3
$\begingroup$

I am trying to answer a couple questions about counting $F[x]$-submodules and I think I am having trouble understanding definitions. I will try to explain how I have been approaching the problem and hope people will point out what I am doing wrong.

Let $F = \mathbb{Z} / p \mathbb{Z}$ be the finite field with $p$ and consider the $F[x]$-module $V = F[x]/(x^2) \oplus F[x]/(x^3)$.

Question 1. How many $F[x]$-submodules with $p$ elements does $V$ have.

I think the answer is 5 submodules which I will list explicitly:

$\{ v \in V : v = (a,0), \,\, a \in \mathbb{Z_p} \}$ $\{ v \in V : v = (0,b), \,\, b \in \mathbb{Z_p} \}$ $\{ v \in V : v = (ax,0), \,\, a \in \mathbb{Z_p} \}$ $\{ v \in V : v = (0,bx), \,\, b \in \mathbb{Z_p} \}$ $\{ v \in V : v = (0,bx^2), \,\, b \in \mathbb{Z_p} \}$

Is this correct or did I miss something?

Question 2. How many cyclic $F[x]$-submodules with $p^2$ elements does $V$ have and how many noncyclic $F[x]$-submodules with $p^2$ elements does $V$ have?

This is where I get confused in constructing modules with $p^2$ elements.

Should I consider sets of the form $\{ v \in V : v = (a+bx,0), \,\, a,b \in \mathbb{Z_p} \}$ will this still be cyclic?

$\{ v \in V : v = (ax,bx^2), \,\, a ,b\in \mathbb{Z_p} \}$ $\{ v \in V : v = (ax,bx), \,\, a,b \in \mathbb{Z_p} \}$ will these have $p^2$ elements and not be cyclic?

Thanks for your help

  • 5
    $\{v\in V\mid v=(a+(x^2),0+(x^3))\ a\in\mathbb{Z}_p\}$ is not an $F[x]$ module under the natural action: what do you get when you multiply $(1+(x^2),0+(x^3))$ by $x$? It's not in the set. Neither are your second and fourth sets (interpreted as pairs of congruence classes) closed under multiplication by $x$, so they are not $F[x]$ modules either.2011-07-30
  • 5
    +1 for showing your work2011-07-30
  • 2
    A cyclic module is one that is generated, under both addition *and* multiplication by elements of $F[x]$, by a single element. E.g., the cyclic $\mathbb{Z}[x]$-submodule of $\mathbb{Z}[x]/(x^3)$ generated by $2+(x^3)$ consists of all $\mathbb{Z}[x]$-multiples of $2+(x^3)$.2011-07-30
  • 0
    Thank you for the input. Is it true that if $M$ is $F[x]$-submodule not equal to zero then $M$ has at least $p$ elements in this case?2011-07-30
  • 0
    I don't understand your response about multiplying $(1+(x^2),0+(x^3))$ by $x$ because submodules do not have a multiplication operation for vectors. For instance $R$ a ring and $M$ an $R$-module than a subset $N$ of $M$ is a submodule of $M$ if and only if $x+ry \in N$ for all $r \in R$ and all $x,y \in N$2011-07-30
  • 1
    @user7980: Recapping Arturo's concern in the case of your first subset - call it $N$. The pair $v=(1,0)$ is in $N$. As $N$ is supposed to be an $F[x]$-submodule, the element $xv$ must also be in $N$. Questions for you: 1) What is $xv$? After all $V$ is an $F[x]$-module, so we can multiply any element of $V$ with $x$. 2) Is it in $N$?2011-07-30
  • 0
    $v= (1 + (x^2), 0 )$ so $ x v = ( x + (x^2) , 0) \in N$ is this correct?2011-07-30
  • 1
    @user7980: The multiplication is correct. Well done. But your first set $N$ only contains elements of the form $(a+(x^2),0)$ for some constant $a\in F$, so the product $xv$ is not in $N$.2011-07-30
  • 0
    Thank you! It has been awhile since I worked with factor rings I keep wanting to think the elements of $R[x]/(x^2)$ are of the form ${a+bx, a,b\in R}$ and that is one of my main confusions2011-07-30
  • 0
    so in general $R[x]/(x^2)$ is not isomorphic to a set of linear combinations of polynomials?2011-07-30
  • 1
    @user7980: I don't understand? A linear combination of polynomials is a polynomial. All the cosets of $(x^2)$ in $F[x]$ have a (unique) representative of the form $a+bx$ for some constants $a,b\in F$. You are right about that. But here we also concern ourselves about the $F[x]$-module structure. So for example $$x(a+bx)=ax+bx^2\equiv ax$$ modulo the ideal $(x^2)$, because that term $bx^2$ is in that ideal, and therefore the cosets $ax+(x^2)$ and $ax+bx^2+(x^2)$ are equal.2011-07-30
  • 2
    A hint: What is the $F[x]$-submodule of $V$ generated by the single element $(x,kx^2)\in V$ for some constant $k\in F$? How many elements does it have?2011-07-30

