I How to show ##p(x)=g(x)x\pm 1\in\Bbb{Q}[x]## is irreducible in ##\Bbb{Q}_{\Bbb{Z}}[x]##?

[elias001]

Background


Notation: $\Bbb{Q}_{\Bbb{Z}}[x]=\Bbb{Z}+x\Bbb{Q}[x]$

Assumed exercises:

(1)(a) Prove that the only units in $\Bbb{Q}_{\Bbb{Z}}[x]$ are $1$ and $-1$.

(b) If $f(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$, show that the only associates are $f(x)$ and $-f(x)$.

(2)(a) If $p$ is prime in $\Bbb{Z}$, prove that the constant polynomial $p$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(b) If $p$ and $q$ are positive primes in $\Bbb{Z}$ with $p\neq q$, prove that $p$ and $q$ are not associates in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(3)(a) Show that the only divisors of $x$ in $\Bbb{Q}_{\Bbb{Z}}[x]$ are the integers (constant polynomials) and first-degree polynomials
of the form $\frac{1}{n}x$ with $0\neq n\in \Bbb{Z}$.

(b) For each nonzero $n\in \Bbb{Z}$, show that the polynomial $\frac{1}{n}x$ is not irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(c) Show that $x$ cannot be written as a finite product of irreducible elements in $\Bbb{Q}_{\Bbb{Z}}[x]$.

Exercise: Prove that $p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ if and only if $p(x)$ is either a prime integer or an irreducible polynomial in $\Bbb{Q}[x]$ with constant term $\pm 1$.

Question:

If I already proved the assumed exercises $(1)$ to $(3)$ above and I want to show one direction for the Exercise above:

$p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ implies $p(x)$ is an irreducible polynomial in $\Bbb{Q}[x]$ with constant term
$\pm 1$

I am not clear on how to show: if an irreducible polynomial
$p(x)\in\Bbb{Q}[x]$ is of the form $p(x)=g(x)x\pm 1,$ where $g(x)\in \Bbb{Q}[x],$ then $p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

Thank you in advance

MENTOR note: Replaced all '$' with double '#' for proper mathjax rendering of the page.

 

[fresh_42]

Background


Notation: $\Bbb{Q}_{\Bbb{Z}}[x]=\Bbb{Z}+x\Bbb{Q}[x]$

Assumed exercises:

(1)(a) Prove that the only units in $\Bbb{Q}_{\Bbb{Z}}[x]$ are $1$ and $-1$.

(b) If $f(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$, show that the only associates are $f(x)$ and $-f(x)$.

(2)(a) If $p$ is prime in $\Bbb{Z}$, prove that the constant polynomial $p$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(b) If $p$ and $q$ are positive primes in $\Bbb{Z}$ with $p\neq q$, prove that $p$ and $q$ are not associates in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(3)(a) Show that the only divisors of $x$ in $\Bbb{Q}_{\Bbb{Z}}[x]$ are the integers (constant polynomials) and first-degree polynomials
of the form $\frac{1}{n}x$ with $0\neq n\in \Bbb{Z}$.

(b) For each nonzero $n\in \Bbb{Z}$, show that the polynomial $\frac{1}{n}x$ is not irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

(c) Show that $x$ cannot be written as a finite product of irreducible elements in $\Bbb{Q}_{\Bbb{Z}}[x]$.

Exercise: Prove that $p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ if and only if $p(x)$ is either a prime integer or an irreducible polynomial in $\Bbb{Q}[x]$ with constant term $\pm 1$.

Question:

If I already proved the assumed exercises $(1)$ to $(3)$ above and I want to show one direction for the Exercise above:

$p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ implies $p(x)$ is an irreducible polynomial in $\Bbb{Q}[x]$ with constant term
$\pm 1$

I am not clear on how to show: if an irreducible polynomial
$p(x)\in\Bbb{Q}[x]$ is of the form $p(x)=g(x)x\pm 1,$ where $g(x)\in \Bbb{Q}[x],$ then $p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

Thank you in advance

MENTOR note: Replaced all '$' with double '#' for proper mathjax rendering of the page.

If $p(x)$ is reducible in $\mathbb{Q}_{\mathbb{Z}}[x]$ then it is also reducible in $\mathbb{Q}[x]$ using the same factor polynomials since $\mathbb{Q}_{\mathbb{Z}}[x]\subseteq \mathbb{Q}[x].$

 

[elias001]

@fresh_42

If $p(x)$ is reducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ then it is also reducible in $\Bbb{Q}[x]$ using the same factor polynomials since $\Bbb{Q}_{\Bbb{Z}}[x]\subset \Bbb{Q}[x]$.

if i go by the contrapositive direction as you suggest. I don't know how to show $p(x)=g(x)x\pm 1$ is reducible. I can prove that the ring $\Bbb{Q}_{\Bbb{Z}}[x]$ are $\{-1, 1\}$ for units. I also can show that $\frac{1}{n}x$ is not irreducible. But I don't know how to show that $g(x)x\pm 1$ is not an unit element of $\Bbb{Q}_{\Bbb{Z}}[x]$. Basically i don't know how to show that $\{-1, 1\}$ are the only units of $\Bbb{Q}_{\Bbb{Z}}[x]$.

 

[fresh_42]

You are right, I forgot the units of $\mathbb{Q}_\mathbb{Z}[x].$ Let's see.
$$1=(a+xu(x))\cdot (b+xv(x))=ab+x(bu(x)+av(x)+xu(x)v(x))$$
means that $a,b=\pm 1$ and $bu(x)+av(x)+xu(x)v(x)=0$ in $\mathbb{Q}[x].$ My only idea is to write out these polynomials and compare their coefficients at $x^n.$ Inserting some values in $x$ like $0$ and $1$ should give $u(x)=v(x)=0.$ E.g., $x=0$ gives immediately $u(0)=\pm v(0)$ for their constant term. The rest should follow accordingly.

I think, it can be shown that $bu(x)=-av(x)$ by repeatedly using $x=0,$ and then $0=xu(x)v(x)$ which means $u(x)=0$ or $v(x)=0$ since $\mathbb{Q}[x]$ is an integral domain. Hence, $u(x)=v(x)=0.$

 

[elias001]

@fresh_42 when you say

"My only idea is to write out these polynomials and compare their coefficients at $x^n.$ Inserting some values in $x$ like $0$ and $1$ should give $u(x)=v(x)=0.$"

From $1=(a+xu(x))\cdot (b+xv(x))=ab+x(bu(x)+av(x)+xu(x)v(x))$, then $x(bu(x)+av(x)+xu(x)v(x))=0,$ if $x=0$ and for $x=1$ should give $ab+(bu(1)+av(1)+u(1)v(1))=1$. How did you get $u(x)=v(x)=0$?

