MHB What Are the Prime and Maximal Ideals in the Ring $\mathbb{Z}_{12}$?

[mathmari]

Hey!

I want to find the prime and maximal ideals of the ring $\mathbb{Z}_{12}$.

Could you give me some hints what we could do to find them? (Wondering)

 

[Deveno]

Ideals are always additive subgroups, so since $\Bbb Z_{12}$ is cyclic under addition, the only candidates for ideals are cyclic subgroups of $\Bbb Z_{12}$, and there is only one of each order. This also means any ideal has to be principal.

These are:

$(0) = \{0\}$.

$(1) = \Bbb Z_{12}$ (any $(a)$ with $\gcd(a,12)$ will be all of $\Bbb Z_{12}$, since these are precisely the units).

$(2) = \{0,2,4,6,8,10\}$,

$(3) = \{0,3,6,9\}$,

$(4) = \{0,4,8\}$,

$(6) = \{0,6\}$.

Are all of these ideals? What do you think?

You can see that $(4) \subset (2)$, and that $(6) \subset (3)$, so if these are ideals, they are not maximal.

Any maximal ideal is prime, so if you've found the maximal ideals, you've surely found *some* of the prime ideals.

To find the prime ideals, you may want to use the following theorem:

$I$ is a prime ideal of $R$ if and only if $R/I$ is an integral domain. Since we have already seen in another thread a finite integral domain is a field, we know that $(a)$ is a prime ideal if and only if $\Bbb Z_{12}/(a)$ is a field.

 

[mathmari]

Ideals are always additive subgroups, so since $\Bbb Z_{12}$ is cyclic under addition, the only candidates for ideals are cyclic subgroups of $\Bbb Z_{12}$, and there is only one of each order. This also means any ideal has to be principal.

These are:

$(0) = \{0\}$.

$(1) = \Bbb Z_{12}$ (any $(a)$ with $\gcd(a,12)$ will be all of $\Bbb Z_{12}$, since these are precisely the units).

$(2) = \{0,2,4,6,8,10\}$,

$(3) = \{0,3,6,9\}$,

$(4) = \{0,4,8\}$,

$(6) = \{0,6\}$.

Are all of these ideals? What do you think?

So, we have to prove that $n\mathbb{Z}_{12}$ is an ideal of $\mathbb{Z}_{12}$, when $n$ divides $12$.

So, we take two elements of $n\mathbb{Z}_{12}$, say $x_1=nz_1$ and $x_2=nz_2$, and an element of $\mathbb{Z}_{12}$, say $y$.

We have that $-x_1=-nz_1=n(-z_1)\in n\mathbb{Z}_{12}$, $x_1+x_2=nz_1+nz_2=n(z_1+z_2)\in n\mathbb{Z}_{12}$, so $n\mathbb{Z}_{12}$ is an additive subgroup of $\mathbb{Z}_{12}$.

Then since $\mathbb{Z}_{12}$ is commutative it is enough to check $n\mathbb{Z}_{12}$ is a left ideal.
Since $x_1y=n_1zy=n(z_1y)\in n\mathbb{Z}_{12}$ it is a left ideal.

Is this correct? (Wondering)

So, all the above are ideals, right? (Wondering)

You can see that $(4) \subset (2)$, and that $(6) \subset (3)$, so if these are ideals, they are not maximal.

Any maximal ideal is prime, so if you've found the maximal ideals, you've surely found *some* of the prime ideals.

To find all the maximal ideals do we have to check which of the above ideals is a subset of an other one and the ones that don't belong to any other are then maximal? (Wondering)

To find the prime ideals, you may want to use the following theorem:

$I$ is a prime ideal of $R$ if and only if $R/I$ is an integral domain. Since we have already seen in another thread a finite integral domain is a field, we know that $(a)$ is a prime ideal if and only if $\Bbb Z_{12}/(a)$ is a field.

Ah ok... How could we check if $\Bbb Z_{12}/(a)$ is a field? (Wondering)

 

[Deveno]

So, we have to prove that $n\mathbb{Z}_{12}$ is an ideal of $\mathbb{Z}_{12}$, when $n$ divides $12$.

