# Why doesn't adjoining $\sqrt{3}$ to $\mathbb{F}_{11}$ return $\mathbb{F}_{11}$?

I am confused about a particular instance where adjoining an element of a field to itself makes it not equal to itself and I am asking for clarification. I can see the result is true, but I can not see why. We are not introducing any new element and we are not setting any new elements equal to zero.

In the finite field $$\mathbb{F}_{11}$$ we adjoin $$\alpha$$ where $$\alpha^2 - 3 =0$$. Because $$(\pm 5)^2 -3 =0$$ the two square roots of $$3$$ are already in $$\mathbb{F}_{11}$$, so we are either adjoining $$5$$ or $$-5$$. We do not know which, although both elements are invertible. However we cannot invert $$\alpha +5$$ because we don't know if $$\alpha +5$$ or $$\alpha-5 = 0$$ so $$\mathbb{F}_{11}[\alpha]$$ is not a field.

By adjoining an ambiguous element of the field to itself I thought maybe we were setting elements equal to zero, but it's not $$5 = -5$$ because then $$10 = 0$$ which makes every element $$0$$.

Source: This was an example in Algebra by Artin where Artin says $$\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$$ is not a field on page 366.

Sorry for edits

Edit 2: If I am understanding what is being said, $$\alpha$$ must assume a specific value in $$\mathbb{F}_{11}[\alpha]$$. So it is $$not$$ true $$\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$$ because the kernel of the evaluation homomorphism $$\phi: \mathbb{F}_{11}[x] \rightarrow \mathbb{F}_{11}[\alpha]$$ is not the ideal $$(x^2-3)$$, but one of the ideals $$(x-5)$$ or $$(x+5)$$.

I will rewrite a comment here hopefully to clarify.

In this image Artin describes $$R'$$ as "obtained by adjoining an element $$\alpha$$ to $$\mathbb{F}_{11}$$". A page earlier Artin defined "$$R[\alpha] = \text{ring obtained by adjoining} \ \alpha \ \text{to} \ R$$". There are also examples of using the evaluation homomorphism to show results such as $$\mathbb{R}[x]/(x^2+1)\simeq \mathbb{Q}$$ and $$R[x,y] \simeq R[x][y]$$.

Artin writes $$$$R' = \mathbb{F}_{11}[x]/(x^2-3)$$$$

In the same paragraph he says "...procedure applied to $$\mathbb{F}_{11}$$ does not yield a field", and "But we haven't told $$\alpha$$ whether to be equal to $$5$$ or $$-5$$. We've only told that its square is $$3$$."

With this wording it sounds like the kernel of $$\phi$$ is not $$(x-5)$$ or $$(x+5)$$, but only $$(x^2-3)$$. Which again confuses me because then $$\mathbb{F}_{11}[\alpha]$$ is not a field.

• I don't understand what your question is... Can you reformulate it in a precise way? Jun 15 '20 at 18:44
• is this better? Jun 15 '20 at 18:52
• Small remark: the fact that you cannot invert one of the two $(\alpha+5,\alpha-5)$, because one of them is $0$, doesn't mean that $\mathbb{F}_{11}[\alpha]$ is not a field. In order to conclude like that, you should find a non-zero element. In this particular case, $\mathbb{F}_{11}[\alpha]$ is indeed a field (which is isomorphic to $\mathbb{F}_{11}[X]/(X-\alpha)\sim\mathbb{F}_{11})$. Jun 15 '20 at 18:55
• And perhaps the title means "why isn't $\mathbb{F}_{11}[X]/(X^2-3)$ isomorphic to $\mathbb{F}_{11}$ eventhough $X^2-3$ has two solutions in $\mathbb{F}_{11}$?"? Jun 15 '20 at 19:02

There are two different rings being discussed here:

• $$\mathbb{F}_{11}[\alpha]$$, the ring obtained by adjoining some specific $$\alpha\in \overline{\mathbb{F}}_{11}$$ with $$\alpha^2 = 3$$;
• $$\mathbb{F}_{11}[X] / (X^2 - 3)$$.

In the first case, $$\mathbb{F}_{11}$$ already contains $$\alpha$$; as you point out, $$\alpha = \pm 5\in \mathbb{F}_{11}$$. In the latter case, we no longer have a field: \begin{align*} \mathbb{F}_{11}[X]/(X^2 - 3) = \mathbb{F}_{11}[X]/(X - 5)\oplus \mathbb{F}_{11}[X]/(X + 5). \end{align*} If $$f\in \mathbb{F}_{11}[X]$$ is an irreducible nonconstant polynomial, then the map $$\mathbb{F}_{11}[X]/(f) \to \mathbb{F}_{11}[\alpha]$$ given by $$X \to \alpha$$, where $$\alpha$$ is a zero of $$f$$ in $$\overline{\mathbb{F}}_{11}$$, is an isomorphism; for then $$\alpha$$ is not a zero of any polynomial of degree less than $$\deg f$$, and comparing dimensions gives the result. That result doesn't hold without the irreducibility assumption, though.