Then "I think, it can be shown that $bu(x)=-av(x)$
and then $0=xu(x)v(x)$
which means $u(x)=0$ or $v(x)=0$ since $\mathbb{Q}[x]$ is an integral domain. Hence, $u(x)=v(x)=0.$".

I am guessing you mean if $x(bu(x)+av(x)+xu(x)v(x))=0$, then $bu(x)+av(x)=-xu(x)v(x)$. But then, how did you draw the rest of the conclusion? I am trying to work out your algebra.

 

[fresh_42]

No. I meant what I wrote. We can use the fact that two rational polynomials are equal if and only if all their coefficients are identical. We may compare by the powers of $x.$ So if
$$
1=ab+x(bu(x)+av(x)+xu(x)v(x))
$$
then $ab=1$ which means $a=b=\pm 1$ since $a,b\in \mathbb{Z}$ and $x\cdot(bu(x)+av(x)+xu(x)v(x))=0$ means that $bu(x)+av(x)+xu(x)v(x)=0$ since $\mathbb{Q}[x]$ is an integral domain and $x\neq 0.$ Next we can compare the coefficients of $bu(x)+av(x)$ with those of $xu(x)v(x)$ for each power of $x$. This should give the zeros we are looking for because there is always a factor $x$ more in the product compared to the sum.

 

[elias001]

@fresh_42 wait, I think I have it, do you mean the following steps: $1=(a+xu(x))\cdot (b+xv(x))=ab+x(bu(x)+av(x)+xu(x)v(x))$ i forgot that i can use $a=b=\pm 1$. Then $x(bu(x)+av(x)+xu(x)v(x))=0$ imply $bu(x)+av(x)=-xu(x)v(x))$ from plugging in $x=0$. Afterwards, we have $bu(0)+av(0)=-0u(0)v(0)=0$ implying $bu(x)=-av(x)$. With $a=-1, b=1$, gives $0=xu(x)v(x)=xu^2(x)=0$. Since $\Bbb{Q}[x]$ is an integral domain, and $x\neq 0$, then $u^2(x)=0$ which implies $u(x)=0$. But $u(x)=v(x), u(x)=v(x)=0$. If my steps are correct,, how does it show that $\pm 1+xu(x)$ is reducible?

 

[fresh_42]

@fresh_42 wait, I think I have it, do you mean the following steps: $1=(a+xu(x))\cdot (b+xv(x))=ab+x(bu(x)+av(x)+xu(x)v(x))$ i forgot that i can use $a=b=\pm 1$. Then $x(bu(x)+av(x)+xu(x)v(x))=0$ imply $bu(x)+av(x)=-xu(x)v(x))$ from plugging in $x=0$. Afterwards, we have $bu(0)+av(0)=-0u(0)v(0)=0$ implying $bu(x)=-av(x)$. With $a=-1, b=1$, gives $0=xu(x)v(x)=xu^2(x)=0$. Since $\Bbb{Q}[x]$ is an integral domain, and $x\neq 0$, then $u^2(x)=0$ which implies $u(x)=0$.

It only implies $bu(0)=-av(0).$ But these coefficients could have been zero anyway. Managing the coefficients properly is painful but not difficult. I think we have to use written-out polynomials and fight our way through the equation system. Maybe assuming a minimal non-zero coefficient can abbreviate that fight.

But $u(x)=v(x), u(x)=v(x)=0$. If my steps are correct,, how does it show that $\pm 1+xu(x)$ is reducible?

You wrote ...

I am not clear on how to show: if an irreducible polynomial
$p(x)\in\Bbb{Q}[x]$ is of the form ... then $p(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

This is equivalent to:

If $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x]$ is reducible, then $p(x)\in \mathbb{Q}[x]$ is reducible.

To prove it, we take a factorization $p(x)=u(x)v(x)$ in $\mathbb{Q}_{\mathbb{Z}}[x].$ This is also automatically a factorization in $\mathbb{Q}[x]$ regardless of the form. We only need to show that the factors didn't become units. The factors, however, were only units if they had already been units in the subring, contradicting our assumption of being reducible in the bigger ring.

 

[elias001]

@fresh_42 in my argument, it is sufficient to consider only the case $a=-1, b=1$ and i don't have to really worry about the other three cases for the choice of different $a$ or $b$ like $a=b=1$. Also at the end, you said "The factors, however, were only units if they had already been units in the subring, contradicting our assumption of being reducible in the bigger ring." How were the factors were not units in the subring? You were referring to the factor $xu(x)\pm 1$?

 

[fresh_42]

@fresh_42 in my argument, it is sufficient to consider only the case $a=-1, b=1$ and i don't have to really worry about the other three cases for the choice of different $a$ or $b$ like $a=b=1$. Also at the end, you said "The factors, however, were only units if they had already been units in the subring, contradicting our assumption of being reducible in the bigger ring." How were the factors were not units in the subring? You were referring to the factor $xu(x)\pm 1$?

We want to show that irreducibility in $\mathbb{Q}[x]$ implies irreducibility in $\mathbb{Q}_{\mathbb{Z}}\,[x].$

This is equivalent to showing that reducibility in $\mathbb{Q}_{\mathbb{Z}}\,[x]$ implies reducibility in $\mathbb{Q}[x].$

So we take a in $\mathbb{Q}_{\mathbb{Z}}\,[x]$ reducible polynomial $p(x),$ i.e. $p(x)=u(x)\cdot v(x)$ where neither factor is a unit in $\mathbb{Q}_{\mathbb{Z}}\,[x].$

Next we look at the very same equation $p(x)= u(x)\cdot v(x)$ but this time in $\mathbb{Q}[x].$ This is a proper factorization in $\mathbb{Q}[x]$ unless, say $u(x)\in \mathbb{Q}[x]$ is a unit. But then $u(x)=\pm 1$ which is also a unit in $\mathbb{Q}_{\mathbb{Z}}\,[x]$ and thus contradicts our assumption that $p(x)\in \mathbb{Q}_{\mathbb{Z}}\,[x]$ was reducible.

That is it written as negation $(A\Longrightarrow B) \Longleftrightarrow (\lnot B\Longrightarrow \lnot A).$ You can also write it as a proof by contradiction, $(A\wedge \lnot B\Longrightarrow \emptyset).$

 

[elias001]

@fresh_42 thank you so much for all your patience. I think your argument bypasses all the messy algebra. This is one of those proofs that it feels better to actually see the messy details. But if you can find a shorter proof that avoids all those messy details, it makes you sleep better at night, but you don't get to see all the beautiful messy details.

