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Assume $P_\mathbb R$ is the set of all polynomials which have only real coefficients and only real roots. Define $0$ as a polynomial with infinitely many real roots and all other constant polynomials as having no roots, thus no complex roots either.

Say we want to find an operation $\star$ such that $(P_\mathbb R,\star)$ is a group.

Let's start looking with some obvious ones. If we select $\star = +,$ ie. addition of polynomials, we immediately see that $(P_\mathbb R,+)$ is not a group. Take as an example $p_1 = 1+x$ and $p_2 = x^2.$ Then clearly $p_i \in P_\mathbb R$ but $p_1+p_2 = 1+ x+ x^2$ has no real roots. However, construction $(P_\mathbb R,+)$ has an identity element $0$ and every member has an inverse, namely inverse of $p$ is $-p$. It also has associativity. Thus $(P_\mathbb R,+)$ is a groupoid.

If we select $\star = \times$, meaning multiplication of polynomials, then clearly $(P_\mathbb R,\times)$ is closed. If $p_1$ has roots $r_1, r_2 \dots r_n$ and $p_2$ has roots $R_1, R_2 \dots R_k$, then $p_1 p_2$ has roots $r_1, r_2 \dots r_n, R_1, R_2 \dots R_k$. These obviously need not be distinct.

So multiplication seems promising. We also have an identity element, namely constant polynomial $1$. $$1\times p = p \times 1 = p.$$ Multiplication has associativity.

But finding an inverse of $p \in P_\mathbb R$ seems difficult, if not impossible. Lets say we want to invert $1+x$. We need a polynomial $p \in P_\mathbb R$ such that $(1+x)p = 1.$ Solving $p$, we get $p = \frac{1}{x+1},$ which is not a polynomial.

Or is it? We have the Taylor series of $\frac{1}{x+1}$ around $0$ given by $\sum_{k=0}^\infty (-1)^k x^k,$ and, indeed, $$(x+1)\times\sum_{k=0}^\infty (-1)^k x^k = 1$$ as long as $-1<x<1.$ But I have serious doubts that $$\sum_{k=0}^\infty (-1)^k x^k \in P_\mathbb R.$$

Looking at above, it is doubtful $(P_\mathbb R, \times)$ can be a group, no matter how the inverse is defined. If not, we get that $(P_\mathbb R, \times)$ is a monoid.

Is there any way to interpret the inverse of a polynomial in the multiplicative case, that also preserves the group structure? The answer seems to be no, as if we look at the polynomial ring over field of real numbers under multiplication and addition, the only elements with multiplicative inverses are constant polynomials with exception of $0$, which has no multiplicative inverse. As this preserves the properties of multiplication in that ring, multiplication doesn't give a group, even if $0$ is removed.

Does there exists a subring of the polynomial ring under the addition and multiplication defined in the field of real numbers, restricted to some $P^* \subset P_\mathbb R$? In other words, is there a restriction to some subset of polynomials with real coefficients and real roots such that this set forms a group under addition and a monoid under multiplication.

Valtteri
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    Note that even if you allow formal power series and not just polynomials, you won’t get inverses for all non-zero elements: $x$ has no inverse. More generally, nothing with constant term $0$ has an inverse. – Brian M. Scott Jun 14 '15 at 21:38
  • The question is, what properties do you want the group operation to have? Since every set has a group structure, it's certainly possible to define a group operation, but it may not have anything to do with the fact that the elements of the group are polynomials. – Jim Belk Jun 14 '15 at 21:55
  • @JimBelk how about something that has something to do with the fact that the elements of the group are polynomials? If we look at the addition, we see the operation doesn't preserve the roots at all. On the other hand, multiplication preserves them completely. So, for a starting point, this mystery operation should keep the root structure somewhat intact; at least the roots of the result should depend on the roots of the components. Would also be nice if the group had some nontrivial subgroups, for example, polynomials with rational coefficients. – Valtteri Jun 14 '15 at 22:04
  • If you change the condition to "only real roots and real poles", then you can just allow division as well as multiplication, but you have to exclude 0. – Matt Samuel Jun 14 '15 at 22:10
  • I changed the question to a hopefully more constructive one – Valtteri Jun 14 '15 at 22:44
  • If you assume AC, then every non-empty set has a group structure. Your particular set has cardinality of the continuum, so you can find a bijection to the real numbers, and then pull back the additive group of the reals to your set. Or the multiplicative group of the non-zero reals, if you pick a bijection to those, or any other group of continuum order (so you could get non-abelian ones, say). – zibadawa timmy Jun 14 '15 at 23:12

1 Answers1

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Not a complete answer, just some remarks too lengthy for a comment.

A non-zero polynomial with all real roots corresponds uniquely to a finite (possibly empty) set of real numbers (the roots) tagged with positive integers (the order of vanishing), together with a non-zero real number (the leading coefficient). That is, $$ c\prod_{k=1}^{N} (x - a_{k})^{m_{k}} \leftrightarrow\bigl\{(a_{1}, m_{1}), \dots, (a_{N}, m_{N}), c\bigr\}. $$ Alternatively, the data on the right may be viewed as a non-zero real number $c$ together with an "order-of-vanishing" function $m$ from the reals to the non-negative integers that is non-zero at only finitely many points. (In the notation of the preceding equation, $m(a_{k}) = m_{k}$ for $k = 1, \dots, N$, and $m(x) = 0$ for all other real $x$.)

The product operation has the pleasant interpretation of multiplying the leading coefficients and adding the order-of-vanishing functions. It should be clear why there are no inverses, however: The identity element has coefficient $c = 1$ and $m \equiv 0$, but $\times$ does not decrease the order of vanishing at any point.

If you're looking for a binary operation that "keeps the root structure somewhat intact", the data $(c, m)$ may be useful for ensuring your set is closed under your operation....

Andrew D. Hwang
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