I Equivalent definitons for primitive polynomials

[elias001]

$\textbf{Definition 1:}$ Let $R$ be an UFD. A nonzero polynomial in $R[x]$ is said to be $\textbf{primitive}$ if the only constants that divides it are the units in $R$.

$\textbf{Definition 2:}$ Let $R$ be an UFD. Hence highest common factors of finite subsets of $R$ exists, by collecting common factors of the unique factorization of the elements into products of irreducibles.A polynomial $f(x)\in R[x]$ is called $\textbf{primitive}$ if the highest common factors of the non-zero coefficients of $f(x)$ is equal to $1$. This implies that if $u\in R$ and $u|f(x)$ then $u$ is a unit in $R.$

$\textbf{Exercise:}$ Prove that a polynomial is $\textbf{primitive}$ if and only if $1_R$ is a greatest common divisor of its coefficients. This property is often taken as the definition of primitive.

$\textit{Hint:}$ Since $c_1c_2\cdots c_mf(x)=g(x),$ each $c_i$ divides $g(x).$ Therefore, $c_i$ is a unit in $R$, because $g(x)$ is primitive.

For $\textbf{Exercise}$, it says the statement in the Exercise can be taken to be the definition of primitive polynomial instead of $\textbf{Definition 1}$. But according to $\textbf{Definition 2}$, it seems that the statement of $\textbf{Exercise}$ implies $\textbf{Definition 1}$. Regardless if it is an implication or an equivalent statements. How does one go about proving either?

Another thing is, how does gcd related to units in a commutative ring, ED, PID or UFD?

Thank you in advance.

 

[fresh_42]

I am confused too, and my book actually defines the statement of the exercise as the definition of a primitive polynomial.

I have difficulty seeing where this statement is any different from definition 2 since the zero coefficients do not change anything.

So do you want to prove the equivalence of the two definitions?

 

[elias001]

@fresh_42 yes, how would one go about proving the equivalence of the two statements. Also i replied to two of my past thread tagging you that you respond to recently.

 

[elias001]

@fresh_42 the posts are here and here. For the second one, I want to know how to correctly generalized the example in my original post and also, how to count the equivalence classes. I asked grad students and they told me they don't know enough about that specific topic. .

 

[fresh_42]

@fresh_42 the posts are here and here. For the second one, I want to know how to correctly generalized the example in my original post and also, how to count the equivalence classes. I asked grad students and they told me they don't know enough about that specific topic. .

You make me dizzy with all these cross references. I'm old. Let us deal with one thread at a time and simple reply in those threads. I will get an automatic notice. No tags necessary.

 

[fresh_42]

@fresh_42 yes, how would one go about proving the equivalence of the two statements. Also i replied to two of my past thread tagging you that you respond to recently.

First list what we have:

We have a polynomial $f(x)=a_0x^n+a_1x^{n-1}+\ldots+a_{n-1}x+a_n.$ For simplicity, say $S=\{a_0,a_1,\ldots,a_n\}$ and be $R^0$ the units of $R.$

primitive (def 1) says: $f(x)$ is primitive iff $\left[\;g(x) \,|\,f(x) \Longrightarrow g(x)\in R^0\;\right]$

Note that $g(x)\in R^0$ means two different things: first, that $g(x)\in R$ and second, that it can be inverted.

primitive (def 2) says: The greatest common divisor of elements in $S$ is $1.$

Note that it should better be said that $\operatorname{gcd}S\in R^0$ since we can always have other units as divisors. But we don't care about units, so the author says $\operatorname{gcd}S=1$ because it is easier not to deal with obsolete elements like units.

I leave it to you to prove:
a) $f(x)$ is primitive (def 1) $\Longrightarrow$ $f(x)$ is primitive (def 2)
and
b) $f(x)$ is primitive (def 2) $\Longrightarrow$ $f(x)$ is primitive (def 1).

 

[elias001]

@fresh_42 i don't get how the gcd of the coefficients of a polynomial in a polynomial ring relates to units in the said polynomial ring. Also for the case of units, it doesn't mean gcd has to be $1$?

