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Let $R$ be a commutative ring with unity. A finite sequence $x=\left< x_0,\dots,x_n\right>$ with elements in $R$ is called to be prime if there exists $a_0,\dots,a_n \in R$ such that $\sum_{i=0}^n a_i x_i =1$.

Prove that for any finite sequences $x,y$, the discrete convolution $x \ast y$ is prime if and only if $x,y$ are both prime.

My thought: If we show that $x$ is prime if and only if a polynomial with coefficient $x$ satisfies a specific 'condition' and the 'condition' is preserved under multiplication, then we can finish as follows: $x \ast y$ is prime if and only if a polynomial with coefficient $x \ast y$ (which is a product of two polynomials with coefficients $x$ and $y$, respectively) satisfies the 'condition' if and only if two polynomials both satisfy the 'condition' if and only if $x,y$ are both prime.

However, I cannot find any possible conditions. Is this approach right?

EDIT: The original problem is proving "or disproving" the given statement.

user26857
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Actually you want to prove the following: $\langle a_0,\dots,a_m\rangle=R$ and $\langle b_0,\dots,b_n\rangle=R$ iff $\langle a_0b_0,a_0b_1+a_1b_0,\dots,a_{m-1}b_n+a_mb_{n-1},a_mb_n\rangle=R$. (Here $\langle S\rangle$ denotes the ideal generated by a subset $S$ of $R$.)

If $f(X)=a_0+a_1X+\cdots+a_mX^m$ is the polynomial associated to the sequence $a_0,\dots,a_m$, then your question reduces to the following:

Let $f,g\in R[X]$. Then $f$ and $g$ are primitive iff $fg$ is primitive.

For a proof of this result see Product of two primitive polynomials.

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user26857
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