 

[fresh_42]

The other way is a bit of a problem, to show that irreducibility in $\mathbb{Q}_{\mathbb{Z}}\,[x]$ means irreducibility in $\mathbb{Q}[x].$ Here is where the form kicks in since we must somehow get that $ab=1$ implies $a=\pm 1.$

 

[elias001]

@fresh_42 I am ok with the other direction. Again, thank you for all your help.

 

[elias001]

@fresh_42

I found the written solution to the first part of the question, the part in my post. The solution is as follows:

Solution: If $p(x)$ is a prime integer then it is prime. Suppose $p(x)$ is irreducible in $\Bbb{Q}[x]$ and has constant term $\pm 1$. If it factors in $\Bbb{Q}_{\Bbb{Z}}[x]$ one of the terms must be of degree $0$. Compare constant terms to show that it must be $\pm 1$, and hence is a unit.

I am not sure what the solution meant by "terms" versus factors. See because if we have a polynomial $f(x)\in \Bbb{Q}[x]$ and it factors as $f(x)=p_1(x)p_2(x)=(a+xm(x))(b+xn(x)), m(x),n(x)\in\Bbb{Q}[x], a,b\in\Bbb{Z}[x]$, and $p_1(x)=(a+xm(x)), p_2(x)=(b+xn(x)).$ Then in the case of the factor $p_1(x)=(a+xm(x))$, by "term", is it referring to: $(a+xm(x)),$ $m(x),$ or $xm(x)$? Also, when it says compare constant terms, what does the author mean matching the degree of factors $\text{deg}(p_1(x))$ or $\text{deg}(xm(x))$? I assume for the constant terms, the author is referring to $a, b$. Sorry for troubling you for further clarification about my question.

 

[fresh_42]

If we write a polynomial as $f(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n$ then the word term usually refers to a summand, so the terms are $a_0,a_1x,\ldots,a_nx^n.$ Summands would be an alternative, but people say terms. The constant term is thus $a_0=f(0)$, which is also the coefficient of the term $a_0x^0.$

I'm not 100% sure whether authors always distinguish between $a_k$ and $a_kx^k$ when they say term, so term could possibly also refer to the coefficient $a_k$ since the $x^k$ part only tells us where the term stands and we usually only need $a_k.$ But term in general means the whole summand $a_kx^k.$

More generally, outside the world of polynomials or series, term refers to a part of an equation. It needs some further specification as to which part of the equation is meant. In the case of polynomials, we speak e.g., of the constant term $a_0,$ the leading term or term of highest degree $a_nx^n,$ the linear term $a_1x$ or the quadratic term $a_2x^2.$

A factor is what it always is. If $f(x)=p_1(x)p_2(x)$ then $p_1(x)$ and $p_2(x)$ are factors of $f(x).$ They do not need to be non-units, or irreducible, or something. Any factorization leads to the word factor. This is important in cases where primality or irreducibility plays a role because we distinguish between a proper factorization where the factors aren't units and irreducible polynomials that do not allow a proper factorization, i.e., where one of the factors has to be a unit.

Theoretically, one could write $f(x)=p_1(x)p_2(x)$ and speak of $p_1(x)$ as the $p_1$ term of the equation. However, I would avoid this by any means in the realm of polynomials, where term is used otherwise, namely as a name for the summands, not the factors, and I have never seen an author who used the word term here for a factor. The $p_1$ part or $p_1$ factor would be the preferred choice in this case. So this is more of a linguistic possibility rather than a mathematical one.

 

[elias001]

@fresh_42 thank you so much for your detail write ups. I really appreciate it. Can I trouble you to take a look at the following post: https://www.physicsforums.com/threa...-dimension-of-quotient-vector-spaces.1064662/

awhile ago. I am wondering if I can hear if the solution looks okay. It is a solution to about the quotient vector spaces and monomials. If you don't have time, I totally understand. I just want to hear if the direction and idea of my argument look okay since i am employing counting argument and it would be nice if I have not made any mistakes over the way the $d_i$s are counted. Thank you in advance.

 

[fresh_42]

I'll have a look, but it'll take me a moment. Btw, do you know whether the vector space dimension and the Krull dimension of $k[x_1,\ldots,x_n]/\mathfrak{a}$ coincide? I suppose so, but I couldn't find a quick reference.

I assume that the $x_k$ are all transcendental over $k.$

 

[elias001]

@fresh_42 hey i just saw this your last post. I know what the krull dimension of a ring is, but I don't know it well enough to comment for quotient out ideals in the context of quotient vector spaces.

I tried the same trick with the other direction, basically trying to prove that if $p(x)\in \Bbb{Q}_{Z}[x]$ is irreducible, then it is irreducible in $\Bbb{Q}[x]$.

What I did is proved by contra-positive. I did the same thing you suggested in the other direction.

So suppose $p(x)\in \Bbb{Q}[x]$ is a non constant reducible polynomial and $p(x)=a(x)b(x), a(x), b(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$. Then $a(x)b(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$ and hence must be reducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ also. Otherwise, either one of $a(x)$ or $b(x)$ must be an unit, say $a(x), \text{and } a(x)=\pm 1$. But $a(x)b(x)\in \Bbb{Q}[x]$ and this would mean $p(x)=a(x)b(x)=\pm 1\cdot b(x)\in \Bbb{Q}[x]$. Then $p(x), b(x)$ are each other's associates with $b(x)$ being not an unit element of $\Bbb{Q}[x]$, which in turn imply that $p(x)$ is irreducible in $\Bbb{Q}[x]$, a contradiction.

Does the proof look alright to you. I know it sounds a bit verbose. I have to keep track which set the $a(x),b(x)$ are in for every statement.

 

[fresh_42]

@fresh_42 hey i just saw this your last post. I know what the krull dimension of a ring is, but I don't know it well enough to comment for quotient out ideals in the context of quotient vector spaces.

I tried the same trick with the other direction, basically trying to prove that if $p(x)\in \Bbb{Q}_{Z}[x]$ is irreducible, then it is irreducible in $\Bbb{Q}[x]$.

What I did is proved by contra-positive. I did the same thing you suggested in the other direction.

So suppose $p(x)\in \Bbb{Q}[x]$ is a non constant reducible polynomial and $p(x)=a(x)b(x), a(x), b(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$.

