Example of $\deg(fg)<\deg(f)+\deg(g)$

Let $R$ be an integral domain and $f,g\in R[X_1,...,X_n]$ where $n>1$. What is an example of a pair $f,g$ such that $\deg(fg)<\deg(f)+\deg(g)$?

Moreover, I have proven that the units of $R[X_1,...,X_n]$ are just the units in $R$. (Proof is done inductively) Is it true?

EDIT

Even though the each term in the underlined summation is nonzero, isn't it possible that the summation is 0? Why is this impossible ?

1 Answer

If $R$ is an integral domain, then in fact for any $f,g$ in $R[X_1,\dots,X_n]$, $\deg(fg)=\deg(f)+\deg(g)$. There can only be an inequality over a ring with zero divisors. For instance, let $R$ be the integers mod $6$. Then if $f=2x^2+1$, $g=3x$, we have $fg=6x^3+3x=3x$, in which you can see the desired inequality of degrees.

If $R$ is an integral domain, then for two polynomials $f,g$ in $R[x_1,\dots,x_n]$, $$\deg(fg)=\deg(f)+\deg(g).$$

Lemma: If $R$ is an integral domain, then $R[x_1,\dots,x_n]$ is an integral domain.

I will omit the proof of this, but more can be read here.

Now let $f,g$ be members of $R[x_1,\dots,x_n]$, with $\deg(f)=k$, $\deg(g)=l$. Define a leading term in $f$ to be a term of degree $k$. For instance, if $h=x_1x_2+x_1^2+x_2$, then $h$ has degree $2$ and has two leading terms, $x_1x_2$ and $x_1^2$. Define $f^*$ as the sum over the leading terms of $f$. Continuing our example, $h^*=x_1x_2+x^2$. Note that if $f\in R[x_1,\dots,x_n]$ and $f\neq 0$, then $f^*\in R[x_1,\dots,x_n]$ and $f^*\neq0$.

Now examining the product $fg$, we can see that $$ fg=f^*g^*+\text{ remaining terms }. $$ Since $R[x_1,\dots,x_n]$ is an integral domain, $f^*g^*$ is not equal to zero. But every term in $f^*g^*$ must have degree $k+l$. Moreover, no term in the $(\text{ remaining terms })$ has degree $k+l$, so no cancellation can occur between these two parts. So there will be at least one nonzero term left of degree $k+l$, and $fg$ will have degree $k+l$.