So, we take two elements of $n\mathbb{Z}_{12}$, say $x_1=nz_1$ and $x_2=nz_2$, and an element of $\mathbb{Z}_{12}$, say $y$.

We have that $-x_1=-nz_1=n(-z_1)\in n\mathbb{Z}_{12}$, $x_1+x_2=nz_1+nz_2=n(z_1+z_2)\in n\mathbb{Z}_{12}$, so $n\mathbb{Z}_{12}$ is an additive subgroup of $\mathbb{Z}_{12}$.

Then since $\mathbb{Z}_{12}$ is commutative it is enough to check $n\mathbb{Z}_{12}$ is a left ideal.
Since $x_1y=n_1zy=n(z_1y)\in n\mathbb{Z}_{12}$ it is a left ideal.

Is this correct? (Wondering)

So, all the above are ideals, right? (Wondering)


To find all the maximal ideals do we have to check which of the above ideals is a subset of an other one and the ones that don't belong to any other are then maximal? (Wondering)

Since we have a finite set of ideals, it's easy enough to make a lattice partially ordered by inclusion. In fact, I recommend it.

Ah ok... How could we check if $\Bbb Z_{12}/(a)$ is a field? (Wondering)

Well, what properties of finite fields do you know? For example, can there be a finite field of order 6?

It might be helpful to know what $\Bbb Z_{12}/n\Bbb Z_{12}$ is isomorphic to, as an abelian group. There are only so many choices for $n$, so you can investigate them all.

 

[mathmari]

Since we have a finite set of ideals, it's easy enough to make a lattice partially ordered by inclusion. In fact, I recommend it.

We have the following:

View attachment 5507 So, the maximal ideals are $(2)$ and $(3)$, right? (Wondering)

Well, what properties of finite fields do you know? For example, can there be a finite field of order 6?

All finite fields have order $p^n$. So, there cannot be a finite field of order $6$.

It might be helpful to know what $\Bbb Z_{12}/n\Bbb Z_{12}$ is isomorphic to, as an abelian group. There are only so many choices for $n$, so you can investigate them all.

We have that $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$.
But now we have $\mathbb{Z}_{12}$ instead of $\mathbb{Z}$.

Suppose that $n=2$, then $2\mathbb{Z}_{12}=\{0, 2, 4, 6, 8, 10\}$. Does $\mathbb{Z}_{12}/2\mathbb{Z}_{12}$ contain then all the elements of $\mathbb{Z}_{12}$ modulo the even ones? (Wondering)

 

[Deveno]

We have the following:

So, the maximal ideals are $(2)$ and $(3)$, right? (Wondering)

You should also have a line going between $(2)$ and $(6)$ since $(6) \subset (2)$, as well.

All finite fields have order $p^n$. So, there cannot be a finite field of order $6$.

Indeed. Nor is there a field or order 12.

We have that $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$.
But now we have $\mathbb{Z}_{12}$ instead of $\mathbb{Z}$.

Suppose that $n=2$, then $2\mathbb{Z}_{12}=\{0, 2, 4, 6, 8, 10\}$. Does $\mathbb{Z}_{12}/2\mathbb{Z}_{12}$ contain then all the elements of $\mathbb{Z}_{12}$ modulo the even ones? (Wondering)

Let's look at an example: $\Bbb Z_{12}/3\Bbb Z_{12}$. We need to look at the cosets of:

$H = (3) = \{0,3,6,9\}$ ($H$ is clearly our first coset). Since $1 \not\in H$, we have $1+H$ is another coset:

$1 + H = \{1,4,7,10\}$. In the same manner, we have a third coset:

$2 + H = \{2,5,8,11\}$. We have now completely partitioned $\Bbb Z_{12}$, so that's all.

Note that for $a \in 1 + H$, and $b \in 2 +H$, we have $a + b \in H$, for example:

$7 + 5 = 0$, and $4 + 11 = 3$.

