Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $F$ be a field. Let $p$ and $q$ be monic members $F[x]$. Let $I = \{p\cdot r : r\in F[x]\}$ and $J = \{q\cdot r : r\in F[x]\}$. I know that if $F[x]/I$ is isomorphic to $F[x]/J$ then ($\operatorname{deg}(p) = \operatorname{deg}(q)$ and (p is reducible if and only if q is reducible)). For cases where we do have $\operatorname{deg}(p) = \operatorname{deg}(q)$ and (p is reducible if and only if q is reducible), is there any faster way to check whether $F[x]/I$ is isomorphic to $F[x]/J$ than seeing if any element of $F[x]/I$ has q as its minimal polynomial?

(In the particular cases I'm interested in, $F$ is very finite.)

share|improve this question
    
For finite fields, factorization is what you need. –  Hurkyl May 16 '11 at 5:19

1 Answer 1

up vote 7 down vote accepted

If $F$ is finite, $p$ and $q$ are irreducible, and have the same dagree $d$, then $F[x]/I$ and $F[x]/J$ are isomorphic $F$-algebras, since there is (up to isomorphism) only one extension field of a finite field of any given degree (see for example Birkhoff & Mac Lane, A survey of Modern Algebra, Section 15.6).

Things get a little more interesting when $p$ and $q$ are not irreducible. Suppose that $p=p_1^{n_1}p_2^{n_2}\ldots$. Then $F[x]/I$ is isomorphic to $\bigoplus_i F[x]/(p_i^{n_i})$. It turns out that, for any perfect field $F$, if $E_i=F[x]/(p_i)$, then $F[x]/(p_i^{n_i})$ is isomorphic to $E_i[t]/(t^{n_i})$ (this result is well-known, if not well-documented; a proof appears here (see Theorem A.19)). Thus for finite fields the general answer to you question is the following:

For finite fields

$F[x]/I$ and $F[x]/J$ are isomorphic if and only if the numerical invariants of the polynomials $p$ and $q$ are the same, in other words, if $p=p_1^{n_1}p_2^{n_2}\ldots p_r^{n_r}$ and $q=q_1^{m_1}q_2^{m_2}\ldots q_s^{m_s}$ are the decompositions of $p$ and $q$ into irreducible factors, then there is a bijection $w:\{1,\ldots,r\}\to \{1,\ldots,s\}$ such that the degree of $q_{w(i)}$ is the same as the degree of $p_i$ and $m_{w(i)}=n_i$ for all $i\in \{1,\ldots,r\}$.

For general perfect fields, this does not give a complete solution. What we get is

For perfect fields

$F[x]/I$ and $F[x]/J$ are isomorphic if and only if when$p=p_1^{n_1}p_2^{n_2}\ldots p_r^{n_r}$ and $q=q_1^{m_1}q_2^{m_2}\ldots q_s^{m_s}$ are the decompositions of $p$ and $q$ into irreducible factors, then there is a bijection $w:\{1,\ldots,r\}\to \{1,\ldots,s\}$ such that $F[x]/(q_{w(i)})$ is isomorphic to $F[x]/(p_i)$ and $m_{w(i)}=n_i$ for each $i\in \{1,\ldots,r\}$.

Thus the problem is reduced to determining whether or not two irreducible polynomials give rise to isomorphic field extensions.

Edit: Ricky Demer pointed out that the "only if" parts of the above assertions are not obvious, and so here is a way out: We can recover the subalgebras $F[x]/(p_i^{m_i})$ from $F[x]/I$ as the indecomposable ideals. It remains to show that if $p_i$ is irreducible then the field $E_i=F[x]/p_i$ and the integer $m_i$ can be recovered from $F[x]/(p_i^m)$. This is not hard:

  1. The field $F[x]/p_i$ can be recovered from $F[x]/(p_i^n)$ as the quotient of the algebra by its radical.

  2. The integer $m_i$ can be recovered as the length of the ring $F[x]/p_i^{m_i}$.

share|improve this answer
1  
How do you get "and only if"? –  Ricky Demer May 16 '11 at 5:48
    
@Ricky: for finite fields I think you can do it by counting the number of roots of various polynomials and induction. Alternately, you can apply the structure theorem to modules over $F[x]$ (since if $\phi : F[x]/I \to F[x]/J$ is an isomorphism of rings you get an isomorphism of $F[x]$-modules by letting $x$ act as $\phi(x)$ on $F[x]/J$.) –  Qiaochu Yuan May 16 '11 at 5:52
    
Oh, you also need to transport $F$, I guess. –  Qiaochu Yuan May 16 '11 at 5:52
1  
re edit: We still need to show that the factors ($E_i$ and $n_i$) can be recovered from $F[x]/I$. –  Ricky Demer May 16 '11 at 6:49
2  
The fact that (for $p(x) \in F[x]$ irreducible) $F[x]/p^n$ is isomorphic to $E[t]/t^n, E = F[x]/p$ is (essentially equivalent to and) a consequence of the fact that the completion of $F(x)$ at the place $p$ is isomorphic to the power series field $E((t))$. This is the well-known characterization of equicharacteristic complete discretely valued fields. This you can find in most books on local fields, e.g., Serre's. –  Felipe Voloch May 16 '11 at 11:13

Your Answer

 
discard

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.