This is what we need to show. You may not assume $a(x),b(x)\in \Bbb{Q}_{\Bbb{Z}}[x].$

The statement is that irreducibilty in $\Bbb{Q}_{\Bbb{Z}}[x]$ implies irreducibilty in $\Bbb{Q}[x].$ This is equivalent to the statement that reducibility in $\Bbb{Q}[x]$ implies reducibility in $\Bbb{Q}_{\Bbb{Z}}[x].$

Reducibility in $\Bbb{Q}[x]$ means that we have a factorization $p(x)=a(x)b(x)$ with factors $a(x),b(x)\in \Bbb{Q}[x].$

The goal now is to modify these factors such that we get a factorization with polynomials $a'(x),b'(x)$ in $\mathbb{Q}_{\mathbb{Z}}[x].$

But there is a problem. $\mathbb{Q}_{\mathbb{Z}}[x]\subsetneq \mathbb{Q}[x]$ is a proper subset. What if $p(x)\not\in \mathbb{Q}_{\mathbb{Z}}[x]$? We have to rule out this case. This means that a contra-positive proof won't do. We need an indirect proof:

given: $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x]$ irreducible.
to show: $p(x)\in \mathbb{Q}[x]$ is irreducible.
assumption: $p(x)\in \mathbb{Q}[x]$ is reducible.
goal: use $p(x)=a(x)\cdot b(x)$ in $\mathbb{Q}[x]$ and construct a factorization $p(x)=a'(x)\cdot b'(x)$ in $\mathbb{Q}_{\mathbb{Z}}[x]$ which contradicts the given condition, i.e. our assumption is false, and what we had to show is true.

The difference with that structure is, that we still have the given fact that $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x]$ other than in the contraposition case where we have to deal with arbitrary polynomials of $\mathbb{Q}[x].$ Consider the polynomial $p(x)=(1/2)x^2 \in \mathbb{Q}[x].$ This polynomial is reducible in $\mathbb{Q}[x]$ but not in $\mathbb{Q}_{\mathbb{Z}}[x].$ It isn't even an element of $\mathbb{Q}_{\mathbb{Z}}[x].$ That's why we need an indirect proof. We must have $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x].$

Then $a(x)b(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$ and hence must be reducible in $\Bbb{Q}_{\Bbb{Z}}[x]$ also. Otherwise, either one of $a(x)$ or $b(x)$ must be an unit, say $a(x), \text{and } a(x)=\pm 1$. But $a(x)b(x)\in \Bbb{Q}[x]$ and this would mean $p(x)=a(x)b(x)=\pm 1\cdot b(x)\in \Bbb{Q}[x]$. Then $p(x), b(x)$ are each other's associates with $b(x)$ being not an unit element of $\Bbb{Q}[x]$, which in turn imply that $p(x)$ is irreducible in $\Bbb{Q}[x]$, a contradiction.

Does the proof look alright to you. I know it sounds a bit verbose. I have to keep track which set the $a(x),b(x)$ are in for every statement.

No. You used what you wanted to show. We need to modify the factors:
$$
p(x)=a(x)\cdot b(x)\in \mathbb{Q}[x] \Longrightarrow p(x)=a'(x)\cdot b'(x) \in \mathbb{Q}_{\mathbb{Z}}[x]
$$
for some new factors $a'(x),b'(x)\in \mathbb{Q}_{\mathbb{Z}}[x].$

Write it down explicitly:
$$p(x)=p_0+p_1x+p_2x^2+\ldots+p_nx^n=(a_0+a_1x+\ldots+a_sx^s)\cdot(b_0+b_1x+\ldots+b_tx^t)$$
where all coefficients $p_0,\ldots,p_n,a_0,\ldots,a_s,b_0,\ldots,b_t \in \mathbb{Q}.$

What do we know about $p_0$ and why?
What do we need $a_0,b_0$ to be from and why?
How can we achieve this goal?

I think your proof is correct, and only formulated a bit clumsy. E.g., instead of an indirect proof, we can add $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x]$ to the list of given conditions, i.e. saying that the contraposition doesn't hold for all rational polynomials, and only for those in $\mathbb{Q}_{\mathbb{Z}}[x].$

I think the explicitly written polynomials help to clarify the principle. But you should definitely strictly distinguish between the two rings. This is very much about logic rather than it is about mathematics, so rigor is crucial.

 

[elias001]

@fresh_42 before I reply to your comment on my proof, I want to ask how many questions can one ask per 24 hours on here. Also, if I have a paid membership, would that affect the number of questions. I am writing up a message to you right now, and get everything together within. So I will reply to your comment and questions about my proof after. There is an attachment of a print to pdf page of a stack exchange post, which is what the message is about and other incidents relating to it.

 

[fresh_42]

@fresh_42 before I reply to your comment on my proof, I want to ask how many questions can one ask per 24 hours on here.

There is no limit. But if it gets too many, then it might be considered trolling, and you will be warned. Many means more than someone could be assumed to handle all at once, especially if they are from different areas. Personally, I think that nobody should handle more than six different threads at a time, but there is no official limit. It will be an on-the-spot decision. As I mentioned, PF is based on dialogues, and how many dialogues can you deal with simultaneously?

Also, if I have a paid membership, would that affect the number of questions. I am writing up a message to you right now, and get everything together within. So I will reply to your comment and questions about my proof after. There is an attachment of a print to pdf page of a stack exchange post, which is what the message is about and other incidents relating to it.

We should start with a clear statement. My suggestion is:

Let $R=\mathbb{Q}[x]$ and $S=\mathbb{Q}_{\mathbb{Z}}[x]=\mathbb{Z}+x\cdot \mathbb{Q}[x]=\mathbb{Z}+xR \subseteq R.$ Then every polynomial $p\in S$ is irreducible in $R$ if and only if it is irreducible in $S.$

Proof: If $p$ is irreducible in $R,$ then $p$ is irreducible in $S\subseteq R$ because we have even fewer possible factors. Now, let $p\in S$ be irreducible (in $S$) and assume we have a factorization in the larger ring $R.$ Then we can write
$$
p=a\cdot b \text{ with } a,b\in R.
$$
$\ldots$
Please, continue.


This statement makes sure that we only talk about polynomials $p$ that are definitely elements of $S$ so that an example like $p=(1/2)x^2$ cannot occur. The abbreviations $S,R$ simply save typing. If we then turn a factorization $p=ab$ in $R$ into a factorization $p=a'b'$ for some polynomials $a',b'\in S$ then we can argue with the irreducibility of $p\in S$ and get that w.l.o.g. $a'\in S$ is a unit etc.... This is the main trick, the transition of a factorization over $R$ into one over $S.$ That's why we need $p\in S$ to begin with.

 

[elias001]

@fresh_42 I am almost done typing up my message. Before I attempt to give the proof write up another try. I want to clarify when you say $\frac{1}{2}x^2\notin\Bbb{Q}_{\Bbb{Z}}[x].$ According to Hungerford's text, it defines $\Bbb{Q}_{\Bbb{Z}}[x]$ as the set of polynomials with rational coefficients and integer constant terms. That would include having constant term $0$, and for $\frac{1}{2}x^2$, it should be in $\Bbb{Q}_{\Bbb{Z}}[x]$, since $x^2=2\cdot \frac{1}{2}x^2$?