This is purely an informal observation, but I believe (because I have faith in you!) you can prove:

$\Bbb Z_{12}/3\Bbb Z_{12} \cong \Bbb Z_3$ (as RINGS!).

Now $\Bbb Z_3$ is indeed a field, so $(3)$ is a prime ideal.

Now in this case, we could have just argued by order:

$|\Bbb Z_{12}/3\Bbb Z_{12}| = \dfrac{|\Bbb Z_{12}|}{|3\Bbb Z_{12}|} = \dfrac{12}{4} = 3$, and any group of order 3 is isomorphic (as a group) to $\Bbb Z_3$, and it is not hard to see that corresponding multiplication of multiplication (mod $3$) in $\Bbb Z_3$ is "the same" as the coset multiplication in $\Bbb Z_{12}/3\Bbb Z_{12}$, that is:

$k + H \mapsto k$ (mod $3$) is a RING-HOMOMORPHISM (and thus an isomorphism since it's bijective).

The case of $(4)$ is an exception, one cannot argue purely by order, because the resulting quotient has order 4, and there do exist fields of order 4. But $(4)$ is not a prime ideal, do you see why?

$(0)$ is another interesting case, one that people often over-look. It's not prime...but why?

Finally, why is $(1)$ (another "edge case") not a prime ideal?

 

[mathmari]

but I believe (because I have faith in you!) you can prove:

$\Bbb Z_{12}/3\Bbb Z_{12} \cong \Bbb Z_3$ (as RINGS!).

We consider the following mapping:
$f:\mathbb{Z}_{12}\rightarrow \mathbb{Z}_3 \\ x\mapsto x\pmod 3$

This mapping is an homomorphism, since
$f(xy)=(xy)\pmod 3 \\ f(x)f(y)=(x\pmod 3 )(y \pmod 3)=(xy)\pmod 3$
so, $f(xy)=f(x)f(y)$. We have that $\ker f =\{x\in \mathbb{Z}_{12} : f(x)=0\}=\{x\in \mathbb{Z}_{12} : x\equiv 0 \pmod 3\}$, right? (Wondering)

Is this equal to $\ker f=3\mathbb{Z}_{12}$ ? (Wondering)

The case of $(4)$ is an exception, one cannot argue purely by order, because the resulting quotient has order 4, and there do exist fields of order 4. But $(4)$ is not a prime ideal, do you see why?

$(0)$ is another interesting case, one that people often over-look. It's not prime...but why?

Why aren't these prime ideals? (Wondering)
I don't really have an idea...

why is $(1)$ (another "edge case") not a prime ideal?

We have that $\mathbb{Z}_{12}/(1)$ is the trivial ring, or not? (Wondering)
Is the trivial ring not a prime ideal? (Wondering)

 

[mathmari]

Do we have the following?

We have that an ideal $I$ is prime iff $\mathbb{Z}_{12}/I$ is an integral domain.
Since each finite integral domain is a field, we have that an ideal $I$ is prime iff $I$ is maximal.

Therefore, the prime ideals of $\mathbb{Z}_{12}$ are $(2),(3)$.

Is this correct? (Wondering)

 

[Deveno]

Yes.

$(0)$ is not a prime ideal, because $\Bbb Z_{12}$ has zero-divisors.

$(1)$ is not a prime ideal, because a field must have at least two elements, and $\Bbb Z_{12}/(1)\cong \{0\}$.

$(4)$ is not a prime ideal, because we have $4 \in (4)$ and $4 = 2\cdot 2$, but $2 \not\in (4)$.

For the same reason, you can see $(6)$ is not a prime ideal, either. Both of these cases are directly related to the fact that:

$\Bbb Z_{12}/(4) \cong \Bbb Z_4$
$\Bbb Z_{12}/(6) \cong \Bbb Z_6$

and both these rings have zero-divisors (and thus cannot be integral domains, much less fields).

As I pointed out earlier, $\Bbb Z_{12}$ is an "easy ring" to investigate, its cyclic (and commutative), which greatly reduces the work we have to do to find ideals.

 

[mathmari]

Thank you for your help! (Mmm)