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Suppose we're in the multivariate polynomial ring $\Bbb C[x_1,x_2,...,x_n]$, and we construct some quotient ring. I'm interested in the following case in particular:

  • $\Bbb C[x_1,x_2,...,x_n]/(x_1^2-1,x_2^2-1,...,x_n^2-1)$

Given an arbitrary polynomial in this quotient ring, is there some general way to see if it factors within the ring?

I don't care if the algorithm is slow, I just want to understand the approach for quotient rings like these. I can't tell if this translates nicely into some easily solvable problem in the ambient complex multivariate polynomial ring, or if something else is needed here.

I'm particularly interested in polynomials where the coefficients are all $\pm 1$.

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  • $\begingroup$ What do you mean by "factor", exactly? Your ring is not a domain, and in fact is just a product of fields, so factorization is not a very natural or interesting concept in it... $\endgroup$ – Eric Wofsey May 4 '17 at 19:41
  • $\begingroup$ By "factor" I just mean finding two other polynomials within the ring such that the polynomial in question is the product of the two, mod the quotient relations. How is it a product of fields? $\endgroup$ – Mike Battaglia May 4 '17 at 19:58
  • $\begingroup$ But you presumably want to rule out some "trivial" factorizations, like $f=1\cdot f$. In a domain there is an obvious choice to consider trivial only the factorizations where one factor is a unit, but in a non-domain this tends to not be very well-behaved. Your ring is a product of fields since it is the tensor product (over $\mathbb{C}$) of $n$ copies of $\mathbb{C}[x]/(x^2-1)$ which is isomorphic to $\mathbb{C}^2$ by the Chinese remainder theorem. $\endgroup$ – Eric Wofsey May 4 '17 at 20:13
  • $\begingroup$ Thanks. One thing though - how are you using the Chinese remainder theorem here? The version of it I've been studying (from here) is basically a statement involving the sum of two or more comaximal ideals, but $\Bbb C[x]/(x^2-1)$ involves only one ideal. $\endgroup$ – Mike Battaglia May 4 '17 at 20:26
  • $\begingroup$ If $I$ and $J$ are comaximal, $R/(I\cap J)\cong R/I\times R/J$. In this case, $R=\mathbb{C}[x]$, $I=(x-1)$, $J=(x+1)$, and $I\cap J=(x^2-1)$. $\endgroup$ – Eric Wofsey May 4 '17 at 20:29
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Well, your example $R=\Bbb C[x_1,x_2,...,x_n]/(x_1^2-1,x_2^2-1,...,x_n^2-1)$ is a very easy to understand ring: it is just isomorphic to $\mathbb{C}^{2^n}$. Explicitly, given a function $e:\{1,\dots,n\}\to \{-1,1\}$, define a homomorphism $R\to\mathbb{C}$ by sending $x_k$ to $e(k)$ for each $k$. This gives $2^n$ homomorphisms $R\to\mathbb{C}$ which together give a homomorphism $R\to\mathbb{C}^{2^n}$ which is an isomorphism. (There are many ways you can show it is an isomorphism--for instance, you can show it is surjective, and then conclude it must be injective as well since $R$ and $\mathbb{C}^{2^n}$ have the same finite dimension as $\mathbb{C}$-vector spaces.)

Assuming that you say $f$ "factors" if you can write $f=gh$ where neither $g$ nor $h$ is a unit, an element $f\in\mathbb{C}^{2^n}$ factors iff at least one of its coordinates is $0$. Indeed, if the $e$th coordinate of $f$ is $0$, then $f$ factors as $fh$ where $h$ is $1$ in every coordinate except the $e$th and $0$ in the $e$th coordinate. Conversely, if all the coordinates of $f$ are nonzero, $f$ is a unit, so $f=gh$ implies $g$ and $h$ are both units.

Explicitly in terms of polynomials, this means that $f\in R$ factors iff $f(\pm1,\pm1,\dots,\pm1)=0$ for some choice of signs.

If by "factors" you mean you can write $f=gh$ where neither $g$ nor $h$ is a unit multiple of $f$, then $f\in\mathbb{C}^{2^n}$ factors iff at least two of its coordinates are $0$. Indeed, if the $e$ and $e'$ coordinates are $0$, let $g$ be $f$ modified so that its $e$ coordinate is $1$, and let $h$ be $0$ on its $e'$ coordinate and $1$ on all others. Conversely, if $f$ has at most one nonzero coordinate and $f=gh$, then one of $g$ and $h$ must have the same set of nonzero coordinates as $f$, and thus be a unit multiple of $f$.

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