tegra97

04-30-2007, 02:23 AM

Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z[i]/I? Thanks

View Full Version : How many elements are in Z[i]/I, for I = <2+2i>?

tegra97

04-30-2007, 02:23 AM

Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z[i]/I? Thanks

daon

04-30-2007, 08:39 AM

Hi I have a quick question. If I=<2+2i> how would you determine how many elements are in Z[i]/I? Thanks

Z[i]/I = (a+bi) + <2+2i>

Also note that (2+2i) + <2+2i> = 0 + <2+2i>, so 2+2i=0. Or... 2=-2i, 1=-i.

So in this particular factor ring, -1 = i. Then squaring both sides we see 1 = -1, i.e. that 2=0.

Therefore when we have something like 9+8i + <2+2i> we actually get 9 + <2+2i> = (8+1)+<2+2i> = 1+<2+2i>.

Counting the number of distinct ones shouldn't be hard now.

Z[i]/I = (a+bi) + <2+2i>

Also note that (2+2i) + <2+2i> = 0 + <2+2i>, so 2+2i=0. Or... 2=-2i, 1=-i.

So in this particular factor ring, -1 = i. Then squaring both sides we see 1 = -1, i.e. that 2=0.

Therefore when we have something like 9+8i + <2+2i> we actually get 9 + <2+2i> = (8+1)+<2+2i> = 1+<2+2i>.

Counting the number of distinct ones shouldn't be hard now.

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