1

Find $\gcd(15-15i,7-i)$ in $\mathbb Z[i]$

I have a trouble with this question because depending on how I factorizes $15-15i$ I get other sollutions:

$$7-i=(1-i)(1+2i)(2-i)$$ 1) $$15-15i=15(1-i)=3\cdot5\cdot(1-i)=3\cdot(2-i)(2+i)(1-i)$$ Then: $ \gcd(15-15i,7-i)=(1-i)(2-i)=1-3i$

2) $$15-15i=15(1-i)=3\cdot5\cdot(1-i)=3\cdot(1-2i)(1+2i)(1-i)$$ Then: $\gcd(15-15i,7-1)=(1-i)(1+2i)=3+i$

Where is the mistake?

Bernard
  • 175,478
MP3129
  • 3,195
  • 1
    The gcd is only unique up to units, in your case they differ by $-i$. So no contradiction. The same happens for, say, polynomials. Also in $\Bbb Z$ it is only unique up to sign. – Dietrich Burde Jan 12 '20 at 20:54
  • 1
    Around the time the previous comment was posted, I posted my answer below. Did you have any questions about my answer? Please let me know. I'll be happy to help. – user729424 Jan 12 '20 at 21:13
  • @AndrewOstergaard everything is a clear, very accurate answer, thank you! – MP3129 Jan 12 '20 at 21:26

1 Answers1

2

You did not make any mistakes.

The GCD of two elements of $\mathbb{Z}[i]$ is only unique up to multiplication by a unit. The units of $\mathbb{Z}[i]$ are $\pm1$, $\pm i$.

Hence for any $a,b\in\mathbb{Z}[i]$, if $d$ is a GCD of $a$, $b$, then so is $-d$, $i\cdot d$ and $-i\cdot d$.

Since $1-3i$ is a GCD of $7-i$ and $15-15i$, $$3-i=-i\cdot(1-3i)$$

is also a GCD of $7-i$ and $15-15i$.

user729424
  • 5,061