# Dimension of quotient vector space

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?

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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

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.