Degree of a product

Builds on Degree of a polynomial Integral domain Leading coefficient

Statement

Let $R$ be an integral domain and let $f, g \in R[x]$ . Then $\deg(f g) = \deg f + \deg g$ , with the convention $\deg 0 = -\infty$ .

Remarks

The proof: if $f$ and $g$ are nonzero with leading coefficients $a$ and $b$ , then the coefficient of $x^{\deg f + \deg g}$ in $f g$ is $a b \neq 0$ . Over every ring, $\deg(f g) \le \deg f + \deg g$ . The equality fails in $(\mathbb{Z}/4\mathbb{Z})[x]$ : $(2x + 1)(2x + 1) = 4x^2 + 4x + 1 = 1$ .

Reading path

To understand this page from the ground up, read these 9 pages in order. The foundations come first.

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Group concept
Ring definition
Commutative ring definition
Polynomial ring definition
Zero divisor definition
Degree of a polynomial definition
Integral domain definition
Leading coefficient definition
Degree of a product statement

Where this leads

31 pages build on this one.

Tag	`0T233AZ1GE`
URI	`m2://0T233AZ1GE`
Kind	statement, proposition
Writing	W0 stub
MSC	13F20
Status	This copy of the site has no build data. Run `m2 status 0T233AZ1GE`.
Uses	Degree of a polynomial, Integral domain, Leading coefficient

Degree of a product

Statement

Remarks

Reading path

Where this leads

Further reading

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