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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?

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  • 2
    $\begingroup$ 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. $\endgroup$ – Don Thousand Nov 3 at 3:22
  • $\begingroup$ @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\ \ $ $\endgroup$ – Bill Dubuque Nov 3 at 3:25
  • $\begingroup$ @BillDubuque Sorry, the expression you've written isn't rendering for me/ $\endgroup$ – Don Thousand Nov 3 at 3:26
  • $\begingroup$ @Don Alternatively $\,\gcd(f,g) = f\gcd(1,f/g) = f.\ $ It renders fine here. $\endgroup$ – Bill Dubuque Nov 3 at 3:30
  • $\begingroup$ That only works when $f\neq0$, no? $\endgroup$ – Don Thousand Nov 3 at 3:31

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