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$\mathrm{GF}( 29^2)$ is created by adjoining the root of the irreducible quadratic $p= x^2 + 7x +15$ to the field $\mathrm{GF}(29)$ . The cubic polynomial $q = Y^3 + (26x + 26)Y^2 + (8x+22)Y +13x+23$ is irreducible over this new field.

a. If the root of the given cubic is adjoined to $\mathrm{GF}(29^2)$ in order to form a larger field extension, how many elements are in the new field?

b. Using the matrix representation find the product of the two polynomials $Y^2+x+1$ and $Y^2+xY+1$

I know how to make fields and how to generate companion matrix and do the multiplication of the given polynomials in part b.

But I am not sure that I am building the matrix correctly. I need help in understanding the question properly.

It is my assignment question I am not looking for a straight answer but for well build explanation which helps me understand the base concept.

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    $\begingroup$ This question is not appropriate for math.overflow. MO is for research-level questions; please consider asking your question in math.stackexchange instead, which welcomes questions at all levels. $\endgroup$
    – Arturo Magidin
    Commented Apr 20, 2014 at 16:09

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Whenever you have a field $F$ and an irreducible polynomial $f$ of degree $d$ in $F[x]$, the field you get by adjoining a root of $f$ to $F$ will be a vector space over $F$ of dimension $d$. You can see this bigger field explicitly as

$$F[x]/f$$

which is a field because $F[x]$ is a principal ideal domain, $f$ is an irreducible polynomial, so $(f)$ is a maximal ideal. A basis for $F[x]/f$ over $F$ is given by $\{1,x,x^2,\ldots,x^{d-1}\}$.

Hence, answer to (a) is $(29^2)^3 = 29^6$.

I don't know what you mean by "matrix representation", but the product $(Y^2 + x + 1)(Y^2 + xY + 1)$ is just $Y^4 + xY^3 + (x+2)Y^2 + (x^2 + x)Y + (x+2)$. If you're working in the larger field of $29^6$ elements, then you're subject to the relations:

$$Y^3+(26x+26)Y^2+(8x+22)Y+13x+23 = 0, x^2 + 7x + 15 = 0$$

Hence you can if you wish reduce the product by replacing $Y^3$ with $-(26x+26)Y^2 - (8x+22)Y - 13x - 23$, so you get: $$Y(-(26x+26)Y^2 - (8x+22)Y - 13x - 23) + x(-(26x+26)Y^2 - (8x+22)Y - 13x - 23) + (x+2)Y^2 + (x^2+x)Y + (x+2)$$

$$...= -(26x+26)Y^3 - (8x+22)Y^2 - (13x+23)Y - (26x^2+26x)Y^2 - (8x^2+22x)Y - (13x^2+23x) + (x+2)Y^2 + (x^2+x)Y + (x+2)$$

Again replacing $Y^3$ by $-(26x+26)Y^2 - (8x+22)Y - 13x - 23$... $$...= -(26x+26)(-(26x+26)Y^2 - (8x+22)Y - 13x - 23) - (8x+22)Y^2 - (13x+23)Y - (26x^2+26x)Y^2 - (8x^2+22x)Y - (13x^2+23x) + (x+2)Y^2 + (x^2+x)Y + (x+2)$$

Now just collect the terms and you're done.

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