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There is a unit u and primes $\pi_{j}=a_{j}+b_{j}i$ in $\mathbb{Z}[i]$ with $a_{j}>0, b_{j}>0$ and $7+i = u\pi_{1}\dots\pi_{k}$

$\mathbb{Z}[i]$ has four units: $i,-i,1,-1$. The product of the primes is also in $\mathbb{Z}[i]$, so : $7+i=u\pi… \pi_{k} = i(a+bi) = -b+ai$. This can't be the case since then it would follow that : $a=1, b=-7$ but since $b_{j}> 0$ i therefore can't be that unit.
For -i : we get -i(a+bi)= -ai+b . With this it follows that $a=-1, b= 7 $, therefore since $a_{j}>0$ this also can not be true.
For 1: 1(a+bi)= a+bi , a=7, b=1. This could be true, so now it is to show that 7+i is prime in $\mathbb{Z}[i]$. We can see directly that 7+i is irreducible over $\mathbb{Z}[i]$ and that means it must also be prime since $\mathbb{Z}[i]$ is a UFD. Therefore 1 is that unit.

I believe this can't be right, because I didn't use the indices at all. Does anybody see the right way? Please do tell.

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$7 + i = (1 + 3i)(1 - 2i)$ –  The Chaz 2.0 Nov 16 '11 at 20:53
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Hint: the norm of $7+i$ is $50$, so its factorization must be into three primes, with $N(\pi_1)=2$, $N(\pi_2)=5$ and $N(\pi_3)=5$. –  Thomas Andrews Nov 16 '11 at 21:04
    
Hint: Two of the primes are $2+i$ and $2+i$. –  André Nicolas Nov 16 '11 at 21:19
    
Aren't there infinite many such prime numbers??? –  VVV Nov 16 '11 at 22:21
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There are only finitely many primes in ${\bf Z}[i]$ with a given norm. Anyway, even if there were infinitely many, it wouldn't matter, as you're only being asked for one way to solve the problem, not for all possible ways. Follow the hints from Thomas and Andre; find the three primes, taking care to have $a_j\gt0$ and $b_j\gt0$, and then find the unit that works. –  Gerry Myerson Nov 16 '11 at 23:22

1 Answer 1

up vote 2 down vote accepted

Your claim that $u=i$ is impossible does not follow.

You know that each $b_i$ is positive; but that does not tell you that the final product $a+bi$ must have $b$ positive.

For example, you can have $(a_1+b_1i)(a_2+b_2i)$ with $b_1,b_2\gt 0$, but with product $(a_1a_2-b_1b_2) + (a_1b_2+b_1a_2)i$ having imaginary part negative. Just take $a_1=0$, $b_1=2$, $a_2=-2$, $b_2=1$. Then you have $(0+2i)(-2+i)$, and the product is $-2-4i$, with negative real and negative imaginary parts. The other conclusions are likewise incorrect.

Finally, your claim that $7+i$ is irreducible is also incorrect. Note that $N(7+i) = 50 = 2\times 5^2$; if it is reducible, you could look for an element with norm $5$ and one with norm $10$, which quickly leads to $7+i = (2+i)(3-i)$. Since neither factor is a unit (their norms are indeed $5$ and $10$), then this shows $7+i$ is indeed reducible. (You can also look for elements with norms $2$ and $25$).

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Arturo Magidin, Thanks. –  VVV Nov 16 '11 at 23:55

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