 

[fresh_42]

@fresh_42 i don't get how the gcd of the coefficients of a polynomial in a polynomial ring relates to units in the said polynomial ring.

The units in $R[x]$ and the units in $R$ are identical. There are no invertible polynomials except constant polynomials, which are already invertible in $R.$ Adding the indeterminate $x$ doesn't change this unless you do not also add $1/x$ to your ring.

Also for the case of units, it doesn't mean gcd has to be $1$?

Forget that unit thing. Just pretend $1$ is the only unit. If we had the polynomial ring $\mathbb{Q}[x]$ over the rational numbers, then any non-zero rational number would be a unit. Strictly speaking, $\dfrac{1}{7}x$ would be primitive according to definition 1 but not according to definition 2. That doesn't make sense. It also doesn't make sense to bother about the unit $\dfrac{1}{7}$ since this makes no significant difference. So definition 2 is only a bit sloppy here if it only allows $1$ as divisor and not $\dfrac{1}{7}.$

Try to concentrate on the logic and not on units.

 

[elias001]

@fresh_42 let's do $(a)$ first.

We assume that $f(x)$ satisfies def 1 for being a primitive polynomial. Let $d$ be the gcd of the coefficients of $f(x)$, meaning $d=\text{gcd}(a_n,a_{n-1},\ldots,a_0).$ Let $S=\{a_n,a_{n-1},\ldots,a_0\},$ and each $a_i\in S$ is being divided by $d$, then $a_i=dc_i$ for some element $c_i\in R$. Then $c_i|a_i,$ for $i=0,1,2,\ldots,n.$ So $c_i|f(x)$, and $f(x)=c_ig(x)$ for some $g(x)\in R[x]$. We have $dc_i-c_i=0$ implying $c_i(d-1)=0$ which gives $d=1$.

I am not sure how to justify that $c_i$ are units in $R$?

 

[fresh_42]

@fresh_42 let's do $(a)$ first.

We assume that $f(x)$ satisfies def 1 for being a primitive polynomial. Let $d$ be the gcd of the coefficients of $f(x)$, meaning $d=\text{gcd}(a_n,a_{n-1},\ldots,a_0).$ Let $S=\{a_n,a_{n-1},\ldots,a_0\},$ and each $a_i\in S$ is being divided by $d$, then $a_i=dc_i$ for some unit $c_i\in R$. Since $c_i|f(x)$, then $f(x)=c_ig(x)$ for some $g(x)\in R[x]$. We have $dc_i-c_i=0$ implying $c_i(d-1)=0$ which gives $d=1$.

I am not sure how to justify that $c_i$ are units in $R$?

Me neither. And you do not need it.

We must show definition 2. So we assume that $\operatorname{gcd}S=d.$ But $d$ divides also $f(x)$ and by definition 1, this means that $d$ is a unit in $R.$ But if $\operatorname{gcd}S=d$ then we also have $\operatorname{gcd}S=d\cdot d^{-1}=1$ since units don't change divisibility properties. And that is what definition 2 requires.

 

[elias001]

@fresh_42 I just edited my post, does it fix the problem. I am not sure I can just do what you suggest: that since $d$ is a gcd, then it is an unit. I try to just show that $d=1$ according to definition 2. I mean that is the requirement for $f$ to be a primitive polynomial? Also Def 1 states: "...if the only constants that divides it are the units in...". So I think it is require that I need to show that $d$ is an unit, since there are might be other elements that divides $f(x)$ that are not units. In def 1, it doesn't state that gcd are units. Am I reading too much into the definition or do I need to find some way to avoid that issue all together.

 

[fresh_42]

@fresh_42 I just edited my post, does it fix the problem. I am not sure I can just do what you suggest, that since $d$ is a gcd, then it is an unit.

No. It is only a common factor of all coefficients. You can use this factor as $g(x)=d$ in definition 1. It then follows that $d$ is a unit since $f$ is primitive (def 1). You don't need the $c_i$ and the equation $dc_i-c_i=0$ doesn't hold.

