The set Z5 is a field, under addition and multiplication modulo 5. To see this, we already know that Z5 is a group under addition.

Zp is a commutative ring with unity. Here x is a multiplicative inverse of a. Therefore, a multiplicative inverse exists for every element in Zp−{0}. Therefore, Zp is a field.

While Z/4 is not a field, there is a field of order four. In fact there is a finite field with order any prime power, called Galois fields and denoted Fq or GF(q), or GFq where q=pn for p a prime.

This shows that algebraic facts you may know for real numbers may not hold in arbitrary rings (note that Z10 is not a field).

p 316, #12 Since it has degree 2, to show that x2 + x + 4 is irreducible in Z11[x] it suffices to show it has no roots in Z11, as Z11 is a field. If f(x) ∈ Zp[x] is irreducible of degree 2, then f(x) = ag(x) for some a ∈ F, a = 0 and g(x) ∈ F[x] irreducible, monic and of degree 2.

In other words, it is the cyclic group whose order is four. It can also be viewed as: The quotient group of the group of integers by the subgroup comprising multiples of . The multiplicative subgroup of the nonzero complex numbers under multiplication, generated by (a squareroot of ).

Z4 × Z4: The elements have orders 1, 2, or 4. The elements of order 2 are (2, 0), (2, 2), and (0, 2). Thus, there is 1 element of order 1 (identity), 3 elements of order 2, and the remainder have order 4, so there are 12 elements of order 4.

So, Z4 is a cyclic group. Since, a group is said to be a subgroup of Zn only if HCF of a and n = 1. The elements belonging to Z4 are [0],[1],[2] and [3]. Among these 4 elements, the numbers whose HCF with 4 is 1 will be called a subgroup of Z4.

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However, if you confine your attention to the units in Zn — the elements which have multiplicative inverses — you do get a group under multiplication mod n. It is denoted Un, and is called the group of units in Zn.

Suppose f = (2x+2) ∈ Z[x] is a degree 1-polynomial. Since g is a unit, we have f=unit⋅degree-1 polynomial. Hence f is irreducible in Z[x] .

Now, from this carachterization it is obvious that x−1 is an irreducible polynomial (over any field indeed). About g:=x2+x+1, suppose it is not irreducible; then there must be a polynomial ax+b∈R[x] (a≠0) such that ax+b∣g, i.e −b/a must be a root for g.

Use long division or other arguments to show that none of these is actually a factor. If a polynomial with degree 2 or higher is irreducible in , then it has no roots in . If a polynomial with degree 2 or 3 has no roots in , then it is irreducible in .

When a polynomial is integer valued, one may appeal to Gauss’s lemma which states that if the coefficients of a non-constant polynomial f are relatively prime and f is irreducible in Z[X], then f is irreducible in Q[X].

[‚ir·ə′dü·sə·bəl i′kwā·zhən] (mathematics) An equation that is equivalent to one formed by setting an irreducible polynomial equal to zero.

A primitive polynomial must have a non-zero constant term, for otherwise it will be divisible by x. Over GF(2), x + 1 is a primitive polynomial and all other primitive polynomials have an odd number of terms, since any polynomial mod 2 with an even number of terms is divisible by x + 1 (it has 1 as a root).

A polynomial is said to be irreducible if it cannot be factored into nontrivial polynomials over the same field.

If p(x) divides the product f(x)g(x) of two polynomials over F then p(x) must divide one of the factors f(x) or g(x). Corollary 17.13. Let p(x) be an irreducible polynomial over a field F.

But if you multiply x by any non-zero polynomial, the result will always contain x or higher powers, so it has no inverse. Consider C[x] the ring of polynomials with coefficients from C. This is an example of polynomial ring which is not a field, because x has no multiplicative inverse.

A polynomial with integer coefficients that cannot be factored into polynomials of lower degree , also with integer coefficients, is called an irreducible or prime polynomial .