If $f$ and $g$ are independent polynomials and $h$ is a nonzero polynomial, then $fh$ and $gh$ are independent

Question

I am doing some problems outside of class and have a couple of questions that I cannot figure out how to start.

1. If $f$ and $g$ are independent polynomials and $h$ is a nonzero polynomial over $F$, show that $fh$ and $gh$ are independent.

I think this is relatively intuitive, but cannot find a proof for it.

Since f and g are independent, that means they should each form a basis for the fields they are over (in terms of polynomials). Consequently, I believe multiplying by a nonzero polynomial is the equivalent of scaling it in the field. So, the only way they would be dependent is if $h=0$ or if $f=g=0$.

Answer 1

The question reveals a few points where your understanding may be shaky. Leading off by pointing them out:

1. For the question to make sense it seems likely to me that we are to discuss linear independence of these polynomials over the field of coefficients $F$.
2. For that to make sense those polynomials must be viewed as elements of some vector space $V$ over $F$. The natural candidate for $V$ is the ring of polynomials $V=F[x]$. But then $V$ is an infinite-dimensional space, so it is impossible for either $f$ or $g$ to form a basis for $V$. We cannot view them as elements of $F$.
3. Linear independence of two polynomials, $p_1$ and $p_2$, in $F[x]$ means that it should be impossible to find constants $c_1,c_2\in F$, not both equal to zero, such that $c_1p_1+c_2p_2=0$. Equivalently (this holds for any two vectors in any vector space) the polynomials $p_1$ and $p_2$ are linearly dependent if and only if one is a scalar multiple of the other. So for example polynomials $p_1(x)=1+x+x^2$ and $p_2(x)=2+2x+2x^2$ are linearly dependent because $2p_1+(-1)p_2=0$ or because $p_2=2p_1$. OTOH the polynomials $p_1(x)=x$ and $p_2(x)=x^2$ are linearly independent because they are of a different degree.

On with the actual exercise. Only a hint here - at least for now. Clearing up items 1-3 is IMO more important. More concepts of abstract algebra are coming up.

We are given that $f,g\in F[x]$ are linearly independent over $f$ and that $h\in F[x]$ is a non-zero polynomial. Linear independence of polynomials $fh$ and $gh$ is asking whether there might be scalars $c_1,c_2\in F$ such that $$ c_1 fh+c_2 gh=0\qquad(1) $$ in the ring $F[x]$. Because $F[x]$ is a ring we can use the distributive law to rewrite $(1)$ in the form $$ (c_1 f+c_2 g)h=0.\qquad(2) $$ Now you need to call upon a special property of $F[x]$ to deduce from this another equation where $h$ no longer appears. I feel that this is the point of this exercise, so I won't take a swing at it (but I did put the ball on the tee for you). You should arrive at an equation that allows you to deduce that we must have $c_1=c_2=0$ for $(1)$ to hold.