I try to just show that $d=1$according to definition 2. I mean that is the requirement for $f$ to be a primitive polynomial?

Yes, but this is nitpicking. If $d$ is a unit in $R,$ which it is by definition 1, then there is an element $d^{-1}\in R.$ Now $d\cdot d^{-1}$ also divides all coefficients, so $\operatorname{gcd}S=d\cdot d^{-1}=1.$

Making that divisor $1$ is just a technical trick.

The other direction is the difficult one.

Say $f$ is primitive (def 2). Then we assume that $g(x)\,|\,f(x).$ Hence, there is a polynomial $h(x)\in R[x]$ such that $f(x)=g(x)\cdot h(x).$

Now, you must show that $g(x)$ is a unit in $R,$ i.e., that it is a constant polynomial, and the constant is invertible. How? We only know that the coefficients of $f$ have no common divisor, or the divisor $1$ to be exact. How can we use this to get information about $g(x)$?

 

[elias001]

@fresh_42 I still don't understand how the portion of Def 1 where it states: "...if the only constants that divides it are the units in..." allows me to infer $d$ is an unit? There are might be other elements that divides $f(x)$ that are not units. I want to get this subtle point sorted out before doing (b).

 

[fresh_42]

@fresh_42 I still don't understand how the portion of Def 1 where it states: "...if the only units that divides it are the units in..." allows me to infer $d$ is an unit? There are might be other elements that divides $f(x)$ that are not units.

Are you sure that is exactly so in the book?

$\textbf{Definition 1:}$ Let $R$ be an UFD. A nonzero polynomial in $R[x]$ is said to be $\textbf{primitive}$ if the only units that divides it are the units in $R$.

I think it should be ...

$\textbf{Definition 1:}$ Let $R$ be an UFD. A nonzero polynomial in $R[x]$ is said to be $\textbf{primitive}$ if the only factors that divides it are the units in $R$.

... since I do not see that units in $R$ differ from the units in $R[x]$ which would make it an empty statement.

 

[elias001]

@fresh_42 I attached a screenshot. Sorry, I made a typo it should say:"...if the only constants that divides it are the units in...". I edited post 1 in Def 1 also to reflect it. Still it doesn't change the fact that I still need to show that $d$ is an unit.

 

[fresh_42]

@fresh_42 I attached a screenshot. Sorry, I made a typo it should say:"...if the only constants that divides it are the units in...". I edited post 1 in Def 1 also to reflect it. Still it doesn't change the fact that I still need to show that $d$ is an unit.


View attachment 363639

$d$ is a unit because definition 1 says so. We have $d\,|\,f(x)$ and definition 1 says that only units divide $f.$

 

[elias001]

@fresh_42 I should read that portion of Def 1 as: if $x|f(x)$ and $x$ is a constant, then $x$ is an unit?

I find the notion of primitive polynomial to be not very intuitive and i am having a hard time relating units of a polynomial to the gcd of its coefficients.

I asked one of the LLMs for an explanation, and here is what it says:

Content of a Polynomial: The GCD of the coefficients of a polynomial $f(x)∈F[x]$ is called the content of the polynomial, denoted $\text{cont}(f).$ If $\text{cont}(f)=d$, then $f(x) = d \cdot g(x) ,$ where $g(x)\in F[x]$ is a polynomial whose coefficients have GCD $1$ (i.e., $g(x)$ is primitive).

Relation to Units: The content $\text{cont}(f)$ is only defined up to a unit in $F$. If $d$ is the GCD of the coefficients, then any other element $d' = u \cdot d$ , where $u∈F^*$ (the group of units in $F$), is also a valid GCD. This is because, in an UFD, if $d$ divides each coefficient, so does $u \cdot d$, and vice versa, since $u$ is invertible.

For example:

In $\mathbb{Z}[x]$ , the units are $\pm 1$, so the GCD of the coefficients is unique up to sign (e.g., $\text{gcd}⁡(2,4)=2$ or $−2$).

In a field like $\mathbb{Q}$ , all non-zero elements are units, so the content of a polynomial is less significant, as any non-zero constant can be factored out and absorbed into a unit.