Another thing is, if I posted something on MSE, and I don't understand the answers given there, can I re-post the question here along with the answer, and ask for clarifications? Do I have to provide the original links? I am asking this, it has to do with the message I will be sending you. I think after reading it and me explaining the situation, I think you will understand why I am asking this.

 

[fresh_42]

@fresh_42 I am almost done typing up my message. Before I attempt to give the proof write up another try. I want to clarify when you say 12x2∉QZ[x]. According to Hungerford's text, it defines QZ[x] as the set of polynomials with rational coefficients and integer constant terms. That would include having constant term 0, and for 12x2 should be in QZ[x]?

You are right. Forget my example, it is also reducible in $\mathbb{Q}_{\mathbb{Z}}[x].$ We still need $p(x)\in \mathbb{Q}_{\mathbb{Z}}[x]$ since the statement makes no sense otherwise.

 

[elias001]

@fresh_42 I am not sure how to deal with PF not rendering certain text mode properly after displaying math notations. So I typed up my attempted solution along with what I think is wrong with the author's solution. If ou go to: https://stackedit.io/ and copy and paste what I wrote, it will render the latex into math automatically.

In the attached solution, I don't think the proof is convincing. In particular from the point where it says "Since $z$ is an integer...." all the way to the end of that paragraph and also in the next paragraph "Then the integer $p(0)$ is not divisible by ....".

In the first part, where it says $a'(x)=(s/v)a(x)$ and $b'(x)=(v/s)b(x)$. Then $a'(0)=r/v$ and .... Well, I don't think this step is necessary, and also it is not convincing, since it seems it is there only for helping with the next part where $p(0)$ is not divisible by any prime number so it must be equal to $\pm 1.$ It is not convincing because in the next part, it doesn't really show why $p(x)=a(x)b(x)\in \Bbb{Q}[x], p(0)=\pm 1$ is irreducible in $\Bbb{Q}[x]$.

So here is what I did, we need to state which elements in both $\Bbb{Q}_{\Bbb{Z}}[x], \Bbb{Q}[x]$ are non-units.

For $\Bbb{Q}_{\Bbb{Z}}[x]$, the non-unit elements are $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$ and $\text{deg}(p(x))\geq 0$, with $p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=a_0\in \Bbb{Z}, a_n,a_{n-1},\ldots a_1\in \Bbb{Q}$.

For $\Bbb{Q}[x],$ because of the requirement for the condition, the only non unit element of $\Bbb{Q}[x]$ is a polynomial of the form $p(x)\in \Bbb{Q}[x], p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=\pm 1, a_n,a_{n-1},\ldots a_1\in \Bbb{Q}$. [the solution never proved this, and I don't know how to prove it. This is the other reasons why i think the author's solution is not convincing because it is missing this vital step.]

If $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x], p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=\pm 1, a_n,a_{n-1},\ldots a_1\in \Bbb{Q},$ then $p(x)\in \Bbb{Q}[x]$ is automatically irreducible.

However if $p(x)\in \Bbb{Q}[x], p(0)=\pm 1, p(x)=a'(x)b'(x)\in \Bbb{Q}[x], a'(x),b'(x)\in \Bbb{Q}[x], p(x)=a'(x)b'(x), a'(x), b'(x)\in \Bbb{Q}_{\Bbb{Z}}[x],$ we can divide the construction of the polynomial $p(x)\in \Bbb{Q}[x]$ into two cases. In both cases, automatically $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$.

Case 1: For any prime integer $p$, $a'(x)=a(x), a(0)=p, b(0)=\frac{1}{p}, b'(x)=pb(x). a'(x)b'(x)=a(x)(pb(x))=p$

Case 2: For any integer $m, n\in \Bbb{Z},$ with either being composite, say $n$, then $n$ can be factor uniquely as $n=s_1^{e_1}s_2^{e_2}\cdots s_k^{e_k}$ where each $s_i$ is some prime integer and $e_i\in \Bbb{Z}_{\geq 0}$ for $i=1,2,\dots k$. We pick an $s_i$ amongst the $s_1,s_2,\ldots s_k$. Then let $a(x),b(x)\in\Bbb{Q}[x], a(0)=n , b(0)=\frac{1}{n},$ $p(x)=a(x)b(x)\in \Bbb{Q}[x], a(x)b(x)=\pm 1,$ and let $a'(x)=a(0)s_i, b'(x)=b(0)$ then $a'(x)b'(x)\in \Bbb{Q}_{\Bbb{Z}}[x], a'(x)b'(x)=s_i or let $a'(x)=a(x), b'(x)=b(x)$ with $a'(0)=n, b'(0)=\frac{1}{n}$.

I think I covered all the cases. But i still don't know how to show for $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x], p(0)=\pm 1$ is a non-unit in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$. Thank you in advance.

 

[fresh_42]

