1. ## 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 :

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?

2. 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 :

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.

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

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 $\displaystyle (X^2)/(X^4).$

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!

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?

4. 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?

5. 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?

6. 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.