I'm working on this question:
Let $F$ be a (commutative) field. Show that the ring $F[x]$ of polynomials with coefficients from F has no zero-divisors. In other words, if $P(x)$ and $Q(x)$ are polynomials in $f[x]$, then $P(x)Q(x)$ cannot be the zero polynomial unless either $P(x)$ or $Q(x)$ is the zero polynomial.
I let $P(x)$ and $Q(x)$ be such polynomials in $F[x]$ such that $$P(x) = a_0x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n$$ and $$Q(x) = b_0x^m + b_1x^{m-1} + ... + b_{m-1}x + b_m$$
Then we have that $$P(x)Q(x) = a_0b_0x^{n+m} + a_0b_1x^{n+m-2}+...$$
I was thinking to split this into $2$ cases, one where I let $a_0 = 0$ and the second where $b_0 = 0$ Is this the right approach?
Taking two nonzero polynomials $P$ and $Q$ such that $\deg P = n$ and $\deg Q = m$, we have $$P(x) = a_0x^n + \dotsb \qquad Q(x) = b_0x^m + \dotsb$$ where $a_0$ and $b_0$ are nonzero. Since $F$ is a field, the product $a_0b_0$ cannot be zero, so the product $PQ$ will be a degree $n+m$ polynomial with leading term $a_0b_0$, so it's not the zero polynomial.