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

Question

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 \begin{equation} R' = \mathbb{F}_{11}[x]/(x^2-3) \end{equation}

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.

Answer 1

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.

Answer 2

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 $f(x)$.

Answer 3

Your sequence

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.

is confusing.

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.

Answer 4

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.