例句與造句

1. Instead the sequence that emerges from the first isomorphism theorem is
2. :: Do you really NEED the First Isomorphism Theorem?
3. Conversely, the first isomorphism theorem in universal algebra.
4. This is the first isomorphism theorem for affine spaces.
5. By the first isomorphism theorem we then have that
6. It's difficult to find first isomorphism theorem in a sentence. 用first isomorphism theorem造句挺難的
7. By the first isomorphism theorem, the isomorphic to the quotient of " A " by this congruence.
8. Then \ mathrm { Im } \ phi and thus the result follows by use of the first isomorphism theorem.
9. By the first isomorphism theorem, \ operatorname { exp } induces the isomorphism \ mathfrak { g } / \ Gamma \ to G.
10. The fundamental theorem on homomorphisms ( or first isomorphism theorem ) is a theorem, again taking various forms, that applies to the quotient algebra defined by the kernel.
11. This is a consequence of the first isomorphism theorem, because is precisely the set of those elements of that give the identity mapping as corresponding inner automorphism ( conjugation changes nothing ).
12. By the first isomorphism theorem this is true if and only if there exists a linear map \ mathcal { L } : \ mathbb { R } \ to \ mathbb { R } such that \ mathcal { L } \ circ D _ y g = D _ y f.
13. Sub-and quotient groups are related in the following way : a subset " H " of " G " can be seen as an injective map, i . e . any element of the target has at most one image of group homomorphisms and the first isomorphism theorem address this phenomenon.
14. As in the theory of associative rings, ideals are precisely the kernels of homomorphisms; given a Lie algebra \ mathfrak { g } and an ideal " I " in it, one constructs the "'factor ( or quotient ) algebra "'\ mathfrak { g } / I, and the first isomorphism theorem holds for Lie algebras.
15. Okay, I think I can see R / I and R / J as R-modules now, but I have no idea how to prove that the isomorphism implies I = J . I tried using the fact that, since R / I and R / J are isomorphic, there exists a bijective homomorphism ? : R ( R / I )-> R ( R / J ), and then by the first isomorphism theorem, R ( R / I ) / ker ( ? ) H " im ( ? ), and since ? is bijective, ker ( ? ) = I and im ( ? ) = R ( R / J ), so R ( R / I ) / I H " R / J, but this looks very suspiciously like not telling me anything useful.