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$F$ is a field. $\langle p(x) \rangle$ is a maximal ideal. So $K = F[x]/\langle p(x) \rangle$ is a field extension.

I am trying to understand what would be the structure of $K[x]$?

$F[x]$ has all polynomials where coefficients of the polynomials are from the Field $F$. So $K[x]$ would be the set of polynomials with their coefficients are from the field $K$.

Every element of $K$ is a coset of the form $a(x) + \langle p(x)\rangle$ where $a(x) \in F[x]$ and degree of $a(x)$ is lesser than that of $p(x)$ and the coset can be denoted as $[a(x)]$. I am not able to understand how a coset can be a coefficient of a polynomial. I think probably the elements of the cosets could be. The element of each coset is apart of the equivalence class $[a]$. So I think the coefficients of the polynomials in $K[x]$ would be the different $a$'s. If that is true, then I think the set of polynomials in $K$ would be exactly the same as the set of polynomials in $F$ - i.e. $K[x]$ would be the same as $F[x]$. Or am I mistaken?

user93353
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  • It's just like how the elements of $\mathbb{Z}/3\mathbb{Z}[x]$ are polynomials with cosets as coefficients. – morrowmh May 03 '22 at 03:52
  • @MichaelMorrow - OK got it, I think that answers the question about coefficients. What about the set of polynomials themselves? I think every polynomial is $F[x]$ would be there in $K[x]$ & vice versa. Also the same for $\mathbb Z[x]$ & $\mathbb Z/3\mathbb Z[x]$ - both would be the same set. Is that right? – user93353 May 03 '22 at 04:00
  • @MichaelMorrow - also there is some diff between $\mathbb Z/3\mathbb Z[x]$ & $F[x]/\langle p(x) \rangle$ - $\mathbb Z/3\mathbb Z$ contains only numbers, so each number can be a coefficient of a polynomial. However $K = F[x]/\langle p(x) \rangle$ is a litte different - it contains both numbers (const polynomials) & other non-constant polynomials. I am assuming the non-const polynomials in $K$ cannot be coefficients of polynomials in $K[x]$ – user93353 May 03 '22 at 04:05
  • Oh I was just using that as an example of how cosets can be coefficients. In general, given any commutative ring $R$, you can make a polynomial ring $R[x]$ where the coefficients are elements of $R$. This still applies here. – morrowmh May 03 '22 at 04:12
  • Please try not to think in terms of numbers and polynomials. Elements of $K[x] $ are polynomials with coefficients in $K$. The elements of $K$ are just elements of $K$ which depend on $F$ as well as $p(x) \in F[x] $. – Paramanand Singh May 03 '22 at 06:15
  • @ParamanandSingh - got it. However, my main question is whether set of polynomials $K[x]$ is the same as the set of Polynomials $F[x]$ or not. I am unable to think of any polynomial which would be in one of those 2 sets but not in the other – user93353 May 03 '22 at 06:19
  • I suggest you work out in detail for yourself what happens in the case $F=\mathbb{R}$, $K=\mathbb{C}=\mathbb{R}[X]/\langle X^2+1\rangle$. Not every real polynomial is a complex polynomial, is it? – ancient mathematician May 03 '22 at 06:33
  • @ancientmathematician - how is $\mathbb C = \mathbb R[x]/\langle x^2 + 1\rangle$? - The extension is a much smaller field than $\mathbb C$. Isn't that the whole point of finding an extension rather than going with $\mathbb C$ which contains roots of all polynomials? – user93353 May 03 '22 at 06:36
  • $\mathbb{C}$ is constructed from $\mathbb{R}$ by "adjoining" a square root of $-1$: that is it is exactly the quotient I wrote down. You've got to understand this relationship before you try fancy stuff about fields in general. – ancient mathematician May 03 '22 at 06:45
  • @ancientmathematician - isn't $\mathbb C$ constructed by adjoining $i$ (i.e. $\sqrt -1$) rather than by quotienting an Ideal - i.e. $\mathbb C = \mathbb R[i]$ & not $\mathbb C= \mathbb R[x]/\langle x^2 + 1\rangle$. i.e. $\mathbb C$ is an extension of $\mathbb R$ but not the way you say it is. – user93353 May 03 '22 at 07:11
  • But what do we mean by "adjoining" the root $i$ of the polynomial $X^2+1$? we mean forming the quotient $\mathbb{R}[X]/\langle X^2+1\rangle$ !!!! ($i$ is the coset $X+\langle X^2+1\rangle$.) – ancient mathematician May 03 '22 at 08:00
  • @ancientmathematician - In the books, they say adjoining $m$ to $R$ is $R[m]= {a + mb}$ where $a, b \in R$ - which seems to be quite a different operation than quotienting – user93353 May 03 '22 at 08:06
  • @ancientmathematician - now that I think about it, it may be the same!! It's a+ib only because it's a quadratic extension - else it would have higher powers – user93353 May 03 '22 at 08:14
  • It depends whether you read old-fashioned books (adjoining) or modern books (quotients). – ancient mathematician May 03 '22 at 08:45
  • @ancientmathematician - thank you very much for your comments - I think I atleast understand a bit more. F[x]/<p(x)> is the same as F[t] where t is the root of p(x). One question I have is that p(x) may have multiple roots which aren't in F - so which of the roots do we add to F using the construction? i.e. let's say there are 3 roots r1, r2 & r3, then will F[x]/<p(x)> give us F[r1], F[r2] or F[r3]? – user93353 May 04 '22 at 02:45
  • For simplicity let's talk about case $F=\mathbb{Q}$ and $p$ is an irreducible cubic. Then we can construct a field $K=F[X]/\langle p(X)\rangle$ where $p$ has a root $\alpha:=X+\langle p(X)\rangle$. But of course it's also the case that $p$ has three roots $r_1,r_2,r_3$ in $\mathbb{C}$ (deep theorem). Each of $F[r_i]$ is isomorphic to $K=F[\alpha]$ but technically none is equal to $K$. You can't sensibly ask "which root do we get". – ancient mathematician May 04 '22 at 06:36
  • You should have a look at this question which I had asked as a beginner in abstract algebra. It deals with similar issues. – Paramanand Singh May 04 '22 at 12:04
  • @ParamanandSingh - thank you - it's very useful. I think I face a similar problem as you - I am also self studying. One question about the answers - I am unable to figure out exactly what "embedding" means in the answers. Is embedding same as subfield - I haven't come across this term. – user93353 May 05 '22 at 02:07
  • Check https://en.wikipedia.org/wiki/Embedding – Paramanand Singh May 05 '22 at 07:20

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