Prove there is a polynomial $d(x) \in \mathbb Q[x]$ that is a gcd of $f(x)$ and $g(x)$ and whose term of minimal degree is $d_rx^r.$ (D&F #9.3.5(a)).

Question

Here is the question I am trying to understand its solution:

Let $R = \mathbb Z + x \mathbb Q[x] \subset \mathbb Q[x]$ be the set of polynomials in $x$ with rational coefficients whose constant is an integer.

$(a)$ Suppose that $f(x), g(x) \in \mathbb Q[x]$ are two nonzero polynomials with rational coefficients and that $x^r$ is the largest power of $x$ dividing both $f(x)$ and $g(x)$ in $\mathbb Q[x],$ (i.e., $r$ is the degree of the lowest order term appearing in either $f(x)$ or $g(x)$). Let $f_r$ and $g_r$ be the coefficients of $x^r$ in $f(x)$ and $g(x),$ respectively ( one of which is nonzero by definition of $r$). Then $\mathbb Z f_r + \mathbb Z g_r = \mathbb Z d_r.$ Prove that there is a polynomial $d(x) \in \mathbb Q[x]$ that is a gcd of $f(x)$ and $g(x)$ and whose term of minimal degree is $d_rx^r.$

$(b)$ Prove that $f(x) = d(x)q_1(x)$ and $g(x) = d(x)q_2(x)$ where $q_1(x)$ and $q_2(x)$ are elements of the subring $R$ of $\mathbb Q[x].$

Here are some thoughts:

I know that: $R$ is an integral domain by the previous problem.

How can I show the existence of such $d(x)$? should I prove that it is a Bezout domain? or should I just find it and say hey this is the gcd and prove that it is a gcd?

For letter (b), I do not know what exactly should I do.

Any help will be greatly appreciated!

Edit:

Here is a way of just giving it:

Let $d(x) \in \mathbb Q[x]$ be the greatest common divisor of $f(x)$ and $g(x)$ in $\mathbb Q[x].$ We must have $x^r | d(x)$ by definition. If $x^{r + 1}| d(x)$ then $x^{r + 1}$ divides both $f(x)$ and $g(x),$ a contradiction. Therefore $x^r$ is the lowest term in $d(x).$ Let $d_t$ be the coefficient of this term in $d(x)$. This $d_t$ must divide both $f_r$ and $g_r,$ so $d_t | d_r$ as $\mathbb Z f_r + \mathbb Z g_r = \mathbb Z d_r$ by the given. On the other hand, $d_t \in \mathbb Z f_r + \mathbb Z g_r,$ so $d_r | d_t$ and therefore $d_t = d_r.$

Note that: Since $x^r$ is the largest power of $x$ dividing both $f(x)$ and $g(x),$ we can write $f(x)= f_r x^r + \dots$ and $g(x)= g_r x^r + \dots.$ Without loss of generality assume that $f_r$ is not zero.

Can I use Bezout identity here?if so why? I know that Dummit & Foote gave it in exercise 7 of section 8.2, but why it is applicable here?

How can I rigorously prove that $d(x) = d_r x^r$ ?if this was a correct guess.

Any help will be greatly appreciated!

Answer 1

The statement $\Bbb Z f_r + \Bbb Z g_r = \Bbb Z d_r$ implies (how?) that $f(x)=d_rx^rP(x)$ and $g(x) = d_rx^rQ(x)$, where $P(x), Q(x) \in R$, and the constant coefficients $\hat p,\hat q$ of $P(x), Q(x)$ are coprime.

Let $h(x) \in R$ be the gcd (in $\Bbb Q[x]$) of $P(x)$ and $Q(x)$ with constant coefficient $1$. $h(x)$ exists and is unique. Set $d(x) = d_r x^r h(x)$. Then the polynomials $f(x)/d(x)$, $g(x)/d(x)$ have constant terms $\hat p$ and $\hat q$ respectively.