Copy of the previous post in order to edit the LaTeX commands into the MathJax version of PF (double sharp instead of dollar for the tags) for better readability. <br /> <br /> <a href="https://www.physicsforums.com/members/572553/" class="username" data-xf-init="member-tooltip" data-user-id="572553" data-username="@fresh_42">@fresh_42</a> I am not sure how to deal with PF not rendering certain text mode properly after displaying math notations. So I typed up my attempted solution along with what I think is wrong with the author&#039;s solution. If ou go to: <a href="https://stackedit.io/" target="_blank" class="link link--external" rel="noopener">https://stackedit.io/</a> and copy and paste what I wrote, it will render the latex into math automatically.<br /> <br /> In the attached solution, I don&#039;t think the proof is convincing. In particular from the point where it says &quot;Since $z$ is an integer....&quot; all the way to the end of that paragraph and also in the next paragraph &quot;Then the integer $p(0)$ is not divisible by ....&quot;. <br /> <br /> In the first part, where it says $a&#039;(x)=(s/v)a(x)$ and $b&#039;(x)=(v/s)b(x)$. Then $a&#039;(0)=r/v$ and .... Well, I don&#039;t think this step is necessary, and also it is not convincing, since it seems it is there only for helping with the next part where $p(0)$ is not divisible by any prime number so it must be equal to $\pm 1.$ It is not convincing because in the next part, it doesn&#039;t really show why $p(x)=a(x)b(x)\in \Bbb{Q}[x], p(0)=\pm 1$ is irreducible in $\Bbb{Q}[x]$.<br /> <br /> So here is what I did, we need to state which elements in both $\Bbb{Q}_{\Bbb{Z}}[x], \Bbb{Q}[x]$ are non-units. <br /> <br /> For $\Bbb{Q}_{\Bbb{Z}}[x]$, the non-unit elements are $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$ and $\text{deg}(p(x))\geq 0$, with $p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=a_0\in \Bbb{Z}, a_n,a_{n-1},\ldots a_1\in \Bbb{Q}$.<br /> <br /> For $\Bbb{Q}[x],$ because of the requirement for the condition, the only non unit element of $\Bbb{Q}[x]$ is a polynomial of the form $p(x)\in \Bbb{Q}[x], p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=\pm 1, a_n,a_{n-1},\ldots a_1\in \Bbb{Q}$. [the solution never proved this, and I don&#039;t know how to prove it. This is the other reasons why i think the author&#039;s solution is not convincing because it is missing this vital step.]<br /> <br /> If $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x], p(x)=\sum_{i=0}^{n}a_ix^i, p(0)=\pm 1, a_n,a_{n-1},\ldots a_1\in \Bbb{Q},$ then $p(x)\in \Bbb{Q}[x]$ is automatically irreducible.<br /> <br /> However if $p(x)\in \Bbb{Q}[x], p(0)=\pm 1, p(x)=a&#039;(x)b&#039;(x)\in \Bbb{Q}[x], a&#039;(x),b&#039;(x)\in \Bbb{Q}[x], p(x)=a&#039;(x)b&#039;(x), a&#039;(x), b&#039;(x)\in \Bbb{Q}_{\Bbb{Z}}[x],$ we can divide the construction of the polynomial $p(x)\in \Bbb{Q}[x]$ into two cases. In both cases, automatically $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$. <br /> <br /> Case 1: For any prime integer $p$, $a&#039;(x)=a(x), a(0)=p, b(0)=\frac{1}{p}, b&#039;(x)=pb(x). a&#039;(x)b&#039;(x)=a(x)(pb(x))=p$<br /> <br /> Case 2: For any integer $m, n\in \Bbb{Z},$ with either being composite, say $n$, then $n$ can be factor uniquely as $n=s_1^{e_1}s_2^{e_2}\cdots s_k^{e_k}$ where each $s_i$ is some prime integer and $e_i\in \Bbb{Z}_{\geq 0}$ for $i=1,2,\dots k$. We pick an $s_i$ amongst the $s_1,s_2,\ldots s_k$. Then let $a(x),b(x)\in\Bbb{Q}[x], a(0)=n , b(0)=\frac{1}{n},$ $p(x)=a(x)b(x)\in \Bbb{Q}[x], a(x)b(x)=\pm 1,$ and let $a&#039;(x)=a(0)s_i, b&#039;(x)=b(0)$ then $a&#039;(x)b&#039;(x)\in \Bbb{Q}_{\Bbb{Z}}[x], a&#039;(x)b&#039;(x)=s_i$ or let $a&#039;(x)=a(x), b&#039;(x)=b(x)$ with $a&#039;(0)=n, b&#039;(0)=\frac{1}{n}$. <br /> <br /> I think I covered all the cases. But i still don&#039;t know how to show for $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x], p(0)=\pm 1$ is a non-unit in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$. Thank you in advance.<br /> <br /> <a href="https://www.physicsforums.com/attachments/361866/" target="_blank">View attachment 361866</a><br /> <br /> <a href="https://www.physicsforums.com/attachments/361867/" target="_blank">View attachment 361867</a>

 

[fresh_42]

Let me answer in steps. First of all, a few general remarks.

PF does not really use LaTeX, it uses a web server version called MathJax. There are several different variations of MathJax on different websites. Our version uses ## as tags for or the $-symbol. See https://www.physicsforums.com/help/latexhelp/

In general, the longer your posts are, the fewer people will read them. So it is always a good idea to concentrate on one question at a time.

First, let me see whether I can understand the proof in the book, since you are referring to it. As a reminder $\mathbb{Q}_\mathbb{Z}[x]=\mathbb{Z}+x\mathbb{Q}[x]=R$ which I will abbreviate with $R$ for easier typing. The notation $\mathbb{Q}_\mathbb{Z}[x]$ is non-standard anyway, so I might as well call it $R$ for ring.

The exercise is about the difference between prime elements and irreducible elements.

A unit $u$ is an element in a ring that has an inverse, i.e. for which the equation $u\cdot x=1$ is solvable. The units in $\mathbb{Z}$ are $\pm 1.$ All other integers cannot be inverted within $\mathbb{Z}.$

An element $f$ is irreducible if it cannot be written as a product of two non-unit factors, and reducible if such a factorization exists. We have to exclude units, since we can always multiply as many units to $f$ as we want without changing its irreducibility or reducibility. $7\in \mathbb{Z}$ is as irreducible as $-7$ is. The minus sign, a factor $u=-1$ does not change anything in terms of factorizations.

An element $p$ is prime if it is no unit (same argument as before, units won't change anything), and if
$$
p\,|\,(a\cdot b) \Longrightarrow p\,|\,a \quad\text{ or }\quad p\,|\,b.
$$
We learn at school that prime numbers are those that have only $1$ and themselves as divisors. However, that is the definition of irreducibility, not the definition of prime. The definition of prime is, that if it divides any product, then it must already divide one of the factors. $6$ divides $3\cdot 4$ but does neither divide $3$ nor $4$ so $6$ cannot be prime. But if $3$ divides any product $n_1\cdots n_k$ then $3$ has inevitably to divide one of the factors $n_i.$

Irreducible elements of the integers are exactly the prime integers. The two definitions are equivalent in the ring $\mathbb{Z},$ which is why the school definition works. (Recommendation: prove this equivalence statement to practice your skills.)

Now, primality and irreducibility are not always the same. Exercise #33 is all about this fact. It can be rewritten as:

33. aa) ($\Longrightarrow$) If $p(x)\in R$ is irreducible, then $p(x)\in \mathbb{Z}$ is a prime integer or irreducible in $\mathbb{Q}[x]$ with a integer unit as constant term, i.e. $\pm 1.$
ab) ($\Longleftarrow$) If $p(x)\in \mathbb{Z}$ is a prime integer or irreducible in $\mathbb{Q}[x]$ with a integer unit as constant term, i.e. $\pm 1,$ then $p(x)\in R$ is irreducible.
b) Every irreducible polynomial $p(x)\in R$ is prime.