Implications for Factorization: The relationship is crucial in Gauss's lemma, which states that for a UFD $F$, the content of a product of polynomials satisfies $\text{cont}(f \cdot g) = \text{cont}(f) \cdot \text{cont}(g)$ (up to units). This implies that if a polynomial has content that is a unit in $F$, it is primitive, and primitivity is preserved under multiplication in $F[x]$.


Let $f(x) = a_n x^n + \cdots + a_0 \in F[x]$ be a polynomial, where $a_i \in F.$ The content of $f,$ denoted $\text{cont}(f),$ is defined as: $\text{cont}(f) = \gcd(a_n, a_{n-1}, \ldots, a_0),$ where the GCD is taken in the UFD $F.$ In an UFD, the GCD exists and is unique up to multiplication by units. An unit in $F$ is an element $u \in F$ such that there exists $v \in F$ with $uv=1,$ where $1$ is the multiplicative identity in $F.$ The set of units in $F$ is denoted $F^*.$

Saying that $\text{cont}(f)$ is "defined up to a unit in $F$" means: If $d \in F$ is a GCD of the coefficients $\{a_n, \ldots, a_0\}$ , then any other element $d' \in F$ of the form $d' = u \cdot d$ , where $u \in F^*$, is also a GCD of the coefficients.
Conversely, if $d'$ is another GCD of the coefficients, then there exists a unit $u \in F^*$ such that $d' = u \cdot d$.

In mathematical notation, this can be expressed as

$\text{cont}(f) \sim d \quad \text{if and only if} \quad d \text{ divides } a_i \text{ for all } i, \text{ and any other } d' \text{ with this property satisfies } d' = u \cdot d \text{ for some } u \in F^*.$
Equivalently, the content is an element of the quotient monoid $\frac{F}{F^*}$ , where two elements $d, d' \in F$ are equivalent if $d' = u \cdot d$ for some $u \in F^*$ .

Does the explanation appears to be ok/clear to you?

 

[fresh_42]

@fresh_42 I should read that portion of Def 1 as: if $x|f(x)$ and $x$ is a constant, then $x$ is an unit?

Not really. It says, that if $f(x)=a_0+a_1x+\ldots+a_nx^n$ and $d\,|\,a_0,a_1,\ldots,a_n$ then $d=1.$
It doesn't say anything about $x.$ The polynomial $f(x)=x$ is primitive, although $x\,|\,f(x)$ and $x$ isn't constant.

Let me quote how my book (van der Waerden) defines it.

The underlying ring $R$ is an integral domain (commutative and without zero divisors) with $1.$ The greatest common divisor $d$ of $a_0,a_1,\ldots, a_n$ is the content of $f(x)=a_0+a_1x+\ldots+a_nx^n.$ If we take separate $d$ by writing
$$
f(x)=d\cdot g(x)
$$
then $g(x)$ has the content $1.$ Both, $g(x)$ and $d$ are uniquely determined up to unit factors. Polynomials of content $1$ are called primitive or unit-forms with respect to $R.$ van der Waerden uses the term unit-forms, but that is a bit outdated.

I find the notion of primitive polynomial to be not very intuitive and i am having a hard time relating units of a polynomial to the gcd of its coefficients.

Whatever you call it, it is all about the greatest common divisor of the coefficients of $f(x),$ not $f(x)$ itself, for which we have the terms prime and irreducible. This greatest common divisor is a ring element, i.e. $d\in R.$ As such, it can be $d=1$ or $d\neq 1.$ Units are more or less irrelevant since we can always multiply any units without changing what it is all about. Units of $R[x]$ are even less relevant in this context. You are obsessed with units. Forget them. They are just irrelevant factors (in this context) and play the same role as $1$ does.

I asked one of the LLMs for an explanation, and here is what it says:

You'd better not. You are already confused by using more than one book. I would concentrate on a single book and only ask questions about that specific book (whichever it is). LLMs tell you what they "think" you want to hear. They are worse than Wikipedia!

