# gcd of $f(x)$ and 0 polynomial

When the integer case,

$$\gcd (a, 0 ) = a$$ for $$a( \neq 0) \in \mathbb Z$$

Plus people generally said $$\gcd(0,0)$$ can't be defined.

Then...

When we expand this consideration by the polynomials ring $$F[x]$$ for a field, $$F$$

What about the case $$\gcd(f(x), 0)$$ ? (Here the $$f (\neq 0) \in F[x]$$, $$0$$ is $$0$$ a polynomial in $$F[x]$$)

plus Could I regard the case $$\gcd(0,0)$$ Can't be defined like the integer case?

• Yea, it's the same thing. I've seen some people define $\gcd(0,0)$ as $0$, implictly viewing $0$ as a divisor larger than any other divisor. But it's a weird construction that doesn't really offer much computational advantage, so most mathematicians don't do that. – Don Thousand Nov 3 at 3:22
• @Don $d\mid f,0\iff d\mid f\$ holds even when $\,f=0.\$ More generally note $\gcd(f,g) = f\,$ when $\,f\mid g\,$ since then: $\ \ d\mid f,g\iff d\mid f\ \$ – Bill Dubuque Nov 3 at 3:25
• @BillDubuque Sorry, the expression you've written isn't rendering for me/ – Don Thousand Nov 3 at 3:26
• @Don Alternatively $\,\gcd(f,g) = f\gcd(1,f/g) = f.\$ It renders fine here. – Bill Dubuque Nov 3 at 3:30
• That only works when $f\neq0$, no? – Don Thousand Nov 3 at 3:31