• Thank you for your answer. In the picture I linked Artin describes $R'$ as "obtained from adjoining an element $\alpha$ to $\mathbb{F}_{11}$ with the relation $\alpha^2-3 = 0$". He later says "we haven't told $\alpha$ whether to equal $5$ or $-5$. We've only told it it's square is $3$". With this wording it sounds like the substitution homomorphism $\phi : \mathbb{F}_{11}[x] \rightarrow \mathbb{F}_{11}[\alpha]$ does not have kernel $(x-5)$ or $(x+5)$, but only $(x^2 - 3)$. It seems in which case $F_{11}[\alpha] \simeq \mathbb{F}_{11}[x] / \ker \phi$ is not a field. Jun 15 '20 at 21:04
• If Artin is describing adjoining a formal element satisfying $x^2 - 3$, he's presumably referring to $\mathbb{F}_{11}[X]/(X^2 - 3)$. Regardless, any ring $R\supset \mathbb{F}_{11}$ containing some $\alpha\not\in \mathbb{F}_{11}$ must not be a field; otherwise, the polynomial $X^2 - 3$ would have at most $2$ solutions. Jun 15 '20 at 21:13

The two square roots of $$3$$ in $$F_{11}$$ are $$5$$ and $$6$$. Adjoining an element of a field that's already in the field does not increase the size of the field, as you seem to understand.

I think perhaps you do not fully understand what $$F[\alpha]$$ means: it means "the smallest ring containing $$F$$ and $$\alpha$$." When $$\alpha$$ i algebraic over $$F$$, it conicides with the smallest field containing $$F$$ and $$\alpha$$ (and it is algebraic because it is a solution to $$X^2-3$$.)

Accordingly, $$F_{11}[\alpha]=F_{11}[-\alpha]=F_{11}[\alpha,-\alpha]=F_{11}$$.

By adjoining an ambiguous element of the field to itself I thought maybe we were setting elements equal to zero,...

I have a hard time deciphering that. When you adjoin an indeterminate (possibly what you mean by "ambiguous element"?) to $$F$$, you get $$F[x]$$, which is a domain but not a field. When $$\alpha$$ is a root of an irreducible polynomial $$f(x)\in F[x]$$, you can talk about the quotient $$F[x]/(f(x))\cong F[\alpha]$$ being a field extension of $$F$$.

Your case is no exception because $$F_{11}[x]/(x-5)\cong F_{11}$$ as we thought.

When $$f(x)$$ is reducible, $$F[x]/(f(x))$$ no longer yields a field extension of $$F$$, so it is useless to compare it to $$F[\alpha]$$ where $$\alpha$$ is a root of $$f(x)$$.

• This is not what Artin means with adjoining an element to the field. If you check in the book, then the relevant chapter is about adjoining a universal solution of a polynomial to a ring, which by definition is the construction $R[x]/(f(x))$. The resulting extension is not a field extension, but it isn't intended to be - it's an extension of rings. Jun 15 '20 at 22:16
• @Wojowu ok: not having the resource, I am at a bit of a disadvantage. I’m a bit surprised he would still define that to be an extension, since it doesn’t correspond to a field extension. Jun 15 '20 at 22:53

Here's why $$\mathbb F_{11}[x]/(x^2-3)$$ is not a field:

The notation $$\mathbb F_{11}[x]$$ denotes all polynomials with coefficients in $$x$$. The $$/(x^2-3)$$ part then means that you consider two polynomials as equivalent iff they differ by some multiple of $$x^2-3$$, and consider the equivalence classes of that equivalence relation.

Now every such class contains a polynomial of the form $$ax+b$$ where $$a,b\in\mathbb F_{11}$$, because in any higher-degree polynomial you can get rid of the highest power by using $$ax^n \equiv ax^n - ax^{n-2}(x^2-3) = 3ax^{n-2}$$, and by repeated application you can eliminate all terms of degree $$2$$ or higher.

What you cannot eliminate is the linear term, as there is no multiple of $$x^2-3$$ that is of degree $$1$$. That is, $$[ax+b]=[cx+d]$$ iff $$a=c$$ and $$b=d$$, where $$[...]$$ denote the equivalence classes of the relation above which contain the enclosed polynomial. In particular, $$[ax+b]=[0]$$ iff $$a=b=0$$.

So now you have a ring of $$11^2=121$$ elements, and we want to show that this is not a field. This is easily done by considering the product $$(x+5)(x-5)$$, which is a product of two non-zero terms: $$(x+5)(x-5) = x^2-5^2 = x^2-3 \equiv 0$$ Thus both $$[x+5]$$ and $$[x-5]$$ are zero divisors, which a field cannot have.

Basically what happened here is that we added a third square root of $$3$$ to our field (and also a fourth because also $$[-x]^2=[3]$$). But a field can only have two square roots of the same number (with the second one being the negative of the first), so adding yet another one leads to a non-field.

However we cannot invert $$\alpha +5$$ because we don't know if $$\alpha +5$$ or $$\alpha-5 = 0$$ so $$\mathbb{F}_{11}[\alpha]$$ is not a field.
Either $$\alpha = 5$$ and in this case $$\alpha+5$$ is invertible... And $$\mathbb F_{11}[\alpha] = \mathbb F_{11}$$ is a field. Or $$\alpha = -5$$ and in that case $$\alpha+ 5=0$$ is not invertible. But in that case still $$\mathbb F_{11}[\alpha] = \mathbb F_{11}$$ is a field.
• But $\mathbb{F}[\alpha]$ isn't a field... Jun 15 '20 at 19:04
• It is!!! How do you define $\mathbb{F}[\alpha]$? Jun 15 '20 at 19:04
• If you check out Artin's book, you will find that $\alpha$ is defined as the residue of $X$ in the quotient $\mathbb F_{11}[X]/(X^2-3)$. So $\alpha$ is a square root of $3$ which is neither $5$ nor $-5$, in fact it doesn't like in (the image of) $\mathbb F_{11}$. Jun 15 '20 at 22:22