Content of a Polynomial: The GCD of the coefficients of a polynomial $f(x)∈F[x]$ is called the content of the polynomial, denoted $\text{cont}(f).$ If $\text{cont}(f)=d$, then $f(x) = d \cdot g(x) ,$ where $g(x)\in F[x]$ is a polynomial whose coefficients have GCD $1$ (i.e., $g(x)$ is primitive).

Relation to Units: The content $\text{cont}(f)$ is only defined up to a unit in $F$. If $d$ is the GCD of the coefficients, then any other element $d' = u \cdot d$ , where $u∈F^*$ (the group of units in $F$), is also a valid GCD. This is because, in an UFD, if $d$ divides each coefficient, so does $u \cdot d$, and vice versa, since $u$ is invertible.

For example:

In $\mathbb{Z}[x]$ , the units are $\pm 1$, so the GCD of the coefficients is unique up to sign (e.g., $\text{gcd}⁡(2,4)=2$ or $−2$).

In a field like $\mathbb{Q}$ , all non-zero elements are units, so the content of a polynomial is less significant, as any non-zero constant can be factored out and absorbed into a unit.

Implications for Factorization: The relationship is crucial in Gauss's lemma, which states that for a UFD $F$, the content of a product of polynomials satisfies $\text{cont}(f \cdot g) = \text{cont}(f) \cdot \text{cont}(g)$ (up to units). This implies that if a polynomial has content that is a unit in $F$, it is primitive, and primitivity is preserved under multiplication in $F[x]$.

Let $f(x) = a_n x^n + \cdots + a_0 \in F[x]$ be a polynomial, where $a_i \in F.$ The content of $f,$ denoted $\text{cont}(f),$ is defined as: $\text{cont}(f) = \gcd(a_n, a_{n-1}, \ldots, a_0),$ where the GCD is taken in the UFD $F.$ In an UFD, the GCD exists and is unique up to multiplication by units. An unit in $F$ is an element $u \in F$ such that there exists $v \in F$ with $uv=1,$ where $1$ is the multiplicative identity in $F.$ The set of units in $F$ is denoted $F^*.$

Saying that $\text{cont}(f)$ is "defined up to a unit in $F$" means: If $d \in F$ is a GCD of the coefficients $\{a_n, \ldots, a_0\}$ , then any other element $d' \in F$ of the form $d' = u \cdot d$ , where $u \in F^*$, is also a GCD of the coefficients.
Conversely, if $d'$ is another GCD of the coefficients, then there exists a unit $u \in F^*$ such that $d' = u \cdot d$.

In mathematical notation, this can be expressed as

$\text{cont}(f) \sim d \quad \text{if and only if} \quad d \text{ divides } a_i \text{ for all } i, \text{ and any other } d' \text{ with this property satisfies } d' = u \cdot d \text{ for some } u \in F^*.$
Equivalently, the content is an element of the quotient monoid $\frac{F}{F^*}$ , where two elements $d, d' \in F$ are equivalent if $d' = u \cdot d$ for some $u \in F^*$ .

Does the explanation appears to be ok/clear to you?

That's more or less exactly how van der Waerden defined it. Again, forget about units. They are only a pollution in uniqueness statements.

It is simple. How would you factor the integer $6$? You would probably answer $6=2\cdot 3.$ This shows that $6$ is reducible. It is also not a prime, since the two terms coincide for integers. However, the reason why $6$ isn't prime is a different one. If $6\,|\,3\cdot 4$ then we cannot conclude $6\,|\,3$ or $6\,|\,4.$ That is why $6$ isn't prime.

Now, what if I were to say my answer to that question is $6=(-2)\cdot (-3)?$ This is certainly true, and, what is more important, wouldn't change any of my previous arguments. We would only need to carry the minus sign through the arguments. $-2$ is a prime number! This is because $-1$ is a unit, and it shows that uniqueness is only given up to factors $\pm 1,$ the units in $\mathbb{Z}.$ And if the ring were a field, we would have many more units. That is why $\mathbb{Q}$ doesn't have any prime elements.

