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My books defines the gcd of a polynomial like this:

Consider the ideal $J = K[x]\cdot p_1(x)+\cdots + K[x]\cdot p_m(x)$ of $K[x]$ generated by the polynomials $p_1, \cdots, p_m(x)$

If $d(x)\in K[x]$ is such that $J=K[x]\cdot d(x)$ then such properties are valids:

a) $\exists r_1(x), \cdots, r_m(x) \in K[x]$ such that $d(x) = r_1(x)\cdot p_1(x)+\cdots+r_m(x)\cdot p_m(x)$

b) $d(x)$ is a common divisor of $p_1(x),p_2(x),\cdots,p_m(x)$

c) if $d'(x)$ is a common divisor of $p_1(x),p_2(x),\cdots,p_m(x)$ then $d'(x)$ is also a divisor of $d(x)$

Then, the book says that a polynomial satisfying $b$ and $c$ is called a gcd of $p_1(x),p_2(x),\cdots,p_m(x)$ in $K[x]$

I understand the intuitive definition of the gcd of a polynomial, but could you explain the case $c$ better? Also, by books says that if $d(x)$ is a gdc, then $a\cdot d(x)$ is also a gcd. Can you give me an example with real polynomials?

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  • $\begingroup$ You might consider creating yourself an example with real polynomials. That's a good way to understand what the statements are saying It also could help to break down the statements above -- what each piece (J, K, d, d'...) since there's a lot there, and it's not presented in the most cohesive manner (in my opinion). $\endgroup$
    – Nate 8
    Commented Dec 30, 2015 at 2:00
  • $\begingroup$ Case (c) is really just saying that $d(x)$ (the GCD of the $p_i$) is a maximal element of the set of divisors of the $p_i$ under the partial ordering $\le$ given by $f(X)\le g(X)$ if $f(X)$ divides $g(X)$. If $d'(X)$ is a common divisor of the $\{p_{i}(X)\}$, then $d'(X)\le d(X)$. That's how I think of it. $\endgroup$
    – Nobody
    Commented Dec 30, 2015 at 2:23
  • $\begingroup$ There is no gcd of a polynomial. There must be at least two! $\endgroup$
    – user26857
    Commented Dec 30, 2015 at 10:19

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