# Rings and modules

• Oct 18th 2009, 03:44 AM
stephi85
Rings and modules
Hi everyone. I'm preparing for a test tomorrow and there are concepts and methods, which I seem to have forgotten or never understood properly.

For eksample there are these exercises, which I'm having difficulties with, if someone could help me throught them, it would be nice (Nod) :

1) Explain why the $\displaystyle \mathbb{Z}$-modules $\displaystyle \mathbb{Z}/8\mathbb{Z}$ and $\displaystyle \mathbb{Z}/36\mathbb{Z}$ have finite lenght and write a chain of composition.

2) Find the associated primes and the support of the above $\displaystyle \mathbb{Z}$-modules

3) k is a field. Explain why $\displaystyle (x)$ is the only prime ideal in $\displaystyle R=k[X]/(X^4)$. Is $\displaystyle (x^2)$ a simple R-module?

4) Are some of the three $\displaystyle k[X, Y]$-modules
$\displaystyle k[X, Y]/(XY)$ , $\displaystyle k[X, Y]/(X^2)$ , $\displaystyle k[X, Y]/(Y^2)$ isomorphic? Are some of them isomorphic as rings?

Now, in 1) I would say that $\displaystyle 8=2^3$ and therefore
$\displaystyle (0)\subset 4\mathbb{Z}/8\mathbb{Z}\subset 2\mathbb{Z}/8\mathbb{Z} \subset \mathbb{Z}/8\mathbb{Z}$ correct?
and $\displaystyle 36=2^2 3^2$ so $\displaystyle \mathbb{Z}/36\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ so $\displaystyle (0)\subset \mathbb{Z}/2\mathbb{Z} \subset \mathbb{Z}/3\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}$? So thus the lenght is three in both cases and thus finite?

In (4) I would say that the two last are isomorphic, because you can make an ismomorphism from $\displaystyle k[X, Y]/(X^2)$ to $\displaystyle k[X, Y]/(Y^2)$ by sending X to Y, Y to X and 1 to 1. Correct?
• Oct 18th 2009, 11:21 AM
NonCommAlg
Quote:

Originally Posted by stephi85
Hi everyone. I'm preparing for a test tomorrow and there are concepts and methods, which I seem to have forgotten or never understood properly.

For eksample there are these exercises, which I'm having difficulties with, if someone could help me throught them, it would be nice (Nod) :

1) Explain why the $\displaystyle \mathbb{Z}$-modules $\displaystyle \mathbb{Z}/8\mathbb{Z}$ and $\displaystyle \mathbb{Z}/36\mathbb{Z}$ have finite lenght and write a chain of composition.

Now, in 1) I would say that $\displaystyle 8=2^3$ and therefore
$\displaystyle (0)\subset 4\mathbb{Z}/8\mathbb{Z}\subset 2\mathbb{Z}/8\mathbb{Z} \subset \mathbb{Z}/8\mathbb{Z}$ correct?
and $\displaystyle 36=2^2 3^2$ so $\displaystyle \mathbb{Z}/36\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ so $\displaystyle (0)\subset \mathbb{Z}/2\mathbb{Z} \subset \mathbb{Z}/3\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}$? So thus the lenght is three in both cases and thus finite?

well, they're both finite modules and thus obviously Artinian (and Noetherian). so they have finite length. your solution for $\displaystyle \mathbb{Z}/36 \mathbb{Z}$ is wrong.

Quote:

2) Find the associated primes and the support of the above $\displaystyle \mathbb{Z}$-modules

recall that $\displaystyle \text{Supp}(R/I)=V(I)=\{P \in \text{Spec}(R): \ \ I \subseteq P \}.$ so $\displaystyle \text{Supp}(\mathbb{Z}/8\mathbb{Z})=\{2\mathbb{Z} \}$ and $\displaystyle \text{Supp}(\mathbb{Z}/36\mathbb{Z})=\{2\mathbb{Z}, \ 3\mathbb{Z} \}.$

let $\displaystyle \text{AP}(M)$ be the set of associated primes of R module M. if $\displaystyle p$ is a prime number, then $\displaystyle \text{AP}(\mathbb{Z}/p^k \mathbb{Z})=\{p\mathbb{Z} \}.$ so $\displaystyle \text{AP}(\mathbb{Z}/8\mathbb{Z})=\{2\mathbb{Z} \}.$

also $\displaystyle \text{AP}(\mathbb{Z}/36 \mathbb{Z})=\text{AP}(\mathbb{Z}/4\mathbb{Z}) \cup \text{AP}(\mathbb{Z}/9\mathbb{Z})=\{2\mathbb{Z}, \ 3\mathbb{Z} \}.$

Quote:

3) k is a field. Explain why $\displaystyle (x)$ is the only prime ideal in $\displaystyle R=k[X]/(X^4)$. Is $\displaystyle (x^2)$ a simple R-module?

