Consider the non-zero polynomials $P(t)=at+b$ and $Q(t)=ct+d$ in $F[t]$ such that at least one of them is non-constant. Then, $(P(t),Q(t))=1$ if and only if $ad-bc\neq0$.
I'm stuck on this problem and any help will be appreciated. Thanks.
Update: Thanks for all the helpful answers! I figured out one direction. Here's a proof that I came up for the other direction:
($\Longrightarrow$) Suppose $(P(t),Q(t))=1$. Assume, for a contradiction, that $ad-bc=0 \Longrightarrow ad=bc$.
By assumption, at least one of the polynomials is non-constant. WLOG, assume that $P(t)$ is non-constant. Then, $a \neq 0$. Now, observe that $dP(t)=bQ(t)$, since $ad=bc$.
If $b=0$, then $P(t)=at$ and $ad=0 \Longrightarrow d=0$, since $a \ne 0$, by assumption. Then, we have $Q(t)=ct$ and since both polynomials are assumed to be non-zero, $c \ne 0$. Then, $t$ divides both $P(t)$ and $Q(t)$, a contradiction.
Therefore, $b \ne 0 \Longrightarrow Q(t)= \left(\frac{d}{b}\right) P(t) \Longrightarrow P(t)$ divides $Q(t)$. Thus, $(Q(t),P(t))=P(t) \ne1$, a contradiction.
Thus, it follows that $ad \ne bc$.
Is this right?
Note first that because $P$ and $Q$ are of degree at most $1$, their only possible divisors are polynomials of degree $1$, or constants.
Consider the vectors $(a,b)$ and $(c,d)$ in $F^2$. They are linearly dependent if and only if the corresponding $2\times 2$ matrix is singular, i.e., if and only if the determinant $ad-bc$ is zero.
When are two vectors linearly dependent? When one is a constant multiple of the other.
So, the determinant is nonzero if and only if neither $P$ divides $Q$, nor $Q$ divides $P$; if and only if their only common divisors are constant.