2018年11月15日 · Then any subring is a #-subring, so the property you've given clearly still holds. As will anything that proves some subset is a subring. The problem is that a #-subring is not necessarily a ring: $\Bbb N$ is a #-subring of $\Bbb Z$, yet $\Bbb N$ is not a ring. Hopefully it is clear to you why a subring has to be a ring for it to be a sensible ...

In abstract algebra, the subring test is a theorem that states that for any ring, a nonempty subset of that ring is a subring if it is closed under multiplication and subtraction. Note that here that the terms ring and subring are used without requiring a multiplicative identity element.

Edit: As pointed out in the comments, you do not need to prove that 0 is the identity since you are proving it is a subring, rather than just proving it is a ring. It should still serve as an example for proving the other parts of the definition are satisfied though.

In the commutative case we identify left and right multiplication and then an ideal is a subring of the ring. However, that is not enough. Since the subring requires closure under multiplication of elements contained in itself while a left or right ideal requires closure under corresponding multiplication by elements of the ring.

2018年5月25日 · (R, +, $\ast$) ( $\ast$ := Multiplication) is a ring. A subset S $\subseteq$ R is called subring when (S, +, $\ast$) is a ring. Show the following: S is a subring from R if the following properties apply: 1.) S $\subseteq$ R . 2.) S $\neq$ $\varnothing$ 3.) $\forall$ r, s $\in$ S : r + s $\in$ S $\land$ r $\ast$ s $\in$ S

2016年11月6日 · $\begingroup$ Any integral domain is a subring of its field of fractions, which is a PID. Therefore the three classes "subrings of fields", "subrings of PIDs" and "integral domains" are the same. $\endgroup$ –

2013年8月5日 · You do need to show that it contains an additive inverse for each of its elements. (For example, $\mathbb{N}$ is not a subring of $\mathbb{Z}$ though it is closed under addition and multiplication.) Provided that you know the subset is nonempty, this together with it being closed under addition will then imply that $0$ is in there.

This relies on the existence of a nontrivial idempotent. Since the identity element of any subring has to be idempotent, you can see that the identity of subrings would have to coincide with the superring's identity if there are only trivial idempotents (like in …

The factors in such a decomposition are naturally quotients of the original ring, but also correspond to an ideal of that ring (namely the kernel of projecting to the other factor(s)); such an ideal is not a subring, but can be made into a ring by choosing the projection of $1$ onto the ideal as new neutral element.

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