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The Division Algorithm

Recall that the division algorithm for integers () says that if a and b are integers with b \gt 0, then there exist unique integers q and r such that a = bq + r, where 0 \leq r \lt b.

The Division Algorithm

Recall that the division algorithm for integers () says that if \(a\) and \(b\) are integers with \(b \gt 0\), then there exist unique integers \(q\) and \(r\) such that \(a = bq + r\), where \(0 \leq r \lt b\). The algorithm by which \(q\) and \(r\) are found is just long division. A similar theorem exists for polynomials. The division algorithm for polynomials has several important consequences. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review at this point.

Example

The division algorithm merely formalizes long division of polynomials, a task we have been familiar with since high school. For example, suppose that we divide \(x^3 - x^2 + 2 x - 3\) by \(x - 2\).

			\(x^2\)	\(+\)	\(x\)	\(+\)	\(4\)
\(x\)	\(-\)	\(2\)	\(x^3\)	\(-\)	\(x^2\)	\(+\)	\(2x\)	\(-\)	\(3\)
			\(x^3\)	\(-\)	\(2x^2\)
					\(x^2\)	\(+\)	\(2x\)	\(-\)	\(3\)
					\(x^2\)	\(-\)	\(2x\)
							\(4x\)	\(-\)	\(3\)
							\(4x\)	\(-\)	\(8\)
									\(5\)

Hence, \(x^3 - x^2 + 2 x - 3 = (x - 2) (x^2 + x + 4 ) + 5\).

Let \(p(x)\) be a polynomial in \(F[x]\) and \(\alpha \in F\). We say that \(\alpha\) is a zero or root of \(p(x)\) if \(p(x)\) is in the kernel of the evaluation homomorphism \(\phi_{\alpha}\). All we are really saying here is that \(\alpha\) is a zero of \(p(x)\) if \(p(\alpha) = 0\).

Corollary

Let \(F\) be a field. An element \(\alpha \in F\) is a zero of \(p(x) \in F[x]\) if and only if \(x - \alpha\) is a factor of \(p(x)\) in \(F[x]\).

Proof

Suppose that \(\alpha \in F\) and \(p( \alpha ) = 0\). By the division algorithm, there exist polynomials \(q(x)\) and \(r(x)\) such that \[\begin{aligned}\end{aligned}\] and the degree of \(r(x)\) must be less than the degree of \(x -\alpha\). Since the degree of \(r(x)\) is less than \(1\), \(r(x) = a\) for \(a \in F\); therefore, \[\begin{aligned}\end{aligned}\]. But \[\begin{aligned}\end{aligned}\] consequently, \(p(x) = (x - \alpha) q(x)\), and \(x - \alpha\) is a factor of \(p(x)\).

Conversely, suppose that \(x - \alpha\) is a factor of \(p(x)\); say \(p(x) = (x - \alpha) q(x)\). Then \(p( \alpha ) = 0 \cdot q(\alpha) = 0\).

Notice the similarity between the proof of and the proof of .

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Ugbu a ị Enweghị kalkulata na-ewepụ nke a, ma ọ bụ na ọ nwere ike ijikwa ya. Jiri otu n'okpuru, mọọbụ pịa nke gị.

Kwụsị ọrụ gị

Akwụkwọ ndebanye n'efu na-egbakwunye ntọala n'ụdị ọmụmụ ihe ọbụla, ndepụta nke ihe ị ga-eme, nsogbu ị ga-ewepụ n'otu ebe, na onye nkụzi ị ga-ajụ banyere ihuakwụkwọ a. Akwụkwọ ndepụta ihe bụ nke ọbụla mepere maka onye ọbụla, ọbụla ma ọ bụ na ọ bụghị.

Akaụntụ Nbanye

Aịkọn ndị ahụ ejirila ebe a

Tinye ọbụla simbụl maka nkọwa zuru ezu, inyogo, na ihe ọbụla ọbụla na-egosi.

Ajụjụ ndị mmadụ jụrụ

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Oge nke ihuakwụkwọ a bụ nke a gbanwere site na Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). E kọwaara ya na ebe a; nsogbu bụ nke anyị.

Oge Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula