What is the Cardinality and Dimension of \mathbb{Z}^{3}_{7}?

jdstokes
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Find the cardinality and dimension of the vector space \mathbb{Z}^{3}_{7} over \mathbb{Z}_{7}.

\mathbb{Z}^{3}_{7} = \{ (a,b,c) \; | \; a,b,c \in \mathbb{Z}_{7} \}.

Then since \mathbb{Z}_{7} is a field 1 \cdot a = a \; \forall \; a, so B = \{ (1,0,0), (0,1,0) , (0,0,1) \} is a basis of \mathbb{Z}^{3}_{7}, so \dim \mathbb{Z}^{3}_{7} = 3. ans = 9, what the?

Thanks

James
 
Last edited:
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Hi again,

Does anyone have any clues on this one? I'm really stuck.

Thanks

James.
 
Are you sure you copied the problem correctly? Because 3 certainly seems to be the right answer.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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