# Thread: Fields and linear algebra

1. ## Fields and linear algebra

What type of argument could be used to show there exists a field with a certain amount of elements, "x".

If your given the integers modulo 4 how can you show it is a field. Ive done the cayley table but how do you interpret that and show the properties of a field how true?

2. Originally Posted by GreenandGold
What type of argument could be used to show there exists a field with a certain amount of elements, "x".

If your given the integers modulo 4 how can you show it is a field. Ive done the cayley table but how do you interpret that and show the properties of a field how true?
Not always can you construct a field with x elements where x is an assinged number.
Also, $\mathbb{Z}_4$ is not a field.
There is no inverse for $2$.

3. how could you show that for say Z_6 and could you elborate more

4. Originally Posted by GreenandGold
how could you show that for say Z_6 and could you elborate more
To be a field you need all non-zero elements to have inverses. This, $\mathbb{Z}_4$ is not a field because what is the inverse of $2$ mod $4$? Just try it out, $2\cdot 1\equiv 2 (\bmod 4)$, $2\cdot 2\equiv 0 (\bmod 4)$, $2\cdot 3\equiv 2(\bmod 4)$. We see that it has no inverse.

And the same with $\mathbb{Z}_6$. See that $2,3,4$ fail to have inverse.

5. so Z_9 fails because 3 and 6 dont have a 1 on the cayley table under multiplication?

6. Originally Posted by GreenandGold
so Z_9 fails because 3 and 6 dont have a 1 on the cayley table under multiplication?

7. Is that enough to say that Z_5 is a field?

8. Originally Posted by GreenandGold
Is that enough to say that Z_5 is a field?
Yes. In general if $p$ is a prime then $\mathbb{Z}_p$ is a field.

To see you need to know if $p\not|a \text{ then }ax\equiv 1(\bmod p)$ is solvable.