 

[elias001]

@fresh_42 for (b)

Assumed fact: every polynomial $m(x)=pq(x)$ where $p$ is the gcd of the coefficients of $m(x)$ and the polynomial $q(x)$ is primitive.

Let $f(x)$ be a primitive polynomial with $\text{content}(f)=1.$ Why do/can I suppose that $g(x)|f(x)?$.

If $g(x)|f(x)$ implies $f(x)=g(x)h(x)$ then the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive. By the assumed fact, $f(x)=g(x)h(x)=\text{content}(f)h(x).$. This implies $g(x)=\text{content}(f)=1.$

Here: "the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive." I want to use one more fact that $\text{content}(f)=\text{content}(gh)=\text{content}(g)\text{content}(h),$ then $1=\text{content}(g)\text{content}(h)$ implying $\text{content}(g)$ is an unit, so $g(x)$ is primitive. But then I don't know how to conclude that "the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive.". Maybe I can do the same argument like with $g(x)$.

By the way, I now know why Def 1 is troubling me. If we are working in $\Bbb{Q}[x]$, the polynomial $3x^2+9x+15$ has gcd of its coefficients equals $3$, but that doesn't satisfy Def 2 since it requires gcd to be $1$.

 

[fresh_42]

@fresh_42 for (b)

Assumed fact: every polynomial $m(x)=pq(x)$ where $p$ is the gcd of the coefficients of $m(x)$ and the polynomial $q(x)$ is primitive.

Let $f(x)$ be a primitive polynomial with $\text{content}(f)=1.$ Why do/can I suppose that $g(x)|f(x)?$.

We have to. We know that $f(x)$ is primitive (def 2). Now, if we want to show $f(x)$ is primitive (def 1), we have to show what definition 1 states. The statement is, if $g(x)\,|\,f(x)$ then $g(x)\in R^0.$ Hence, we must assume that $g(x)\,|\,f(x)$ since these are the only relevant cases. All others don't matter.

If $g(x)|f(x)$ implies $f(x)=g(x)h(x)$ then the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive. By the assumed fact, $f(x)=g(x)h(x)=\text{content}(f)h(x).$. This implies $g(x)=\text{content}(f)=1.$

Let me sort this out. We know that $f(x)$ is primitive (def 2), i.e., the content of $f(x)$ is $1.$ We also know that $f(x)=g(x)\cdot h(x).$ What we need next is that the content of a product is the product of the contents of the factors. This should be proven. Given that we have
$$
1=\operatorname{content}(f(x))=\operatorname{content}(g(x))\cdot \operatorname{content}(h(x)).
$$
But this is an equation in $R,$ so $\operatorname{content}(g(x))$ and $\operatorname{content}(h(x))$ are units by definition of a unit. Hence, $g(x)$ and $h(x)$ are primitive (def 2). The contents are not exactly $1$ but units, which is good enough. If $g(x)=d\in R$ then $d\in R^0$ which proves that $f(x)$ is primitive (def 1).

Here: "the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive." I want to use one more fact that $\text{content}(f)=\text{content}(gh)=\text{content}(g)\text{content}(h),$ then $1=\text{content}(g)\text{content}(h)$ implying $\text{content}(g)$ is an unit, so $g(x)$ is primitive. But then I don't know how to conclude that "the $\text{content}(f)$ is the same as the content of $h(x),$ hence $h(x)$ is primitive.". Maybe I can do the same argument like with $g(x)$.

By the way, I now know why Def 1 is troubling me.

Me, too. I made a mistake. It is really what you said. It has to be

definition 1: $f(x)$ is primitive iff $d\,|\,f(x) \wedge d\in R \Longrightarrow d\in R^0.$

Since $x+x^2$ is primitive, and $x\,|\,(x+x^2)$ we cannot conclude $x\in R^0.$ We must demand $R\ni d\,|\,f(x).$

If we are working in $\Bbb{Q}[x]$, the polynomial $3x^2+9x+15$ has gcd of its coefficients equals $3$, but that doesn't satisfy Def 2 since it requires gcd to be $1$.

