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Let $n,k > 1$ be positive integers. Define the reduced polynomial rings :

$g^k_n = \Bbb R[X_n]/(G^k_n(X_n))$ where $G^k_n$ is a real polynomial of degree $n$ (that keeps the ring reduced). (k is just an index , not a power or such)

The question is how many nonisomorphic reduced polynomial rings of degree $n$ are there ? Lets call that $f(n)$.

With nonisomorphic I mean that the elements of the rings (between any 2 rings) are not linearly dependant.

For instance $ \Bbb R[X]/((X)^3-1)$ and $ \Bbb R[Y]/((Y)^3+1)$ are not linearly dependant and thus nonisomorphic. (one of those 2 rings is algebraicly closed , while the other is not hence they cannot be linearly dependant)

Is $f(n)=n$ ?

Im not sure if isomorphic is the correct terminology for what I want. If not, what is ? And is it still a kind of morphism ?

Thanks in advance.

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  • $\begingroup$ Note that $X^3-1=(X-1)(X^2+X+1)$ and $Y^3+1=(Y+1)(Y^2-Y+1)$ are reducible. $\endgroup$ – Servaes Jun 11 '14 at 13:00
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    $\begingroup$ Also, it is not clear to me what you mean by "elements of the rings (between any $2$ rings) are not linearly dependent.". The word 'nonisomorphic' already has a very clear meaning. $\endgroup$ – Servaes Jun 11 '14 at 13:01
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    $\begingroup$ Neither $\Bbb{R}[X]/(X^3-1)$ or $\Bbb{R}[Y]/(Y^3+1)$ is algebraically closed; they aren't even fields. Moreover they are isomorphic as rings, by the map given by $X\ \longmapsto\ -Y$. So what do you mean by 'nonisomorphic' and 'linearly dependent'? $\endgroup$ – Servaes Jun 12 '14 at 8:46
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    $\begingroup$ And it is still not clear to me what the role of $k$ is in this whole question. $\endgroup$ – Servaes Jun 12 '14 at 8:47
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    $\begingroup$ This is a train wreck, wow. $\endgroup$ – blue Jun 27 '14 at 21:21
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Any polynomial $f\in\Bbb{R}[X]$ with $\deg(f)>2$ is reducible over $\Bbb{R}$, so there are no reduced polynomial rings of degree $n>2$. If $\deg(f)=2$ and $f$ is irreducible, then we have $\Bbb{R}[X]/(f)\cong\Bbb{C}$. Hence there is only one reduced polynomial ring of degree $2$, up to isomorphism, and there are no reduced polynomial rings of degree $n$ for $n>2$. Put differently, we have $$f(n)=\left\{\begin{array}{ll}1&\text{ if }n=2\\0&\text{ if }n>2\end{array}\right..$$

EDIT: As the question has been modified, I'll add another answer:

The quotient $\Bbb{R}[X_n]/(G_n(X_n))$ is reduced if and only if $G_n(X_n)$ is a product of distinct irreducible factors. Suppose $P_1,\ldots,P_m\in\Bbb{R}[X_n]$ are distinct irreducible polynomials such that $$G_n(X_n)=\prod_{i=1}^mP_i(X_n),$$ then by the Chinese remainder theorem we have $$\Bbb{R}[X_n]/(G_n(X_n))\cong\prod_{i=1}^m\Bbb{R}[X_n]/(P_i(X_n))\cong\Bbb{R}^r\times\Bbb{C}^s,$$ where $r$ and $s$ are the numbers of linear resp. quadratic irreducible factors of $G_n$. Hence the number $f(n)$ of non-isomorphic reduced polynomial rings of degree $n$ is precisely the number of ways we can write $n=r+2s$ with $r,s\in\Bbb{Z}_{\geq0}$, which is precisely $\lfloor\tfrac{n}{2}\rfloor+1$.

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  • $\begingroup$ Thanks for your answer. Im sorry I made an awful mistake. I edited the question. I hope it is better now. I added my doubts about the use of the word " isomorphism " and removed the silly restriction on the polynomial $G^k_n$ that it should be irreducible. Maybe you can take a look at it again now ? Thanks. $\endgroup$ – mick Jun 11 '14 at 19:17

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