# Prove that if $a$ is a root of $fg$, $a$ is a root of $f$ or $a$ is a root of $g$.

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!

• 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 $R[x]$.
– D_S
Nov 24 '19 at 4:04
• So, $2$ is a root of $f(x)=x^2-x-2$. What do you conclude? Nov 24 '19 at 4:06
• From the point $a |fg$, which makes no sense, you have not done anything right. Nov 24 '19 at 4:12
• @TedShifrin $f(2)$ = 0. Nov 24 '19 at 4:22
• Well, sure. You are supposed to conclude more. Say something about factors of $f(x)$. Nov 24 '19 at 4:25

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 $$f(x)$$, $$g(x)$$.

• 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 $g$." Nov 25 '19 at 0:33
• @yagayeet: your argument is fine; reductio ad absurdum is not necessary. Cheers! Nov 25 '19 at 1:41

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!).

• 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. Nov 24 '19 at 5:41