# Definition of gcd for a polynomial and $\gcd(3,x) = 1$

I'm following this post about the definition of gcd for a polynomial:

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 valid:

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?

For the $a)$ affirmation, my book says that it comes directly from the equality:

$$K[x]\cdot d(x) = K[x]\cdot p_1(x) + \cdots +K[x]\cdot p_m(x)$$

Why is it? I think that it's because I could choose the polynomial $1$ in the left, and any other polynomials in the right, to end up with something like this:

$$1\cdot d(x) = r_1(x)\cdot p_1(x)+\cdots r_m(x)\cdot p_m(x)$$

but how can I know that the polynomial $1$ isindeed in $K[x]$?

The latter ($c$) is the definition of gcd: we find a common divisor $d(x)$ of the set we want to take the gcd, and them if this polynomial has the property that every other common divisor $d'(x)$ of the set also divides it, then $d(x)$ is our common divisor.

It then proves that the domain $A=\mathbb{Z}[x]$ is not a principal ideal domain by considering that if it were we'd have:

$$A\cdot d(x) = A\cdot 2 + A\cdot x$$

which would imply that $d(x)$ is the gcd of $2,x$ but we know that it must be $1$ because $2$ is prime and does not divide $x$. It then says that we cannot have $p(x)$ and $q(x)$ with integer coefficients such that $1 = 2p(x) + xq(x)$ because the independent term of the expression $2p(x)+xq(x)$ is always even. (WHY?????).

Later, the exercise asks me to prove that

$$1 = \gcd(3,x)$$

In the first case we had that $J = K[x]\cdot p(x)$ being an ideal and being generated by a $d(x)$ implied $d(x)$ was the gcd. I can't use the converse here, that is, not having $J$ being generated by $d(x)$ will imply that it does not have gcd (or it is $1$). I also can't just say that the only polynomial that divides $3$ and $x$ is $1$ because I'm asked to prove it. In the proof above, it used the fact that $2$ and $x$ have a gcd of $1$. How can I do it?

Note that your book says that the greatest common divisor of two (or more) polynomials only needs to satisfy $(b)$ and $(c)$, without mentioning $(a)$ at all. And since $(b)$ and $(c)$ do not involve anything more than the simple concept of divisibility, you are allowed to prove that $1 = \gcd(3,x)$ just by showing that $1$ and $-1$ are their only common divisors.
As for your other question, note that every element of $(2,x)$ is of the form $2\cdot p(x) + x\cdot q(x)$, with $p(x),q(x)\in\mathbb{Z}[x]$. The independent term of $x\cdot q(x)$ is clearly $0$ (if $q(x)=\sum_{i=0}^na_ix^i$, then $$x\cdot q(x) = x\cdot\sum_{i=0}^na_ix^i = \sum_{i=0}^na_ix^{i+1} = a_nx^{n+1} + a_{n-1}x^n + \dots +a_ox,$$ which makes the statement pretty obvious). Now, given that the independent term of $2\cdot p(x)$ is even as well (it is twice the independent term of $p(x)$), deducing that every element of the ideal has an even independent term should not be difficult.
From here, showing that $(2,x)$ is not principal is easy as well. If it were principal, there would exist $d(x)\in (2,x)$ such that every element of the ideal can be expressed as a product of $d(x)$ with an element of the ring (i.e. $d(x)$ divides every element of the ideal). In particular, this holds for $2$ and $x$ since they are both elements of $(2,x)$. It is not hard to see why this yields a contradiction, if you keep in mind everything we have said so far.
• but how do I show that $1$ and $-1$ are the only ones? – Guerlando OCs Jan 10 '16 at 19:11
• You are working in $\mathbb{Z}[x]$, where $\deg(p(x)\cdot q(x)) = \deg(p(x))+\deg(q(x))$. This tells you that the degree of the divisors of a certain polynomial is less or equal to the degree of the polynomial itself. But $\deg(3) = 0$, so any polynomial that divides it must have degree $0$ as well. Does this help you in seeing what the only divisors of $3$ in $\mathbb{Z}[x]$ are? Once you are there, deducing that $3$ and $x$ have no other common divisors but $1$ and $-1$ is immediate. – Labba Jan 10 '16 at 19:24