I think I can also prove that $h$ is a $\Bbb Z$-linear combination of $P(x)$ and $Q(x)$, but that's not asked in the problem.

Answer 2

(a)
1. First let’s find out how the equation $\Bbb{Z}f_r+\Bbb{Z}f_r=\Bbb{Z}d_r$ is obtained (if this is clear, omit the step 1.). We have $f_r,g_r∈\Bbb{Q}$, so that $f_r=\frac{a_1}{b_1} ,g_r=\frac{a_2}{b_2}$ for some $a_1,b_1,a_2,b_2∈\Bbb{Z}$. Consider the subgroup $〈f_r,g_r 〉=〈\frac{a_1}{b_1},\frac{a_2}{b_2}〉$ of $\Bbb{Q}$ generated under addition “+” by these elements (the notation “$〈\frac{a_1}{b_1},\frac{a_2}{b_2}〉$” here means precisely the subgroup of $\Bbb{Q}$ generated by $\frac{a_1}{b_1},\frac{a_2}{b_2}$ under addition, not the ideal $“(\frac{a_1}{b_1},\frac{a_2}{b_2})”$ of $\Bbb{Q})$. Obviously, $〈\frac{a_1}{b_1} ,\frac{a_2}{b_2}〉≤〈\frac{1}{b_1 b_2}〉$, so that $〈\frac{a_1}{b_1} ,\frac{a_2}{b_2}〉$ is cyclic as a subgroup of a cyclic subgroup $〈\frac{1}{b_1 b_2}〉$ of $\Bbb{Q}$. Therefore, $〈\frac{a_1}{b_1} ,\frac{a_2}{b_2}〉=〈\frac{c}{b_1 b_2}〉$, where $\frac{c}{b_1 b_2}=\sum_{i=1}^c \frac{1}{b_1 b_2} $ (so that $c∈\Bbb{Z}$) is an element of $〈\frac{1}{b_1 b_2}〉$ — generator of $〈\frac{a_1}{b_1},\frac{a_2}{b_2}〉$. Denoting $\frac{c}{b_1 b_2}=d_r$, we get that $〈f_r,g_r 〉=〈d_r〉 \;⇔\; \{k_1 f_r+k_2 g_r \;|\; k_1,k_2∈\Bbb{Z}\}=\{k_3 d_r \;|\; k_3∈\Bbb{Z}\} \;⇔\; $ $ \{k_1 f_r\; |\; k_1∈\Bbb{Z}\}+\{k_2 g_r \;|\; k_2∈\Bbb{Z}\}=\{k_3 d_r \; |\; k_3∈\Bbb{Z}\} \;⇔\; \Bbb{Z}f_r+\Bbb{Z}f_r=\Bbb{Z}d_r,$
where $f_r=\frac{a_1}{b_1}, \; g_r=\frac{a_2}{b_2}, \; d_r=\frac{c}{b_1 b_2}$, so that there indeed exists $d_r∈\Bbb{Q}$ satisfying the equation given.

2. Since $\Bbb{Q}$ is a field, $\Bbb{Q}[x]$ is a Euclidean Domain, hence $\Bbb{Q}[x]$ is a Principal Ideal Domain, P.I.D. Hence the ideal $(f(x),g(x)) $ in $\Bbb{Q}[x]$ is a principal ideal, so that $(f(x),g(x))=(δ(x))$ for some $δ(x)∈\Bbb{Q}[x]$. According to its definition in the “language of ideals”, a g.c.d. of $f(x),g(x)$ is a generator of the unique smallest principal ideal, containing the ideal $(f(x),g(x))$. Hence, $δ(x)=g.c.d.(f(x),g(x))$ (this proves existence of a g.c.d. of $f(x),g(x)$ in $\Bbb{Q}[x]$).

