2

I know that if p is a prime number, then the quotient ring $$\frac {\mathbb{Z}_p[x]}{ \langle x^n-1 \rangle }$$ is a vector space over $\mathbb Z_p$. What if p is not a prime? Certainly, $\mathbb{Z}_m[x]$ is not a Euclidean domain while m is not prime. Is it necessary to be a Euclidean domain in order to obtain a vector space?

faith
  • 1,006
  • 1
    No. It is not a vector space. It is a free $\Bbb{Z}_4$-module of a finite rank $n$, though. Sometimes that is good enough, but not always. – Jyrki Lahtonen May 11 '14 at 13:49
  • 1
    Let $A$ be a ring. Then $A[X]/(X^n-1)$ will be isomorphic as an $A$-module to the module $A_n[X]$ of polynomials with coefficients in $A$ of degree at most $n$. If $A$ is a field, $A_n[X]$ is a vector space of dimension $n$. If not, you still have an $A$-module with basis $1,x,x^2,\ldots,x^n$. – Pedro May 11 '14 at 13:50
  • 1
    I am a bit uncertain what you mean with the last sentence. The ring $\Bbb{Z}_4[x]$ certainly has quotient rings that vector spaces. For example $\Bbb{Z}_4[x]/\langle 2, p(x)\rangle$ for some monic polynomial $p(x)$. So it is not necessary for a ring to be domain for some of its quotient rings to be vector spaces. – Jyrki Lahtonen May 11 '14 at 13:51
  • In fact, I wanted to know whether the quotient ring $\mathbb Z_m[x]/\langle x^n-1 \rangle$ has a basis when m is not prime. Now I see, I made a big mistake. Of course it cannot be a vector space over a ring which is not a field. Thanks for your answers. – faith May 11 '14 at 13:59
  • A linearly independent generator set of a free module can be called basis. It's not such a stretch. Depends on the context. – Jyrki Lahtonen May 11 '14 at 15:57

1 Answers1

1

The problem is that you want the vector space axioms to hold. In particular, you want to be able to multiply by scalars from a field. If $p$ is not prime, then it is not obvious which field should act on the ring.

Fredrik Meyer
  • 20,228