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Splitting Fields

Let F be a field and p(x) be a nonconstant polynomial in F[x]. We already know that we can find a field extension of F that contains a root of p(x).

Splitting Fields

Let \(F\) be a field and \(p(x)\) be a nonconstant polynomial in \(F[x]\). We already know that we can find a field extension of \(F\) that contains a root of \(p(x)\). However, we would like to know whether an extension \(E\) of \(F\) containing all of the roots of \(p(x)\) exists. In other words, can we find a field extension of \(F\) such that \(p(x)\) factors into a product of linear polynomials? What is the smallest extension containing all the roots of \(p(x)\)?

Let \(F\) be a field and \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\) be a nonconstant polynomial in \(F[x]\). An extension field \(E\) of \(F\) is a splitting field of \(p(x)\) if there exist elements \(\alpha_1, \ldots, \alpha_n\) in \(E\) such that \(E = F( \alpha_1, \ldots, \alpha_n )\) and \[\begin{aligned}\end{aligned}\]. A polynomial \(p(x) \in F[x]\) splits in \(E\) if it is the product of linear factors in \(E[x]\).

Example

Let \(p(x) = x^4 + 2x^2 - 8\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has irreducible factors \(x^2 -2\) and \(x^2 + 4\). Therefore, the field \({\mathbb Q}( \sqrt{2}, i )\) is a splitting field for \(p(x)\).

Example

Let \(p(x) = x^3 - 3\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has a root in the field \({\mathbb Q}( \sqrt[3]{3}\, )\). However, this field is not a splitting field for \(p(x)\) since the complex cube roots of \(3\), \[\begin{aligned}\end{aligned}\], are not in \({\mathbb Q}( \sqrt[3]{3}\, )\).

Theorem

Let \(p(x) \in F[x]\) be a nonconstant polynomial. Then there exists a splitting field \(E\) for \(p(x)\).

Proof

We will use mathematical induction on the degree of \(p(x)\). If \(\deg p(x) = 1\), then \(p(x)\) is a linear polynomial and \(E = F\). Assume that the theorem is true for all polynomials of degree \(k\) with \(1 \leq k \lt n\) and let \(\deg p(x) = n\). We can assume that \(p(x)\) is irreducible; otherwise, by our induction hypothesis, we are done. By , there exists a field \(K\) such that \(p(x)\) has a zero \(\alpha_1\) in \(K\). Hence, \(p(x) = (x - \alpha_1)q(x)\), where \(q(x) \in K[x]\). Since \(\deg q(x) = n -1\), there exists a splitting field \(E \supset K\) of \(q(x)\) that contains the zeros \(\alpha_2, \ldots, \alpha_n\) of \(p(x)\) by our induction hypothesis. Consequently, \[\begin{aligned}\end{aligned}\] is a splitting field of \(p(x)\).

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

ఇప్పుడుమీరు ఏ గణనపరికరం ఈ వర్తిస్తుంది, కానీ దానిలోని భాగాలు పోబుల్ చేయదగినవి. క్రింద ఒకటి ప్రయత్నించండి లేదా మీ స్వంత టైపు చేయండి.

మీ సొంత పని ఉంచుకోండి

మీరు ఏ పాఠం నేర్చుకున్నారో, ఆ విషయం గురించి ఒక రిపోర్టులో మీరు తీసుకున్న సమస్యలను, మీ సమస్యలను ఒక స్థలంలో పరిష్కరించిన సమస్యలను, మీరు ఈ పేజీ గురించి అడగవచ్చు.

నమోదు చేసుకోండి ప్రవేశించు

ఇక్కడ వుపయోగించిన చిహ్నాలు

మొదటి శతాబ్దంలోని క్రైస్తవులు కూడా అదే విధంగా భావించారు.

ప్రశ్నలు

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.

ఈ పేజీలోని భాగాలు సవరింపుగా ఉన్నాయి Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). ఇక్కడ సమీకరించబడి మరియు తిరిగి ఉద్భవింపచేయబడిన తప్పులు మావి.

ఇంకా 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