I think that we can agree that 33. aa) is correct, so far as exercise 16 is. If $p$ isn't a prime number, but irreducible in $\mathbb{Q}[x],$ and we write $p(x)\in R$ as a product, then it is also a product in $\mathbb{Q}[x].$ But $p(x)$ is irreducible there, so one of the factors must be a unit. The units in $\mathbb{Q}[x]$ are the rational numbers, so one factor is a rational number. Hence, we have
$$
p(x)=\pm 1 + x\cdot f(x) = r\cdot (s+x\cdot g(x))=rs +rxg(x)
$$
and $p(0)=\pm 1=rs.$ But $r,s\in \mathbb{Z}$ so $r,s=\pm 1$ and in particular, $r=\pm 1$ is a unit in $R.$

So we are left with the cases 33. ab) and 33. b).

I will answer this in a new post.

 

[elias001]

@fresh_42 oh thank you for replying to my message. I have been feeling under the weather for this past week, again sorry for all my late replies. As for issue of latex, sometimes I type out stuff, and PF renders any tex into what you see in the first line in the below screenshot. The text are all messed up, even though the tex syntax are all correct. I thought it might be easier if I just tex it out using dollar sign and you can just check what i wrote using stackedit.io.

As to the question, I don't have a problem proving the last bit about prime elements. I just don't think in the other direction, Hungerford's solution made a convincing argument that $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x], p(0)=\pm 1$ is a non unit. I mean that is what both direction would imply as a consequence., also if we just drop the non-unit issue for second, proving that $p(x)\in \Bbb{Q}[x], p(0)=\pm 1$ is irreducible in assuming that $p(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$ is not convincing, and it doesn't completely constructive especially in the last bit showing how the constant term has to be $\pm 1$.

 

[fresh_42]

I am confused, too, by all these various variables. It always helps to sort out things. Remember, that the most important question in algebra is always: what if not?

Given: $p(x) \in R = \mathbb{Q}_\mathbb{Z}[x]=\mathbb{Z}+x\mathbb{Q}[x]$ is irreducible in $R.$

If $p(x)\in \mathbb{Z}\subseteq R$ then $p(x)=z$ for some number $z\in \mathbb{Z}.$ We need to show that $z$ is prime. If it were not prime, then there would be a proper factorization $z=a\cdot b$ which is also a factorization of $p(x)$ in $R,$ which is impossible since $p(x)$ is irreducible.

Hence, we may assume that $\deg p(x) \geq 1.$ We have ...

to show ... that $p(x)$ is also irreducible in the larger ring $\mathbb{Q}[x]\supseteq R$ and that $p(0)=\pm 1.$

This means that even with more available factors in $\mathbb{Q}[x]$ we should still be unable to find a proper factorization. What if not? If not, then there is a factorization $p(x)=a(x)\cdot b(x)$ with polynomials $a(x),b(x)$ from $\mathbb{Q}[x].$ Our goal must be to use this factorization and transform it into a factorization in $R$ from which we know it does not exist. Such a contradiction would imply that "what if not" was wrong, i.e. $p(x)=a(x)b(x)$ couldn't exist unless one of the polynomials would be a unit.

I find the proof in the book also a bit confusing, especially where the $z$ comes from. However, I think your calculation is a bit too complicated. We have
$$
p(x)=a(x)\cdot b(x) \Longrightarrow p(0)=a(0)\cdot b(0)\in \mathbb{Z}
$$
$a(0)$ and $b(0)$ are rational numbers, say $a(0)=\dfrac{r}{s}$ and $b(0)=\dfrac{u}{v}.$ We know that $\dfrac{r}{s}\cdot \dfrac{u}{v}$ is an integer since $p(x)\in R.$ We also know (per assumption that the quotients are fully cancelled) that $(r,s)$ and $(u,v)$ are coprime. So all prime factors of $v$ must be prime factors of $r$ and all prime factors of $s$ must be prime factors of $u.$ Hence $v\,|\,r$ and $s\,|\,u,$ or in other words $\dfrac{r}{v}\in \mathbb{Z}$ and $\dfrac{u}{s}\in \mathbb{Z}.$ This means that we can define two new polynomials
\begin{align*}
a'(x)&=\dfrac{s}{v}\cdot a(x)=\dfrac{s}{v}\cdot (a(0)+xa_1(x))=\dfrac{s}{v}\cdot \dfrac{r}{s}+xa_2(x)=\dfrac{r}{v}+xa_2(x)\in R\\
b'(x)&=\dfrac{v}{s}\cdot b(x)=\dfrac{v}{s}\cdot (b(0)+xb_1(x))=\dfrac{v}{s}\cdot \dfrac{u}{v}+xb_2(x)=\dfrac{u}{s}+xb_2(x)\in R\\
\end{align*}
for some rational polynomials $a_i,b_i.$ The point is that we now have two new polynomials $a'(x)$ and $b'(x)$ which are both in $R.$ Finally,
$$
a'(x)\cdot b'(x)= \dfrac{s}{v}\cdot a(x) \cdot \dfrac{v}{s}\cdot b(x)= a(x)\cdot b(x) = p(x).
$$
Hence, we have transformed our assumed factorization over $\mathbb{Q}[x]$ into a factorization over $R.$ But such a proper factorization does not exist.

So one of the factors, say $a'(x),$ must be a unit of $R.$ Now the only units, invertible elements in $R$ are $\pm 1.$ But if $a'(x)=\pm 1$ then $p(x)=\pm b'(x)$ and $p(0)=\pm b'(0)=\dfrac{v}{s}b(0)=\dfrac{u}{v}.$ Now $p(0)\in \mathbb{Z}$ by definition of $p(x).$ This means $v\,|\,u$ which we assumed to be coprime. That means $v=\pm 1$ and $p(0)=\pm u.$

Now, what is $u$? We need to show that $u=\pm 1.$ What if not? If not, then there is a prime $q\in \mathbb{Z}$ such that $q\,|\,u.$ We can certainly write $p(x)=q\cdot (1/q)\cdot p(x).$ Let's look closer at the second factor. We have $\left(\dfrac{1}{q}\cdot p(x)\right)(0)=\dfrac{1}{q}\cdot p(0)=\pm \dfrac{u}{q}\in \mathbb{Z}$ since $q$ is a prime divisor of $u.$ But this is a proper factorization in $R,$ which is not possible by the choice of $p(x).$ So "what if not" was wrong, i.e., there is no such prime $q.$ But if there is no such prime, then $u=p(0)=\pm 1$ what had to be shown.

This was the proof for 33. ab). It was along the lines of the book, just a bit more detailed and without this strange $z.$ Books are usually right, at least up to typos.

I suggest that you look at the rest, 33. b) (every irreducible element of $R$ is prime) as I did. Take the book as a guide and always ask What if not?

Given in this case (33. b)): $p(x)\in R$ is irreducible (and now we know what it looks like) plus an equation
$$
p(x)\cdot a(x)=c(x)\cdot d(x) \quad\text{ for some polynomials }a(x), c(x), d(x)\in R
$$
plus an assumption that $p(x)$ does not divide $d(x).$

To be shown: $p(x)$ divides $c(x).$

But I would first prove what is claimed in the text.

