How do we construct a field \(GF(p^q)\)?
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How do we construct a field \(GF(p^q)\)?
We take the field \(GF(p)[x]_{m(x)}\) where \(m(x)\) is an irreducible polynomial of degree \(q\).
Then \(GF(p)[x]_{m(x)}\) has \({|F|}^q\) polynomials in it, as all of degree less than \(q\) are coprime to \(m(x)\), by definition of irreducible.
And this field is isomorphic to \(GF(p^q)\).
Then \(GF(p)[x]_{m(x)}\) has \({|F|}^q\) polynomials in it, as all of degree less than \(q\) are coprime to \(m(x)\), by definition of irreducible.
And this field is isomorphic to \(GF(p^q)\).
Example: The field \(GF(2)[x]\) \({x^2 + x + 1}\) is isomorphic to \(GF(2^2 = 4)\).
We can see this is the case as \(GF(2)[x]_{x^2 + x + 1}\) has \(4\) elements, \(\{0, 1, x, x + 1\}\), which we can basically map to \(GF(4)\) as \(\{0, 1, 2, 3\}\).
Indeed \(1 + x = x + 1\) in \(GF(2)[x]_{x^2 + 1 + 1}\) and \(1 + 2 = 3\) which is \(x + 1\) in the isomorphism.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | How do we construct a field \(GF(p^q)\)? | |
| Back | We take the field \(GF(p)[x]_{m(x)}\) where \(m(x)\) is an irreducible polynomial of degree \(q\).<br><br>Then \(GF(p)[x]_{m(x)}\) has \({|F|}^q\) polynomials in it, as all of degree less than \(q\) are coprime to \(m(x)\), by definition of irreducible. <br>And this field is isomorphic to \(GF(p^q)\).<br><br><div> <strong>Example</strong>: The field \(GF(2)[x]\) \({x^2 + x + 1}\) is isomorphic to \(GF(2^2 = 4)\). </div><div>We can see this is the case as \(GF(2)[x]_{x^2 + x + 1}\) has \(4\) elements, \(\{0, 1, x, x + 1\}\), which we can basically map to \(GF(4)\) as \(\{0, 1, 2, 3\}\).</div><div><br></div><div>Indeed \(1 + x = x + 1\) in \(GF(2)[x]_{x^2 + 1 + 1}\) and \(1 + 2 = 3\) which is \(x + 1\) in the isomorphism.</div> |