3
$\begingroup$

The question I have is from a past exam paper:

Identity if it is an integral domain, whether it admits a finite basis over a coefficient ring, or whether it is isomorphic to another ring.

The first one is $$R = \mathbb{Z}[x]/(6x^2 -1, 2x-4)$$

My first angle of attack is to find if we can simplify the ideal $I=(6x^2-1,2x-4)$ in hopes of finding a monic irreducible polynomial by taking combinations of elements of the ideal to end up with $I = (f)$ to apply a substitution isomorphism, maybe this is too hopeful.

We start out by making a few computation on elements of the ring

\begin{align*} 6x^2-1 &= 0\\ 2x*3x-1&=12x-1 = 0\\ \end{align*} So we can use $12x-1=0$ and $2x-4$ to perhaps find if this ring is isomorphic to some modular ring.

$$6(2x-4) = 12x - 24 = 1-24 = 0 = 23 \in I$$

and

$$R = \mathbb{Z}[x]/(23,6x^2-1,2x-4) \cong \mathbb{Z}_{23}[x]/(6x^2-1,2x-4)$$

Next we can use this modulo $23$ in the following way to perform further computations:

$$4(6x^2-1) = 24x^2-4=x^2-4=0$$

We then can reduce the ideal to

$$I = (x^2-4,2x-4)$$

If $x^2 = 4$ and $2x=4$ we must have that $x = 2$ and $I = (x-2)$

We then write $$R \cong \mathbb{Z}_{23}[x] /(x-2) \cong \mathbb{Z}_{23}[2] = \mathbb{Z}_{23}$$

Also, I know that this ring isn't an integral domain because we can take $23,1 \in R$ and get $23*1 = 0$, so therefore $23$ is a zero divisor.

Let me know if my working is correct, I feel a little bit iffy about reducing the ideal, thanks.

EDIT: A finite basis for $R$ would just be $\{a|a \in \mathbb{Z}\}$

  • 1
    Saying $1•23=0$ isn't an example of a pair of nonzero zero divisors since $23=0$ in this ring. $\mathbb Z_{23}$ is a field and therefore a domain. Otherwise, looks ok. – 2017-02-18
  • 0
    What about $23*2=46 = 0$? I know that $ \mathbb{Z}_p$ is a field for $p$ a prime, but I don't understand why that result is true. – 2017-02-18
  • 1
    Using $23$ at all will be useless since $23=0$ in this ring. You would be searching for a pair of **nonzero** zero divisors. – 2017-02-18
  • 0
    That makes sense, and since you cant decompose 23 into a product of two non zero numbers in the ring, it isnt a zero divisor. But how about $\mathbb{Z}_{21} = R$, taking $7*3 = 21 = 0$ in R would give a zero divisor and therefore, $R$ would not be an integral domain. – 2017-02-18
  • 0
    Why do you claim $x = 2$? I guess you mean $x-2$ is in $I$. If that's what you mean, it's OK since $x-2$ _is_ in $I$, but you need to justify the claim. – 2017-02-18
  • 1
    @jamesmartini what you said is correct. $\mathbb Z_{21}\cong \mathbb Z_7\times\mathbb Z_ 3$ is not a domain. – 2017-02-18
  • 0
    Once you show $R$ is isomorphic to $Z_{23}$, then it's automatic that $R$ is a field. – 2017-02-18
  • 0
    Alternatively, to show R is a field, it suffices to show that $I$ is a maximal ideal of $\mathbb{Z}[x]$. – 2017-02-18
  • 0
    @rschwieb So is it often useful to decompose a modulo ring into a product like that using idempotents? How can you reason that $\mathbb{Z}_{7} \times \mathbb{Z}_{3}$ is not a domain, and how can you perform the same argument about $\mathbb{Z}_{7} \times \mathbb{Z}_6 \cong \mathbb{Z}_{23}$ to conclude that $\mathbb{Z}_{23}$ is a domain – 2017-02-18
  • 0
    @quasi I claim that by performing reductions on the ideal using operations constrained by the quotient. I know that its quite easy to say $x-2 \in I$ but saying $(x-2) = I$ is a different story. I suppose my argument for that would be that $(x-2) \subset (6x^2-1,2x-4)$. Showing that $(6x^2-1,2x-4) \subset (x-2)$ would use the fact that you could write $k \in (6x^2-1,2x-4)$ as $k = a(6x^2 -1) + b(2x-4), a,b \in \mathbb{Z}_{23}$, and showing that $k \in (x-2)$, by writing $c(x-2) = a(6x^2 -1) + b(2x-4)$ would conclude by giving my result. – 2017-02-18
  • 1
    @jamesmartini the product of two nonzero rings is never a domain. $(1,0)(0,1)=(0,0)$ is always a pair of zero divisors in such a ring. – 2017-02-18

2 Answers 2

1

I juste realized how wrong. Is this better? Let $f=6x^2-1$ and $g=2x-4$. The division algorithm gives $f=g(3x-6)+23$. We have $6x^2-1\equiv 23\mod g$ and $\mathbb{Z}[X]/(g)=\mathbb{Z}[x]/(x-2)\mathbb{Z}[x]\cong \mathbb{Z}$. Modding successively gives $\mathbb{Z}[x]\to \mathbb{Z}\to \mathbb{Z}/23\mathbb{Z}$.

  • 0
    Yes, this is at least a rough version of a complete solution after the edits. – 2017-02-20
1

You're mixing up things and arrive to a wrong conclusion.

Your strategy is good, it just needs some adjustment in wording.

Let $\xi$ denote the image of $x$ in the quotient ring; then, by assumption, $6\xi^2-1=0$ and $2\xi-4=0$. Therefore $$ 1=3\xi\cdot 2\xi=3\xi\cdot 4=6\cdot 2\xi=24 $$ In particular, $24-1=0$ in the quotient ring, which means $23\in I$.

The next step is noting $4\cdot 6\xi^2=4$, so $\xi^2=4$. Together with $2\xi=4$, this implies $\xi^2=2\xi$. Also $$ 1=6\xi^2=6\cdot2\xi=12\xi $$ and we conclude that $2$ is invertible in the quotient ring. From $2\xi=4$, we conclude $\xi=2$.

So, we have proved that $23\in I$ and $x-2\in I$. Now we can do the final step: proving $I=(23,x-2)$. Clearly, $2x-4\in(23,x-2)$. Also $$ 6x^2-1=(6x+12)(x-2)+23\in(23,x-2) $$ Thus your ring is $$ \mathbb{Z}[x]/(23,x-2)\cong(\mathbb{Z}/23\mathbb{Z})[x]/(x-2) \cong\mathbb{Z}/23\mathbb{Z} $$ is a field.

  • 0
    That's the same conclusion I came to. I did however come to the wrong conclusion regarding whether it was an integral domain or not. – 2017-02-19
  • 0
    @jamesmartini In general, from $2a=4$ you cannot deduce $a=2$. Here it's possible because $2$ is invertible, which you didn't remark. It's one of the reasons for distinguishing between $x$ and $\xi$, so it's clearer where you're working at any time. – 2017-02-19