Differing polynomial GCDs over the field $\mathbb{Z}_{11}$

Question

I am trying to find a mistake in my reasoning or calculations.

I have $$f(x)= x^4 + x^3 +6x^2 +x +1 \text{ and } g(x)= x^4 + 9x^3 + 6x^2 + x +3\\ \text{over the field } \mathbb{Z}_{11}.$$

My calculations are $$ f(x)=g(x)+3x^3 +6x +9\\ g(x)=(x^3+2x+3)(x+9) + 4x^2 +2x +9\\ x^3+2x+3 = (4x^2+2x+9)(3x+4) + 0 $$ resulting into $\gcd(f(x),g(x))=4x^2+2x+9$.

Yet an online calculator gives a different result $x^2+6x+5$ and I don't know what goes wrong.

Multiply your answer by $3$ (which is $4^{-1}$ mod $11$). What do you get? — Viktor Vaughn, Mar 23 '20 at 17:52
Welcome to Mathematics Stack Exchange. $4(x^2+6x+5)\equiv 4x^2+2x+9\bmod11$ — J. W. Tanner, Mar 23 '20 at 17:52
@RichardD.James oups… I been resolving it for so long I lost all ability to notive things... Thank you very much! — Dknot, Mar 23 '20 at 18:00

score 0 · Answer 1 · edited Mar 23 '20 at 19:22

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As it was kindly noticed in comments, my calculations were alright, but I totally forgot that I can get different answers because of the mod[11] (should've multiplied my own answer by 3 to get the same result as the online calculator).

edited Mar 23 '20 at 19:22

user26857

52,094

answered Mar 23 '20 at 18:04

Dknot

515

1

See here for a general discussion of unit normalization of gcds (I can't link it in the dupe list of the question since it has no upvotes yet) – Bill Dubuque Mar 23 '20 at 18:35

Differing polynomial GCDs over the field $\mathbb{Z}_{11}$

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