I've been given the question:
Choose a Ring which is not an integral domain. Find non-zero polynomials $f,g \in R[x]$ such that $fg$ is non zero and $\deg(fg) \neq \deg(f) + \deg(g)$.
So to my understanding, if $a$ and $b$ are two non-zero elements of a ring R such that $ab = 0$, then $a$ and $b$ are called zero divisors. And an integral domain is a ring without said divisors.
Therefore for a ring to be a non-integral domain, I assume it would then contain these divisors, such that $ab = 0 \neq a$ and $0 \neq b$.
However I get stuck here, The question states that $f,g \neq 0$ and $fg \neq 0$, I can clearly see that you can have cases where (since it is not an integral domain) we could have $fg = 0 $ but $f,g \neq 0$ and be done with it. However I'm unsure how to show that these polynomials exist without this. As I can't think of any examples where $\deg(fg) = \deg(f) + \deg(g)$ doesn't hold
Best Answer
Well, work over the ring ${\Bbb Z}_6$, which has units $1,5$ and zero divisors $2,3,4$.
Take $f=2x+1$ and $g=3x+1$. Then $fg= 6x^2 + 2x+3x + 1 = 5x+1$ in ${\Bbb Z}_6[x]$.
Here is the generalization: Let $R$ be a commutative ring with zero divisors. Take the polynomials $$f = a_nx^n+\ldots+a_1x+a_0,\quad a_n\ne 0,$$ and $$g = b_mx^m+\ldots+b_1x+b_0, \quad b_m\ne 0,$$ in $R[x]$ with degrees $m,n\geq 1$. Then the product has the form $$fg = a_nb_m x^{n+m}+\mbox{(terms of order $<n+m$ in $x$)}.$$ If $a_nb_m=0$ (i.e., both zero divisors), then $\deg(fg)<n+m = \deg(f)+\deg(g)$.