Show that a polynomial ring in one indeterminate over a field doesnt have zero divisors.

Check my proof please.

We have a polynomial ring $K[X]$ over a field $K$. Let $p,q\in K[X]$ two non-zero polynomials, i.e. $p\neq 0$ and $q\neq 0$. Then we must prove that $p q\neq 0$.

We define polynomials as formal power series

$$p:\Bbb N_0\to K,\quad n\mapsto p_n$$

with a finite number of non-zero values, then if $p\neq 0$ exist some $n\in\Bbb N_0$ with $p_n\neq 0$ such that for all $k>n$ we have $p_k=0$. Then $n$ is the degree of $p$ and we write $\deg(p)=n$.

The multiplication of polynomials is defined as

$$pq=z\implies z_n=\sum_{k=0}^n p_k q_{n-k}$$

Now suppose that $\deg(p)=n$ and $\deg(q)=m$ and without lose of generality $n\ge m$, and we define $z=pq$. Then we have that

$$z_{n+m}=\sum_{k=0}^{n+m}p_k q_{n+m-k}=p_n q_m$$

Then it is enough to show that a field dont have zero divisors, i.e. doesnt exist $a,b\in K\setminus \{0\}$ such that $ab=0$, other way we have that $a^{-1}ab=0$ what implies $b=0$, contradicting that $b\neq 0$.

Then $p_nq_m\neq 0$ what implies that $pq\neq 0$.

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    $\begingroup$ This proof is fine. $\endgroup$ – Matt Samuel Sep 21 '16 at 1:18
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    $\begingroup$ A commutative ring without zero divisors is called an integral domain. The univariate polynomials over a field are even more special than that, forming a Euclidean domain. $\endgroup$ – hardmath Sep 22 '16 at 19:54

The crux of your proof is to show the product $z = pq$ of nonzero polynomials has a nonzero coefficient (and thus is not the zero polynomial). You summarize this fact with the line:

$$ z_{n+m}=\sum_{k=0}^{n+m}p_k q_{n+m-k}=p_n q_m $$

A bit more detail about why all but one of the terms in that sum disappear would be welcome, but the argument is correct. It might also be worth noting, though not essential to the argument, that this nonzero coefficient is the leading coefficient of the product $z$.


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