1 Answers 1

5

I think it's appropriate in this case to work out most of the first problem. Hopefully this will get the wheels turning.

We have projections $p_1\colon V \to F[x]/(x^2)$ and $p_2\colon V \to F[x]/(x^3)$. If $W$ is a submodule of $V$, then $p_1(W)$ and $p_2(W)$ are submodules of $F[x]/(x^2)$ and $F[x]/(x^3)$, respectively. Some thought will reveal that an $F[x]$-submodule of $F[x]/(x^2)$ is the same thing as an ideal of the ring $F[x]/(x^2)$. The latter sort correspond to ideals of $F[x]$ containing $(x^2)$, which are easy to list because $F[x]$ is a principal ring: $(x^2), (x), (1)$. These correspond to submodules of $F[x]/(x^2)$ of sizes $1$, $p$, and $p^2$, respectively.

Similarly, $F[x]/(x^3)$ has submodules corresponding to the $F[x]$-ideals $(x^3), (x^2), (x), (1)$; these submodules have sizes $1$, $p$, $p^2$, and $p^3$.

Allow me to commit the mild sin of identifying $x$ with its image in these quotient rings/modules. Now, the size of $p_1(W)$ gives us a lower bound on the size of $W$. So $p_1(W) = 0 \text{ or } (x)$, and $p_2(W) = 0 \text{ or } (x^2)$. Clearly $(x) \oplus 0$ and $0 \oplus (x^2)$ are two possibilities.

We need to find out whether we can have at once $p_1(W) = (x)$ and $p_2(W) = (x^2)$. I claim that this can happen in multiple ways. Each of the $p$ elements of $(x)$ (remember that I'm writing this for $(x)/(x^2) = {ax + (x^2) : a \in F}$). will correspond to a unique element of $(x^2)$, and it follows that we merely need to specify an element $(x, ax^2) \in W$, where $a \in F^*$. This gives us $p - 1$ more possibilities. There are certainly some assertions to check, here!

  • 0
    I have no idea how to determine the ideals using the method you listed. If $W$ is a submodule of $V$ then a the how do you caluclate the image without knowing something about the structure of $W$.2011-07-30
  • 1
    @user7980 I'll try to write some more. The point is that the image, whatever it is, is some submodule, and in your situation it's easy to list all of the possibilities.2011-07-30
  • 0
    Thank you very much for elaborating the projection notation was confusing me a little but now it makes alot more sense.2011-07-30
  • 0
    I still have a feeling I am counting elements incorrectly does the submodule $M$ of $F[x]/(x^2)$ corresponding to $(x^2)$ have $p^2$ elements because $M$ is isomorphic to $\{a+bx : a, b \in \mathbb{Z}_p \}$2011-07-30
  • 2
    That should have one element (it's the trivial submodule). On the other hand, that quotient has order $p^2$ for the reason you mentioned.2011-07-30