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In $\mathbb Z[i]$, prove that $5$ is not irreducible.
In $\mathbb Z[\sqrt{-3}]$, factor $4$ into irreducibles in two distinct ways.

I am completely stumped on how to do this. I really need all the help I can get and a possible walkthrough.

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    $\begingroup$ The title is no replacement for the question. Also both questions shouldn't be asked in a single post. Ask separate questions instead. In addition to that, both of them are quite standard so searching a little should give you a duplicate. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – AlexR May 8 '15 at 17:07
  • $\begingroup$ I have searched and nothing has helped me what so ever $\endgroup$ – B ry May 8 '15 at 17:08
  • $\begingroup$ Then you should include what you read and how that didn't help you (i.e. what you didn't understand in the existing answers) $\endgroup$ – AlexR May 8 '15 at 17:09
  • $\begingroup$ i read someone say that (1+2i)(1-2i) is a factorization of 5 in the gauss. integers but I don't know how they concluded that $\endgroup$ – B ry May 8 '15 at 17:11
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    $\begingroup$ But $5=(1+2i)(1-2i)$ already proves that $5$ isn't irreducible. Coming up with this factorisation is quite natural if you notice that $(a+bi)(a-bi) = a^2+b^2$ for any $a,b$, thus the sum of perfect squares is reducible in $\mathbb Z[i]$. $5=4+1=2^2+1^2$ is such a sum. $\endgroup$ – AlexR May 8 '15 at 17:17
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For the first assignment you correctly found $$5 = (1+2i)(1-2i)\ .$$ This already shows that $5$ is reducible in $\mathbb Z[i]$, as claimed.
To provide a little intuition on how to find these factors, consider the third binomial rule for the product $$(a+bi)(a-bi) = a^2 - (bi)^2 = a^2+b^2$$ So any sum of perfect squares is reducible with such factors. Now it should be easy to see that $5$ fulfills this criteria: $5=4+1 = 2^2 + 1^2$. This immediately gives rise to two factorisations of $5$ in $\mathbb Z[i]$, one for each choice of $(a,b)$: $$5 = (1+2i)(1-2i) = (2+i)(2-i)$$


Here is a starter on the second assignment:

$$(a+b\sqrt{-3}) \cdot (c+d\sqrt{-3}) = ac - 3bd + (ad+bc)\sqrt{-3}$$

So you need to find two distinct integer solutions to the equations $$\begin{align*} ac - 3bd & = 4 \\ ad+bc & = 0 \end{align*}$$ One solution is $(a,b,c,d) = (1,-1,1,1)$ corresponding to $$4 = (1-\sqrt{-3})(1+\sqrt{-3})$$ Now you must find another solution to get the second factorisation.

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  • $\begingroup$ would another factorization be (2,0,2,0) $\endgroup$ – B ry May 8 '15 at 21:14
  • $\begingroup$ or would (4,-2,4,-2) work? $\endgroup$ – B ry May 8 '15 at 21:16
  • $\begingroup$ (2,0,2,0) is correct but (4,-2,4,-2) is no solution. $\endgroup$ – AlexR May 8 '15 at 21:17
  • $\begingroup$ never mind I think both of those are reducible and therefore do not meet the qualification for the answer….. $\endgroup$ – B ry May 8 '15 at 21:18
  • $\begingroup$ so would (2 + 0 \sqrt[-3])(2+ 0 \sqrt[3]) be irreducible? $\endgroup$ – B ry May 8 '15 at 21:21
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The concept of being irreducible in $\mathbb{Z}[i]$ is similar to the idea of being prime in regards to $\mathbb{R}$. In this case, we must check if 5 can be written as $(a+b i)(c+d i)$ where $a,b,c,d \in \mathbb{Z}$. As you found, it can be written as $5 = 1 + 4 = 1 - 4 i^2 = (1 + 2 i)(1-2 i)$ and thus is not irreducible.

We apply the same concept to show that 4 is reducible in $\mathbb{Z}[\sqrt{-3}]$. We look for $a,b,c,d \in \mathbb{Z}$ such that $4 = (a + b \sqrt{-3})(c + d \sqrt{-3})$. Fairly obviously, we see that $4 = 1 + 3 = 1 - (\sqrt{-3})^2 = (1+\sqrt{-3})(1-\sqrt{-3})$.

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  • $\begingroup$ NB The task is to find two factorisations of $4$, you give 1 here. I also ask you not to give the other so the OP can do this himself and learn a bit from it. $\endgroup$ – AlexR May 8 '15 at 17:35
  • $\begingroup$ I am aware that the question asked for two reductions of 4, hence why I only gave the obvious one. The other solution definitely requires some playing around with the algebra. $\endgroup$ – Sean Henderson May 8 '15 at 17:43
  • $\begingroup$ SO would another distinct solution for (a,b,c,d) be (2,0,2,0) $\endgroup$ – B ry May 8 '15 at 21:07

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