If $f(x) \in K[x]$ and $g(x) \notin K[x]$, can $f(x)g(x) \in K[x]$?

Question

Let $K$ be a subfield/ring of a field/ring $F$. Is there a nontrivial example of the product of a polynomial over $K$ and a polynomial not over $K$ giving a polynomial over $K$?

For a simple version is there a polynomial with integer coefficients that multiplied buy a polynomial with some real coefficients gives integer coefficient?

I'm guessing the case that $K$ is a field is more difficult because we have inverses unlike $\Bbb Z.$ Is there a counterexample with $K=\Bbb Q$ and $F=\Bbb R$ or $\Bbb C$?

@lulu Perfect, thank you! I'm guessing the case that $K$ is a field is more difficult because we have inverses unlike $\mathbb{Z}$. Is there a counterexample with $K=\mathbb{Q}$ and $F=\mathbb{R}$ or $\mathbb{C}$? — A.B, Feb 02 '21 at 18:33
@A.B The product of an irrational and a rational will always yield an irrational, so you're not going to find product of a polynomial with rational coefficient and a polynomial with irrational coefficients that is a polynomial with rational coefficients. Similarly in the case where $F = \mathbb{C}$ in order to get a multiple of a polynomial with imaginary coefficients to have rational coefficients you'll need both terms of the product to have imaginary coefficients. So in short, no — Noah Solomon, Feb 02 '21 at 18:39
Also in your question by real coefficients do you mean coefficients in $\mathbb{R}\setminus\mathbb{Z}$? Otherwise there are lots of trivial examples — Noah Solomon, Feb 02 '21 at 18:45
@NoahSolomon Thank you. Is there a way to see that it's not possible to have cancellation of the irrational factors after multiplication? Naively I thought it might be possible for (let's say) a term of degree 5 with irrational coefficient to be cancelled by the product of a degree 2 degree 3 term, or a degree 1 degree 4 term etc. — A.B, Feb 02 '21 at 18:48
@NoahSolomon yes I meant $\mathbb{R} \setminus \mathbb{Z}$ but what if only some of the coefficients are in $\mathbb{R}/\mathbb{Z}$? — A.B, Feb 02 '21 at 18:51
@A.B this question's been closed now, but the linked question is very general. The short answer to your specific question is that as long as at least one of the coefficients is not in $\mathbb{R}\setminus\mathbb{Z}$ then the product will have irrational coefficients. To see this you can think about applying the binomial formula along with the fact that any nontrivial $\mathbb{Q}$-linear combination of irrationals must be irrational. Try proving this for yourself if you're not familiar with that result, it's a good short proof. — Noah Solomon, Feb 02 '21 at 19:02
See also here. If any questions remain unanswered then please elaborate. — Bill Dubuque, Feb 02 '21 at 19:07

If $f(x) \in K[x]$ and $g(x) \notin K[x]$, can $f(x)g(x) \in K[x]$?

0 Answers0