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Let $f (x)$ and $g(x)$ be nonzero polynomials in $Q[x]$. Consider the gcds of $f (x)$ and $g(x)$ in $Q[x]$, $R[x]$ and $C[x]$. Must these gcds be the same, or can they be different?

My attempt

Clearly gcd in $Q[x]$ is a divisor in $R[x]$ and $C[x]$.

But if we consider $f(x) = x^3-12$ . Clearly it is irreducible in $Q[x]$. Thus only divisors are units and itself. Then if we take g(x) in such a way that g(x) belongs to Q[x]. And g(x) contains the real root of f(x) and some other roots in a way that g(x) is not equal to f(x). Then we are done. Then the gcds need not be equal.

However , I cant get any such g(x). Does such g(x) exists? or the gcds are same?

Lalit Tolani
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Nick Diaz
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    The gcd can be found using the Euclidean Algorithm. Think about the division algorithm. Will you ever need any number that is not rational? – Arturo Magidin Aug 03 '21 at 16:43
  • There is no need. Thanks. So the GCDs are the same – Nick Diaz Aug 03 '21 at 16:56
  • Insofar as any polynomial of $\Bbb Q[x]$ can be the same as a polynomial over $\Bbb R[x]$. Most people accept this without discussion, me among them. But technically, if one is to be embarrassingly pedantic, they have no elements in common. – Arthur Aug 03 '21 at 17:12
  • @Arthur doesn't that depend on how you define the sets involved? – Asinomás Aug 03 '21 at 17:39
  • @Yorch It does. But I know of no standard concrete construction of the reals and of the rationals for which we truly have $\Bbb Q\subseteq \Bbb R$. However, you might go ahead and define both abstractly through properties like "A Dedekind-complete ordered field, and its smallest subfield". – Arthur Aug 03 '21 at 17:45

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