Since $x^r∣f(x), \; x^r∣g(x)$, by the definition of a g.c.d., $x^r∣δ(x)$, hence $δ(x)$ cannot have terms of degree $<r$. Following the statement of the exercise, the term of minimal degree of at least one polynomial, say, $f(x)$, is of degree $r$, while the degree of the term of minimal degree of the other polynomial $g(x)$ is $≥r$. Since $δ(x)∣f(x), \; f(x)=δ(x)γ(x)$ for some $γ(x)∈\Bbb{Q}[x]$, and it follows that the term of minimal degree of $δ(x)$ cannot be $>r$, since then the degree of the term of minimal degree of $f(x)$ would be $>r$, a contradiction. Thus, the degree of the term of minimal degree of $δ(x)$ can only be $r$, i.e., $δ(x)=δ_n x^n+δ_{n-1} x^{n-1}+⋯+δ_r x^r$ with $δ_i∈\Bbb{Q}$.

A g.c.d. is unique up to associates (i.e., up to multiplication by a unit), because for any unit $u$ we have $(δ(x))=(uδ(x))$, i.e., $uδ(x)$ is also a generator of our unique smallest principal ideal containing $(f(x),g(x))$. Since $\Bbb{Q}$ is a field, every its element is a unit. So, $u=δ_r^{-1} d_r$ is a unit in $\Bbb{Q}$ hence in $\Bbb{Q}[x]$, therefore, $uδ(x)=uδ_n x^n+uδ_{n-1} x^{n-1}+⋯+d_r x^r$ is also a $g.c.d.(f(x),g(x))$. Assuming $uδ(x)=d(x)$, we get that the polynomial $d(x)=uδ_n x^n+uδ_{n-1} x^{n-1}+⋯+d_r x^r$ is a $g.c.d.(f(x),g(x))$ in $\Bbb{Q}[x]$, whose term of minimal degree is $d_r x^r$ with $d_r$ satisfying the equation $\Bbb{Z}f_r+\Bbb{Z}g_r=\Bbb{Z}d_r$.

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(b)
Since $d(x)=g.c.d.(f(x),g(x))$, it follows that $d(x)∣f(x), \; d(x)∣g(x)$, so that $f(x)=d(x) q_1 (x)$ and $g(x)=d(x) q_2 (x)$ for some $q_1 (x),q_2 (x)∈\Bbb{Q}[x]$.

The degree of the term of minimal degree of $f(x)$ is $r$, as before in (a). Then, since the degree of the term of minimal degree of $d(x)$ is also $r$, it can only be that $q_1 (0)≠0$, so that $f_r=d_r q_1 (0)$. The equation $\Bbb{Z}f_r+\Bbb{Z}g_r=\Bbb{Z}d_r \; ⇔ \; \{k_1 f_r\; |\; k_1∈\Bbb{Z}\}+\{k_2 g_r \;|\; k_2∈\Bbb{Z}\}=\{k_3 d_r \; |\; k_3∈\Bbb{Z}\}$ implies that we can assume $k_1=1, \; k_2=0$, so that $1f_r+0g_r=kd_r$ for some $k_3=k∈\Bbb{Z}$. So we have: $f_r=d_r q_1 (0), \; f_r=d_r k$, therefore, $d_r q_1 (0)=d_r k \; ⇔ \; q_1 (0)=k$, hence $q_1 (0)∈\Bbb{Z}$, therefore, $q_1 (x)∈R=\Bbb{Z}+x\Bbb{Q}[x]$.

The degree of the term of minimal degree of $g(x)$ is $≥r$, as before in (a). If it is equal to $r$, the argument analogous to the just given for $f(x)$ leads to the same result, so that $q_2 (x)∈R$. If it is $>r$, then necessarily $q_2 (0)=0$ so as not to get a nonzero term $d_r q_2 (0) x^r$ of degree $r$ in $g(x)$. Since, obviously, $0∈\Bbb{Z}$, again $q_2 (x)∈R$.

⦁ Thus, $f(x)=d(x) q_1 (x), \; g(x)=d(x) q_2 (x)$ with $q_1(x), q_2(x) ∈R$.