2
$\begingroup$

Let $R$ be an integral domain, $f, g \in R[x], a \in R$. Prove that if $a$ is a root of $fg$, then $a$ is a root of $f$ or $a$ is a root of $g$.

My attempt (edited):

pf. Let $R$ be an integral domain, $f, g \in R[x], a \in R$. If $a$ is a root of $fg$, then $fg(a)$ = 0. Since $R$ is an integral domain, a nonzero non-unit $r \in R$ is irreducible if for all $a, b \in R$ such that $r = ab$, either $a$ or $b$ is a unit. Hence if $a$ is a root of $fg$, $b$ must be a unit. Thus, $b^-1 \in R$ and so $b^-1fg = abb^-1$. Hence, $a = b^-1fg$. Thus, $fg \vert a$ so $a$ must divide $f$ or $g$.

I am not too sure about the last couple of statements. Would appreciate any feedback!

$\endgroup$
10
  • 1
    $\begingroup$ I'm sorry to say your proof is way off. You may want to check the definition of what it means for an element $c \in R$ to be a root of a polynomial $h$ in $R[x]$. $\endgroup$
    – D_S
    Nov 24 '19 at 4:04
  • 1
    $\begingroup$ So, $2$ is a root of $f(x)=x^2-x-2$. What do you conclude? $\endgroup$ Nov 24 '19 at 4:06
  • 1
    $\begingroup$ From the point $a |fg$, which makes no sense, you have not done anything right. $\endgroup$ Nov 24 '19 at 4:12
  • $\begingroup$ @TedShifrin $f(2)$ = 0. $\endgroup$
    – yagayeet
    Nov 24 '19 at 4:22
  • $\begingroup$ Well, sure. You are supposed to conclude more. Say something about factors of $f(x)$. $\endgroup$ Nov 24 '19 at 4:25
2
$\begingroup$

If $a \in R$ is a root of

$f(x)g(x) \in R[x], \tag 1$

then

$f(a)g(a) = fg(a) = 0; \tag 2$

now if

$f(a) \ne 0 \ne g(a), \tag 3$

then since $R$ is an integral domain we have

$f(a)g(a) \ne 0; \tag 4$

but this contradicts (2); thus (3) is false and hence

$f(a) = 0 \; \text{or} \; g(a) = 0; \tag 5$

that is, $a$ is a root of at least one of $f(x)$, $g(x)$.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for your detailed response. Would I necessarily need a proof by contradition though? After stating that $fg(a)$ = 0, could I simply state "Since $R$ is an integral domain, $f(a)g(a)$ = 0. Hence, either $f(a)$ = 0 or $g(a)$ = 0. Therefore $a$ is a root of $f$ or $a$ is a root of $g$." $\endgroup$
    – yagayeet
    Nov 25 '19 at 0:33
  • $\begingroup$ @yagayeet: your argument is fine; reductio ad absurdum is not necessary. Cheers! $\endgroup$ Nov 25 '19 at 1:41
0
$\begingroup$

Let $R$ be an integral domain, $f,g\in R[x],$ $a\in R.$ If $a$ is a root of $fg,$ then $fg(a) = 0.$ Since $R$ is an integral domain, a nonzero non-unit $r\in R$ is irreducible if for all $a,b\in R$ such that $r=ab,$ either $a$ or $b$ is a unit.

All of this is fine.

Hence if $a$ is a root of $fg$, $b$ must be a unit.

A stylistic comment: at this point, you haven't defined what $b$ is, but you start talking about it as if your reader will know what it is. How does $b$ relate to $f,$ $g,$ and $a$? You defined what it means for an element $r\in R$ to be irreducible in general, but now you seem to be applying it without explaining what you're assuming is irreducible.

Thus, $b^{−1}\in R$ and so $b^{-1}fg = abb^{-1}$. Hence, $a=b^{-1}fg$.

Where did the equation $b^{-1}fg = abb^{-1} = a$ come from? This is equivalent to saying that $fg = ab,$ which was not an assumption at any point of the problem. Remember that at the beginning of the problem, we assume that $fg(a) = f(a)g(a) = 0$ (remember also that if $p\in R[x]$ is a polynomial and $r\in R,$ then $p(r)$ does not mean $p\cdot r$ (the product of the polynomial $p$ and the constant polynomial $r$) but rather the evaluation of $p$ at $r$ (what you get when you replace all the $x$'s in $p(x)$ with $r$'s).

Thus, $fg\mid a$ so $a$ must divide $f$ or $g$.

Indeed, $a = b^{-1}fg$ implies that $fg$ divides $a.$ However, to conclude from this that $a$ divides $f$ or $g$ requires that $a$ be a prime element, which again is not an assumption of the problem. Moreover, even if you could conclude that $a$ divides $f$ or $g,$ this does not show that $a$ is a root of $f$ or of $g,$ because the definition of $a\in R$ being a root of $f$ is not that $a$ divides $f,$ but rather that $f(a) = 0.$

In order to actually prove the statement, you should proceed as described in Robert Lewis' answer. Either try a proof by contradiction assuming that $a$ is not a root of $f$ or $g,$ so that $f(a)\neq 0$ and $g(a)\neq 0$ and then use the fact that $R$ is an integral domain to conclude that $fg(a)\neq 0,$ so that $a$ is not a root of $fg$ (contradiction!).

$\endgroup$
1
  • $\begingroup$ Hello, thank you for your thorough answer. When I posted my question, I started off by saying "If $a$ is a root of $fg$, then $a \vert fg$." Shortly later after @Ted Shifrin commented I changed that statement to: "If $a$ is a root of $fg$, then $fg(a) = 0$." Thus, the latter statements do not make sense. $\endgroup$
    – yagayeet
    Nov 24 '19 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.