Karteikarten
Rings: Homomorphisms
Karten in diesem Stapel (10)
- 1Let R and S be two rings. \(f:R \to S\) is called a ring homomorphism ifFor every \(r_1,r_2\in R\),
\[a)f(r_1r_2)=f(r_1)f(r_2)\]
\[b)f(r_1+r_2)=f(r_1)+f(r_2)\] - 2If R and S are rings with unity and \(f(1_R)=1_S\) then f is calledA unital ring homomorphism
- 3The zero homomorphism isThe homomorphism which maps every element to the zero element of its codomain
- 4Let R and S be integral domain with \(f: R \to S\) being a nonzero homomorphism. Prove that f is unital.\(1_Sf(1_R)=f(1_R)=f(1_R^2)=f(1_R)f(1_R)\) so by cancellation: \(f(1_R)=1_S\)
- 5The kernel of a (ring) homomorphism isthe set of elements mapped to 0
- 6Prove that the kernel of Ring Homomorphism is SubringLet \(\phi: R_1 \to R_2\) be a ring homomorphism. \[(ker(\phi),+_1) \le (R_1, +_1)\]
Let \(x,y \in ker(\phi)\)
\[\phi(xy)=\phi(x)\phi(y)=00=0\] so \(x,y \in ker(\phi)\) - 7Prove that if R and S be rings, a homomorphism \(\phi\) is injective iff \(ker(\phi)=0\)\[\rightarrow\] Assume \(\phi\) is injective. So \(\phi(x)=0 \Rightarrow \phi(x)=0=\phi(0)\) therefore \(x=0 \Rightarrow ker(\phi)=\{0\}\)
\[\leftarrow\] Assume \(ker(\phi)=\{0\}\). If \(\phi(x)=\phi(y)\) then \(0=\phi(x)-\phi(y)=\phi(x)+\phi(-y)=\phi(x-y)\) hence it follows that \(x-y \in ker(\phi)\). So x-y=0 and hence x=y so \(\phi\) is injective - 8Let F,E be fields. Take f, a function from F to E that is a nonzero homomorphism. Then f is injective and \(f(x)^{-1} =f(x^{-1})\)Since fields are integral domains, f(1)=1. Let x be in F. Then \(f(x^{-1})f(x)=f(x^{-1}x)=f(1)=1\). So it follows that \(f(x)^{-1}=f(x^{-1})\). So \(f(x) \neq =0\). So ker(f)=0
- 9A ring isomorphism isA ring homomorphism that is bijective
- 10A ring automorphism isA ring isomorphism from a ring to itself
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- Undergrad Math Karteikarten am Rings: Homomorphisms, erstellt von Vincent Tembo am 26/07/2016.
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- Vincent Tembo
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