Feature (mathematical)

In abstract algebra, the characteristic of a ring R{displaystyle R} is defined as the smallest positive integer n{displaystyle n} such as 1R+...... nsums+1R=0{displaystyle 1_{R}+{overset {n{text{ sumndos}}}{ldots}} }+1_{R}=0}. If there is no such n{displaystyle n}, it is said that the characteristic R{displaystyle R} It's 0.

Alternatively and equivalently, we can define the characteristic of the ring R{displaystyle R} as the only natural number n{displaystyle n} such as R{displaystyle R} contain an isomorphic ring to the quotient ring Z/nZ{displaystyle mathbb {Z} /nmathbb {Z} }.

The Case of Rings

Yeah. R{displaystyle R} and S{displaystyle S} are rings and there is a homomorphism of rings

R→ → S{displaystyle Rrightarrow S},

then the characteristic of S{displaystyle S} divides the feature R{displaystyle R}. This can sometimes be used to exclude the possibility of certain ring homomorphism. The only ring with characteristic 1 is the trivial ring, which contains only one element 0=1. If not trivial ring R{displaystyle R} They don't have any zero divider, then their feature is 0 or cousin. In particular, this applies to every body, to every domain of integrity and to every ring of division. All characteristic ring 0 is infinite.

The ring Z/nZ{displaystyle mathbb {Z} /nmathbb {Z} } of the entire modules n{displaystyle n} has characteristic n{displaystyle n}. Yeah. R{displaystyle R} It's a subanillo. S{displaystyle S}, then R{displaystyle R} and S{displaystyle S} They have the same characteristic. For example, if q(X){displaystyle q(X)} is a polynomial cousin with coefficients in the body Z/pZ{displaystyle mathbb {Z} /pmathbb {Z} } where p{displaystyle p} It's cousin, then the factor ring (Z/pZ)[chuckles]X]/(q(X)){displaystyle (mathbb {Z} /pmathbb {Z})[X]/(q(X)} is a characteristic body p{displaystyle p}. Since complex numbers contain rationals, their characteristic is 0.

If a switching ring R{displaystyle R} has a characteristic p{displaystyle p}, then you have to (x+and)p=xp+andp{displaystyle (x+y)^{p}=x^{p}+y^{p}} for all elements x{displaystyle x} e and{displaystyle and} in R{displaystyle R}.

The application

f(x)=xp{displaystyle f(x)=x^{p}}

defines a ring homomorphism

R→ → S{displaystyle Rrightarrow S},

This is called the endomorphism of Frobenius. Yeah. R{displaystyle R} is a domain of integrity this is injective.