# Polynomial Ring Z[x]/(x^2)

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

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

Would appreciate help with this exercise

Peter
• Apr 4th 2013, 04:15 PM
Gusbob
Re: Polynomial Ring Z[x]/(x^2)
Quote:

Originally Posted by Bernhard
Describe the ring structure of $\mathbb{Z} [x]/(x^2)$

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

Would appreciate help with this exercise

Peter

Can you see that:

The former is isomorphic to the ring $\mathbb{Z}[\gamma]=\{a+b\gamma | a,b\in \mathbb{Z}, \gamma^2=0\}$

The latter is isomorphic to the ring of Gaussian integers $\mathbb{Z}[i]=\{a+bi|a,b\in \mathbb{Z}, i^2=-1\}$
• Apr 4th 2013, 08:19 PM
Bernhard
Re: 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 $\mathbb{R}[x]/$ where $\mathbb{R}[x]$ is the ring of polynomials with real co-efficients.

Then $\mathbb{R}[x]/ = \{ g(x) + | g(x) \in \mathbb{R}[x] \}$

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

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

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

Thus $g(x) + = q(x)(x^2) + r(x) + $

= $r(x) + $ since the ideal $$ absorbs the term $q(x)(x^2) + r(x)$

= $ax + b + $

Thus $\mathbb{R}[x]/ = \{ ax + b + | a, b \in \mathbb{R} \}$

Now, by a similar argument we can demonstrate that

$\mathbb{R}[x]/ = \{ ax + b + | a, b \in \mathbb{R} \}$

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

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

How do we rigorously demonstrate that

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$ and

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$

Peter
• Apr 4th 2013, 08:41 PM
Gusbob
Re: Polynomial Ring Z[x]/(x^2)
Quote:

Originally Posted by Bernhard

How do we rigorously demonstrate that

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$ and

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$

Write a surjective homomorphism $\mathbb{Z}[x] \to \mathbb{Z}[i]$ with kernel $\langle x^2 +1 \rangle$. The first isomorphism theorem for rings gives an isomorphism between $\mathbb{Z}[x]/\langle x^2+1\rangle$ and $\mathbb{Z}[i]$.

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

But I am still not sure how you rigorously demonstrate that

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$

Can you elaborate?

Peter
• Apr 4th 2013, 08:58 PM
Gusbob
Re: Polynomial Ring Z[x]/(x^2)
Quote:

Originally Posted by Bernhard
Thanks Gusbob

But I am still not sure how you rigorously demonstrate that

$\mathbb{Z}[x]/ = \{ ax + b + | a, b \in \mathbb{Z} \}$

Can you elaborate?

Peter

Basically the same as the one I showed you.

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

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

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

Peter
• Apr 4th 2013, 09:22 PM
Gusbob
Re: Polynomial Ring Z[x]/(x^2)
Quote:

Originally Posted by Bernhard
Sorry to be slow but what is $\gamma$ exactly?

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

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

Sorry for the confusion. I was referring to my previous notation $\mathbb{Z}[\gamma]= \{a+b\gamma |a,b\in \mathbb{Z}, \gamma^2=0\}$. You can check that this is a ring by itself. Here, $\gamma$ is meant to be a variable used to distinguish it from $x$. When I say we introduce a non-trivial element, what I actually meant was adding an element to $\mathbb{Z}$. I apologise for the ambiguity. In general, when we want to add an element satisfying some polynomial property to a ring $R$, we add it by forming the associated polynomial ring, then quotient out by the relation it satisfies. For example, to add $i$ satisfying $i^2+1=0$, we write $\mathbb{R}[x]/(x^2+1)\cong \mathbb{R}[i]\cong \mathbb{C}$
It is true that every element of $\mathbb{Z}[x]/\langle x^2\rangle$ are cosets, but I was writing a ring homomorphism from that ring to $\mathbb{Z}[\gamma]$. If you wish, $\gamma$ does correspond to $\{0+\langle x^2 \rangle\}$, in which case your third sentence is correct.