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Integral Domain

Divisors 7 is prime 〖[a]〗_7 〖[b]〗_7=〖[0]〗_7∈Z_7 〖[ab]〗_7=〖[0]〗_7 ab∈[0]_7 ab is multiple of 7 a∈[0]_7 and b ∈ 〖[0]〗_7 If 〖[a]〗_7=〖[0]〗_7 and 〖[b]〗_7=〖[0]〗_7,then Z_7 has no zero divisors. Did I prove non zero divisors correctly. How do I prove Multiplicative identity(unity) and Multiplication is commutative?

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I think you have the right idea: Z7 is an integral domain if its a ring and ab = 0 implies a = 0 or b = 0.

The key is that 7 is a prime number. Recall that "x = 0" in Z7 means x is a multiple of 7: $7 | x$.

So you can now use euclids lemma: $7 | ab$ implies $7 | a$ or $7 | b$ to show that you have an integral domain.

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    $\begingroup$ can you please help me out with the multiplicative identity and multiplication communative $\endgroup$ – Michael Anthony Pryor Feb 23 '16 at 2:48
  • $\begingroup$ @MichaelAnthonyPryor, sure $\endgroup$ – Brennan.Tobias Feb 23 '16 at 14:55
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as for identity: 1 is the identity and since $1x = x$ in integers the same is true when you move to Z7: $[1][x] = [x]$.

Commutativity is the same: In Z: $xy = yx$ so in Z7: $[x][y] = [y][x]$.

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