the same as the one I showed you.

$\displaystyle f:\mathbb{Z}[x] \to \mathbb{Z}[\gamma]$ defined by $\displaystyle a\mapsto a \quad \forall a\in \mathbb{Z}, x\mapsto \gamma$ is surjective with kernel $\displaystyle \langle x^2 \rangle$ since $\displaystyle \gamma^2=0$ - Apr 4th 2013, 09:08 PMBernhardRe: Polynomial Ring Z[x]/(x^2)
Sorry to be slow but what is $\displaystyle \gamma $ exactly?

Further you say that "one introduces a non-trivial element $\displaystyle \gamma$ which squares to zero" - but the elements of $\displaystyle \mathbb{Z}[x]/<x^2> $ are cosets.

Do you mean $\displaystyle \gamma$ is the coset $\displaystyle \{ 0 + <x^2> \} $? Then you would bbe saying that every element of this like, for example $\displaystyle x^4 + x^2 $ is nilpotent?

Please correct me if necessary.

Peter - Apr 4th 2013, 09:22 PMGusbobRe: Polynomial Ring Z[x]/(x^2)
Sorry for the confusion. I was referring to my previous notation $\displaystyle \mathbb{Z}[\gamma]= \{a+b\gamma |a,b\in \mathbb{Z}, \gamma^2=0\}$. You can check that this is a ring by itself. Here, $\displaystyle \gamma$ is meant to be a variable used to distinguish it from $\displaystyle x$. When I say we introduce a non-trivial element, what I actually meant was adding an element to $\displaystyle \mathbb{Z}$. I apologise for the ambiguity. In general, when we want to add an element satisfying some polynomial property to a ring $\displaystyle R$, we add it by forming the associated polynomial ring, then quotient out by the relation it satisfies. For example, to add $\displaystyle i$ satisfying $\displaystyle i^2+1=0$, we write $\displaystyle \mathbb{R}[x]/(x^2+1)\cong \mathbb{R}[i]\cong \mathbb{C}$

It is true that every element of $\displaystyle \mathbb{Z}[x]/\langle x^2\rangle $ are cosets, but I was writing a ring homomorphism from that ring to $\displaystyle \mathbb{Z}[\gamma]$. If you wish, $\displaystyle \gamma$ does correspond to $\displaystyle \{0+\langle x^2 \rangle\}$, in which case your third sentence is correct.