18.703 Modern Algebra, Special Domains
ثبت نشده
چکیده
Let R be an integral domain. Note some obvious facts. Every ele ment a of R divides 0. Indeed 0 = 0 · a. On the other hand, 0 only divides 0. Indeed if a = q · 0, then a = 0 (obvious!). Finally every unit u divides any other element a. Indeed if v ∈ R is the inverse of u, so that uv = 1 then a = a · 1 = (av)u. Lemma 19.3. Let R be an integral domain and let p ∈ R. Then p is prime if and only if p is not a unit and whenever p divides ab then either p divides a or p divides b, where a and b are elements of R.
منابع مشابه
18.703 Modern Algebra, The Isomorphism Theorems
We have already seen that given any group G and a normal subgroup H, there is a natural homomorphism φ : G −→ G/H, whose kernel is H. In fact we will see that this map is not only natural, it is in some sense the only such map.
متن کامل18.703 Modern Algebra, Presentations and Groups of small order
i Given any word w, the reduced word w associated to w is any word obtained from w by reduction, such that wi cannot be reduced any further. Given two words w1 and w2 of A, the concatenation of w1 and w2 is the word w = w1w2. The empty word is denoted e. The set of all reduced words is denoted FA. With product defined as the reduced concatenation, this set becomes a group, called the free group...
متن کامل18.703 Modern Algebra, Permutation groups
Proof. (5.2) implies that the set of permutations is closed under com position of functions. We check the three axioms for a group. We already proved that composition of functions is associative. Let i : S −→ S be the identity function from S to S. Let f be a permutation of S. Clearly f ◦ i = i ◦ f = f . Thus i acts as an identity. Let f be a permutation of S. Then the inverse g of f is a perm...
متن کامل