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In communication theory classes I recall this sort of extension of $\mathbb{Z}/2\mathbb{Z}$ where an imaginary $\alpha$ is defined so that $\alpha^2+\alpha+1 = 0$. Then more such imaginaries must be added so that other polynomials have solutions.

Can anyone remind me what this set is called?

Myself
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J. C. Salomon
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  • By $\mathbb{Z}_2$ do you mean the field of 2 elements, or the $2$-adic integers? – bzc Jan 29 '12 at 00:35
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    Given the context of communication theory, I think the word you're looking for and the object you're describing is a finite field, or Galois field. (In particular, it's a field.) – Myself Jan 29 '12 at 00:42
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    I'm sorry, but a communication theory class with (group-theory) tag makes this more than likely OP is talking about what mathematicians call $\mathbb{Z}/2\mathbb{Z}$. jcsalomon, the notation $\mathbb{Z}_2$ is used in algebraic number theory (or somewhere around there) to refer to the 2-adics and shouldn't be confused with integers and addition/multiplication modulo 2. – anon Jan 29 '12 at 00:43
  • The unique extension of degree $2$ of the field of two elements is called "the field with $4$ elements" or "the Galois field of order $4$", or $GF(4)$. – Arturo Magidin Jan 29 '12 at 00:45
  • Yup. These are Galois fields. If you scroll half-way down this answer, you will see a construction of $GF(8)$ together with various useful ways of describing its elements. The links there take you to places, where a construction of $GF(256)$ is given. That field is used in AES-crypto and in error-correcting codes on CD-ROMs and such. It is, perhaps, a bit non-standard to call the quantities like $\alpha$ here 'imaginaries', but whatever ;-) – Jyrki Lahtonen Jan 29 '12 at 14:56
  • Re notation: That’s the nice thing about standards; so many to choose from! Yes, by ℤ₂ I meant GF(2). – J. C. Salomon Jan 30 '12 at 20:39

2 Answers2

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The "algebraic closure" of the field $\mathbb{Z}_2$ is the smallest field containing $\mathbb{Z}_2$ in which all polynomials with coefficients in the field have zeroes in the field.

There's a simple proof that the algebraic closure must be infinite: if you have only finitely many members $a$, then $$f(x) = 1+\prod_{a} (x-a)$$ is a polynomial with no roots within the field.

Michael Hardy
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The name you are looking is a "field extension" that is generated by the root of the polynomial $x^2 + x + 1$ ; another notation for this is $GF(4)$ due to the fact that there exists only one field up to isomorphism of degree $p^n$ when $n \ge 1$ and $p$ is a prime, which we denote $GF(p^n)$ (here we're looking at $2^2 = 4$, so $p=2$ is prime and $n=2$). This is a result from Galois Theory, if you're interested in reading more about it.

Hope that helps,

EDIT : After noticing what Michael Hardy said, I didn't read the question completely. If you want to continue "adding roots" to allow other polynomials to "split" too (i.e. have their roots in the field), then that is called the "algebraic closure" of your field, i.e. the field that contains $\mathbb Z / 2 \mathbb Z$ as a subfield and contains the roots of every polynomial with coefficients in itself.

Patrick Da Silva
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  • I wouldn't call the unicity of a finite field of order q "a result from Galois Theory", since it's much more basic than that. – Myself Jan 29 '12 at 00:49
  • But the OP said "more such imaginaries must be added so that other polynomials will have solutions." So it's not just any field extension. – Michael Hardy Jan 29 '12 at 00:49
  • Hm. Then I guess I just read part of the question. You're right, this answer is not complete. – Patrick Da Silva Jan 29 '12 at 00:58
  • @Myself : Maybe it comes from pure field theory but I learned this in a Galois Theory course and my memory is having trouble distinguishing such things. Also note that $GF$ stands for "Galois Field", thus maybe my confusion. – Patrick Da Silva Jan 29 '12 at 01:01