I think you have a tendency to complicate things. But it could also be that I am making assumptions about you based on myself.

 

[elias001]

@fresh_42 After reading your explanation of Hungerford's solution, would it not be simpler if I show that $\sum_{i=1}^{n}a_ix^i\pm 1$ is not an unit in both $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$.

Here is my idea:

If $\sum_{i=1}^{n}a_ix^i\pm 1$ is an unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$, then we know that $\sum_{i=1}^{n}a_ix^i$ is an non unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$, while $\pm 1$ is an unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$ and $\text{deg}(\sum_{i=1}^{n}a_ix^i\pm 1)\geq 0.$

Since $\sum_{i=1}^{n}a_ix^i\pm 1$ is an unit element means there exists some inverse element $b\in \Bbb{Z}$, $(\sum_{i=1}^{n}a_ix^i\pm 1)b=(\sum_{i=1}^{n}a_ix^i)b +(\pm 1)b=1$, which imply that $\sum_{i=1}^{n}a_ix^i\pm 1$ is an inverse. Then either $(\sum_{i=1}^{n}a_ix^i)b=1, (\pm 1)b=0$ or $(\sum_{i=1}^{n}a_ix^i)b=0, (\pm 1)b=1$. In the first case, is not possible since $\sum_{i=1}^{n}a_ix^i$ is not an unit. In the latter case, $(\sum_{i=1}^{n}a_ix^i)b=1$ would mean that $\text{deg}(\sum_{i=1}^{n}a_ix^i\pm 1)=0$ so we have a contradiction.

Does the proof look okay?

 

[fresh_42]

@fresh_42 After reading your explanation of Hungerford's solution, would it not be simpler if I show that $\sum_{i=1}^{n}a_ix^i\pm 1$ is not an unit in both $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$.

I don't see where units play a role at all. We are discussing factorizations and irreducibility. That has little to do with units. It isn't an either-or. Units are only a nasty factor that can be everywhere without changing irreducibility or primality. They disturb uniqueness, but that is all.

Look at the integers: the units (invertible integers) are $\pm 1,$ which means that we cannot distinguish between the prime number $3$ and the prime number $-3.$ They are basically the same; associated is the technical term. Also, we have prime (irreducible) elements and non-prime (reducible) elements. Not every integer different from $\pm 1$ is prime, or irreducible.

The exercise is about the factorization in two different rings, $R\subseteq S,$ and the question is whether irreducibility in $R$ implies irreducibility in $S.$ No unit anywhere. Here is another example: $x^2+1$ is irreducible in $\mathbb{Z}[x]$ but $x^2+1=(x+i)(x-i)$ so it is reducible in $\mathbb{Z}[ i ][x]$ or $\mathbb{C}[x].$ This is an example in which irreducibility in the smaller ring $\mathbb{Z}[x]$ collapses in the larger ring $\mathbb{Z}[ i ][x]$ or $\mathbb{C}[x].$ This is the subject of the exercise.

Here is my idea:

If $\sum_{i=1}^{n}a_ix^i\pm 1$ is an unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$, then we know that $\sum_{i=1}^{n}a_ix^i$ is an non unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$, while $\pm 1$ is an unit element in $\Bbb{Q}_{\Bbb{Z}}[x]\cap \Bbb{Q}[x]$ and $\text{deg}(\sum_{i=1}^{n}a_ix^i\pm 1)\geq 0.$

Since $\sum_{i=1}^{n}a_ix^i\pm 1$ is an unit element means there exists some inverse element $b\in \Bbb{Z}$,

Why that? If $p(x)\in \mathbb{Q}_\mathbb{Z}[x]$ is a unit, then there is another polynomial $q(x)\in \mathbb{Q}_\mathbb{Z}[x]$ such that $p(x)\cdot q(x) = 1.$

$(\sum_{i=1}^{n}a_ix^i\pm 1)b=(\sum_{i=1}^{n}a_ix^i)b +(\pm 1)b=1$, which imply that $\sum_{i=1}^{n}a_ix^i\pm 1$ is an inverse. Then either $(\sum_{i=1}^{n}a_ix^i)b=1, (\pm 1)b=0$ or $(\sum_{i=1}^{n}a_ix^i)b=0, (\pm 1)b=1$. In the first case, is not possible since $\sum_{i=1}^{n}a_ix^i$ is not an unit. In the latter case, $(\sum_{i=1}^{n}a_ix^i)b=1$ would mean that $\text{deg}(\sum_{i=1}^{n}a_ix^i\pm 1)=0$ so we have a contradiction.

Does the proof look okay?

No, because $b\in \mathbb{Z}$ is a wrong assumption. It is far easier than that. Write $p(x)=p_0+xp'(x)$ and $q(x)=q_0+xq'(x).$ Then we get
$$
1= 1+x\cdot 0 +x^2\cdot 0 = p(x)\cdot q(x)= p_0q_0+ x(q_0p'(x)+p_0q'(x))+x^2p'(x)q'(x).
$$
Now compare the polynomials on both sides. We get from this comparison
\begin{align*}
1&=p_0q_0\\
0&=q_0p'(x)+p_0q'(x)\\
0&=p'(x)q'(x)
\end{align*}
Both rings are integral domains, so the last equation forces one of the factors to be zero. Say $p'(x)=0.$ Plugging this into the second equation, we get $p_0q'(x)=0.$ But the first equation rules out that $p_0=0.$ Hence, $q'(x)=0.$ This means $p(x)=p_0$ and $q(x)=q_0.$

Up to this point, we have only used that there are no zero divisors in our rings. Now we make the distinction between $\mathbb{Q}_\mathbb{Z}[x]$ and $\mathbb{Q}[x].$

In the case of $\mathbb{Q}_\mathbb{Z}[x],$ the elements $p_0,q_0$ are integers. The only integers which solve the first equation are $\pm 1.$ So $p(x)=\pm 1$ are the only units in $\mathbb{Q}_\mathbb{Z}[x].$

In the case of $\mathbb{Q}[x],$ the elements $p_0,q_0$ are rationals. All rationals different from zero solve the first equation with $q_0=p_0^{-1}.$ So $p(x)= p_0 \in \mathbb{Q}\setminus \{0\}$ are the units in $\mathbb{Q}[x].$

We don't need the polynomials to be written out, or their degrees.

We also cannot change the rings like we want. Units, irreducibility, and primality depend on the ring. So we must always know where we are in. Try to think simple from step to step, and always keep the ring $\mathbb{Z}$ in mind as a reservoir of examples.