Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A doubt: let $k$ be an algebraically closed field and let $p(x) \in k[x]$ be a polynomial of degree $n \geq 1$ without repeated roots. The quotient $k[x]/(p(x))$ is isomorphic, as a vector space, to $k^{n}$ right? because we can factor $p(x)$ as a product of linear factors and then use the chinese remainder theorem. Is this correct?

share|cite|improve this question
To use the chinese remainder theorem, the factors needs to be comaximal. If $p(x) = (x - a)^n$, then each of the factors are not comaximal. If $p(x)$ is separable, then this is true. – William May 22 '12 at 4:55
@William: just added that $f$ has no repeated roots. – user31509 May 22 '12 at 4:56
In that case, yes. – Arturo Magidin May 22 '12 at 4:59
Correct. But as a vector space $k[x]/\langle p(x)\rangle$ is isomorphic to $k^n$ irrespective of whether $p(x)$ has repeated factors and irrespective of whether $k$ is algebraically closed. This is because each coset has a unique polynomial of degree $<n$ in it, and those form a vector space od dimension $n$. – Jyrki Lahtonen May 22 '12 at 5:01
Also note that $k[x] \slash (p(x))$ is thought of as a quotient of rings. – William May 22 '12 at 5:01
up vote 2 down vote accepted

Regardless of whether $p(x)$ is squarefree or not, and whether $k$ is algebraically closed or not, $k[x]/\langle p(x)\rangle$ has dimension $\deg(p)$: this follows from the division algorithm.

If $k$ is algebraically closed and $p(x)$ is squarefree, then you actually get that $k[x]/\langle p(x)\rangle$ is isomorphic to $k^n$ as rings: the map into the product is a ring homomorphism.

For general $k$ and $p(x)$ squarefree, you get a product of extensions of $k$, and again an isomorphism as rings.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.