Computing the order in a polynomial quotient ring

[Mr Davis 97]

Homework Statement

Consider the quotient ring $F = \mathbb{Z}_3 [x] / \langle x^2 + 1 \rangle$. Compute the order of the coset $(x+1) + \langle x^2 + 1 \rangle$ in the group of units $F*$.

Homework Equations

The Attempt at a Solution

I was thinking that I just continually compute powers of (x+1) until I reach a polynomial with $x^2+1$ as a factor, but this is cumbersome. I guess my question, after each computation, how do I reduce the expression to get a simpler description of the coset? Do I just subtract $x^2 + 1$ as many times as a please?

 

[andrewkirk]

The question, as stated, presupposes that $(x+1)+\langle x^2+1\rangle$ is a unit of $F$. Is that true? Why, or why not?

 

[Mr Davis 97]

The question, as stated, presupposes that $(x+1)+\langle x^2+1\rangle$ is a unit of $F$. Is that true? Why, or why not?

Well $F$ is a field, since $\langle x^2 + 1 \rangle$ is a maximal ideal since $x^2 + 1$ is irreducible in $\mathbb{Z}_3$, so $(x+1)+\langle x^2+1\rangle$ must be a unit of $F$ since all non-zero elements are units by definition of a field.

 

[andrewkirk]

Well $F$ is a field, since $\langle x^2 + 1 \rangle$ is a maximal ideal since $x^2 + 1$ is irreducible in $\mathbb{Z}_3$, so $(x+1)+\langle x^2+1\rangle$ must be a unit of $F$ since all non-zero elements are units by definition of a field.

Good.

But in this latest post you wrote $\mathbb Z_3$, implying that the parent polynomial ring is $\mathbb Z_3[x]$, whereas in the OP you wrote $\mathbb Z[x]$. Which is it?

 

[Mr Davis 97]

Good.

But in this latest post you wrote $\mathbb Z_3$, implying that the parent polynomial ring is $\mathbb Z_3[x]$, whereas in the OP you wrote $\mathbb Z[x]$. Which is it?

Oops, it's $\mathbb{Z}_3 [x]$. Sorry.

 

[andrewkirk]

That makes the problem make more sense.

You can shortcut all the polynomial multiplication by making use of the fact that $\mathbb R[x]/\langle x^2+1\rangle\cong \mathbb C$ via the field isomorphism that maps 1 to 1 and $x$ to $i$. It would seem to be a natural guess then that the quotient field here is isomorphic to the Gaussian Integers modulo 3, ie $\mathbb Z_3+i\mathbb Z_3$. What is the lowest positive integer to which we need to raise $1+i$ to get a real number?

 

[Mr Davis 97]

That makes the problem make more sense.

You can shortcut all the polynomial multiplication by making use of the fact that $\mathbb R[x]/\langle x^2+1\rangle\cong \mathbb C$ via the field isomorphism that maps 1 to 1 and $x$ to $i$. It would seem to be a natural guess then that the quotient field here is isomorphic to the Gaussian Integers modulo 3, ie $\mathbb Z_3+i\mathbb Z_3$. What is the lowest positive integer to which we need to raise $1+i$ to get a real number?

Since $(1+i)^4 = -4$, the answer is 4.

 

[andrewkirk]

Not so fast. Firstly, the question should have been 'what is the lowest positive integer $m$ to which we need to raise $1+i$ to get a number that is 1 mod 3.' My mistake, but we wouldn't want the hints on here to be too accurate would we, or where would be the satisfaction for the students?

Secondly, having found $m$, since we have only guessed that the isomorphism holds, we need to verify that $(x+1)^m=1\mod (x^2+1)$ in $\mathbb Z_3[x]$. Taking the appropriate row of Pascal's Triangle will help us do this.

 

[Mr Davis 97]

Okay, so $(1+i)^8 = 16 = 1 \pmod 3$. Also, rather than using pascal's in this example, is there a more general way? For example, what if we have an irreducible quadratic of the form $ax^2 + bx + c$?

 

[andrewkirk]

Provided the roots of the quadratic are in the Gaussian Rationals $\mathbb Q+i\mathbb Q$ we can proceed the same way. We find the smallest $m$ such that $(f+id)^m=1\mod 3$ where $f+dx$ represents the coset we are considering.

If we have proven the isomorphism, it ends there. If not, we need to divide $(f+dx)^m$ by the irreducible polynomial, in $\mathbb Z_3[x]$ and show that the remainder is $1\mod 3$.