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Given $F$ is integral domain, prove $F[X]$ is integral domain

Need to prove:(I did not use the condition: $F$ is integral domain)

Proof: $f(x)g(x) = 0 \Leftrightarrow f(x) = 0 \text{ or } g(x) = 0 $

Here is my proof, can anyone check whether it make scene or not $$ \begin{align*} \deg(f(x)g(x)) &= \deg(f(x)) + \deg(g(x)) \\ f(x)g(x) &= 0 \\ \implies \deg(f(x)g(x)) &= \deg(f(x)) + \deg(g(x)) = \deg(0) \\ \because \deg(f(x)) + \deg(g(x)) &= - \infty \\ \implies \deg(f(x)) &= -\infty \text{ or } deg(g(x)) = -\infty \\ \implies f(x) &= 0 \text{ or } g(x) = 0 \end{align*} $$

Aron Lee
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1 Answers1

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You need that $F$ is integral domain when you use $\deg (fg)=\deg(f) + \deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= \mathbb Z/ 4\mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-\infty =\deg(fg)\neq \deg(f)+\deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $f\neq 0$ and $g\neq 0$. If $f = a_nx^n + \cdots+a_0$ and $g(x)= b_mx^m + \cdots + b_0$, where $a_n\neq 0$ and $b_m\neq 0$, then we have $$fg = a_nb_m x^{n+m}+\cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m \neq 0$ and, therefore, $$\deg(fg) = n+m = \deg(f)+\deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-\infty$.

Hugo
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  • The rest of your proof is correct. – Hugo Jul 15 '18 at 01:58
  • I was studying your answer. I know that deg $(0)$ is undefined. But why deg $(0) = -\infty$? Can I write deg $(0) = \infty$? –  Dec 27 '21 at 15:35