$\displaystyle k[X]$ is a PID. so a prime ideal of R is in the form $\displaystyle (f(X))/(X^4),$ where $\displaystyle f(X)$ is an irreducible element of $\displaystyle k[X]$ and $\displaystyle (X^4) \subseteq (f(X)).$ but $\displaystyle (X^4) \subseteq (f(X))$ if and only if $\displaystyle f(x) \mid X^4,$ which is possible

only if $\displaystyle f(X)=X$ because $\displaystyle f(X)$ is irreducible. the answer to the second part of your question is no because $\displaystyle (X^3)/(X^4)$ is a non-zero submodule of $\displaystyle (X^2)/(X^4).$

Quote:

4) Are some of the three $\displaystyle k[X, Y]$-modules

$\displaystyle k[X, Y]/(XY)$ , $\displaystyle k[X, Y]/(X^2)$ , $\displaystyle k[X, Y]/(Y^2)$ isomorphic? Are some of them isomorphic as rings?

In (4) I would say that the two last are isomorphic, because you can make an ismomorphism from $\displaystyle k[X, Y]/(X^2)$ to $\displaystyle k[X, Y]/(Y^2)$ by sending X to Y, Y to X and 1 to 1. Correct?

the only possible isomorphism here is that "as rings" we have $\displaystyle k[X,Y]/(X^2) \cong k[X,Y]/(Y^2).$ the isomorphism sends $\displaystyle X$ to $\displaystyle Y$ and $\displaystyle Y$ to $\displaystyle X.$ to see that there are no other isomorphisms let

$\displaystyle R=k[X,Y], \ I=(X^2), \ J=(Y^2), \ K=(XY).$ suppose $\displaystyle f: R/I \longrightarrow R/K$ is a "ring" isomorphism. let $\displaystyle f(X+I)=p(X,Y) + K.$ then $\displaystyle 0=(f(X+I))^2=(p(X,Y))^2+K.$

so $\displaystyle (p(X,Y))^2 \in K$ and thus $\displaystyle p(X,Y) \in K.$ (why?) therefore $\displaystyle f(X+I)=0$ and, since $\displaystyle f$ is an isomorphism, we get $\displaystyle X+I=0.$ that means $\displaystyle X \in I=(X^2),$ which is clearly false. so there's no

such isomorphism. let me show you how to prove that $\displaystyle R/I$ and $\displaystyle R/K$ are not isomorphic as "R-mouldes": suppose $\displaystyle g: R/I \longrightarrow R/K$ is an R-module isomorphism. let $\displaystyle g(q(X,Y)+I)=X+K.$

then $\displaystyle 0=Yg(q(X,Y)+I)=g(Yq(X,Y)+I).$ so $\displaystyle Yq(X,Y) \in I=(X^2)$ and hence $\displaystyle q(X,Y) \in (X^2)=I.$ therefore $\displaystyle q(X,Y)+I=0,$ which gives us $\displaystyle X+K=g(q(X,Y)+I)=0.$ that means

$\displaystyle X \in K=(XY),$ which is obviously nonsense!
• Oct 18th 2009, 11:59 AM
stephi85

Quote:

Originally Posted by NonCommAlg
your solution for $\displaystyle \mathbb{Z}/36 \mathbb{Z}$ is wrong.

Allright, but could you please explain what's wrong with it then? Is it because I only have to include one of the fields in the chain?
• Oct 18th 2009, 12:28 PM
NonCommAlg
Quote:

Originally Posted by stephi85

Allright, but could you please explain what's wrong with it then? Is it because I only have to include one of the fields in the chain?

here's a composition series for $\displaystyle \mathbb{Z}/36\mathbb{Z}: \ (0) \subset 12 \mathbb{Z}/36\mathbb{Z} \subset 4\mathbb{Z}/36\mathbb{Z} \subset 2\mathbb{Z}/36\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}.$ do you see how i found this series?
• Oct 18th 2009, 12:52 PM
stephi85
Ok, thanks. I think I see the pattern now; You say $\displaystyle 36=2^2 3^2$ and then take combinations of the powers $\displaystyle 12=2^2 3^1$, $\displaystyle 4=2^2 3^0$, $\displaystyle 2=1^2 3^0$ and arrange the fields after decreasing number value.

But how come, then, that 6 and 3 aren't included?
• Oct 18th 2009, 01:02 PM
NonCommAlg
Quote:

Originally Posted by stephi85
Ok, thanks. I think I see the pattern now; You say $\displaystyle 36=2^2 3^2$ and then take combinations of the powers $\displaystyle 12=2^2 3^1$, $\displaystyle 4=2^2 3^0$, $\displaystyle 2=2^1 3^0$ and arrange the fields after decreasing number value.

But how come, then, that 6 and 3 aren't included?

the quotient of any two consecutive modules in our series has to be a simple moulde. that's the definition of a composition series. so you can't just put anything you want in there.

another composition series for $\displaystyle \mathbb{Z}/36\mathbb{Z}$ is: $\displaystyle (0) \subset 18 \mathbb{Z}/36\mathbb{Z} \subset 9\mathbb{Z}/36\mathbb{Z} \subset 3\mathbb{Z}/36\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}.$ can you find another one?

so a module can have more than one composition series but the length of all composition series of a module, if they exist, are equal.