No. If we are working in $\mathbb{Q}[x]$ then any non-zero number is a unit. There is no greatest common divisor since all numbers are divisors, and up to units it is $1$ which is as good as any other non-zero number.

Your argument is only valid in $\mathbb{Z}[x],$ in which case $3x^2+9x+15$ isn't primitive whether you use definition 1 or 2.

That is why van der Waerden added "... primitive with respect to $R$" and why I repeat the mantra of saying which ring you refer to.

 

[elias001]

@fresh_42 for (b) how do I put it all together. I am a little confused over our back and forth as to what can be used and what I need to do more justifications. I am using two facts:

Fact 1. Every polynomial in a polynomial ring can be written as a product of its content and a primitive polynomial.

Fact 2. $\text{content}(f(x))=\text{content}(g(x))\cdot\text{content}(h(x)).$

So if I have $f(x)=g(x)h(x)\quad (*)$, and we also have $f(x)=\text{content}(f)h(x)\quad (**)$, then $f(x)=g(x)h(x)=\text{content}(f)h(x),$ and $\text{content}(f)=1$, so in $(**)$ I can conclude that $h(x)$ is primitive from Fact 1? Also since $\text{content}(f)=g(x)=1$, then I can also conclude that $g(x)$ is also primitive?

let's leave the issue of Def 1 with the issues of contestants and units for now until after (b) is resolved, since i think that might affect whether the argument in (a) need to be modified to take into account which ring one is speaking about.

 

[fresh_42]

@fresh_42 for (b) how do I put it all together. I am a little confused over our back and forth as to what can be used and what I need to do more justifications.

Sorry, my bad. Let's clean up the mess a bit.
_________________________________________
We start with $R$ being a UFD and particularly an integral domain, so no zero divisors except $0.$ The units in $R$ will still be denoted as $R^0.$
_________________________________________
Definiton 1: $f(x)\in R[x]$ is primitive iff $\left[d\,|\,f(x)\wedge d\in R\Longrightarrow d\in R^0\right].$

Definiton 2: $f(x)=a_0+a_1x+\ldots+a_nx^n\in R[x]$ is primitive iff $\operatorname{gcd}(a_0,\ldots,a_n)=1.$
_________________________________________
a) Definition 1 $\Longrightarrow$ Definition 2:

Let $d=\operatorname{gcd}(a_0,\ldots,a_n).$ Then $d\in R$ and $d\,|\,f(x).$ By definition 1, this implies that $d\in R^0.$ Thus, it is invertible, i.e., $d\cdot d^{-1}=1$ with $d^{-1}\in R^0.$ Finally, if $d=\operatorname{gcd}(a_0,\ldots,a_n)$ then so is $1=d\cdot d^{-1}=\operatorname{gcd}(a_0,\ldots,a_n)$ proving definition 2.

_________________________________________
b) Definition 2 $\Longrightarrow$ Definition 1:

To prove definition 1, we need an element $d\,|\,f(x)$ with $d\in R.$ This means that $d$ is a divisor of all coefficients of $f(x).$ Therefore, $d$ divides the greatest of all common divisors, i.e.,
$$
d\,|\,\operatorname{gcd}(a_0,\ldots,a_n)=1
$$
by definition 2. But if $d\,|\,1$ then it is an element from $R^0$ by definition of units. If $d\in R^0$ then this is exactly what definition 1 requires to be shown.
_________________________________________

Fact 1 is simply the distributive law in $R.$ Fact 2 seems obvious, but I suggest you try to prove it formally.

 

[elias001]

@fresh_42 The way you rewrote (a) accounts for different polynomial rings: $\Bbb{Q}[x], \Bbb{Z}[x].$

For Fact 1, Hungerford gave a weird direct proof. Screenshots:

Here is a proof by contradiction from a YT video

Can I ask you about the other post about that localization example?

 

[fresh_42]

Can I ask you about the other post about that localization example?

Sure, but do it there or even better in a new thread.

 

[elias001]

@fresh_42 I replied tagging you. I did @fresh.... is that call tagging?