About isomorphisms, minimal polynomials and roots

Question

I am taking a class on algebra (or abstract algebra if you are outside of europe) and I have two questions concerning Isomorphisms between Fields.

1) Let $\phi:L \rightarrow L'$ be a field isomorphism and $h \in L[X]$. $h$ then is of the form $h=a_0+a_1 X+a_2 X^2 +...+a_nX^n$. Define $$\phi h := \phi(a_0)+\phi(a_1)X+\phi(a_2)X^2+...+\phi(a_n)X^n.$$ If $\alpha$ is a root of $h$, i.e. $h(\alpha)=0$, then $\phi(\alpha)=\alpha'$ is a root of $\phi h$, i.e. $(\phi h)(\alpha')=0$. Why is that the case?

2) Let $\phi$ and $h$ be like in 1) but now $h$ is a minimal polynomial of $\alpha$. Is $\phi h$ a minimalpolynomial of $\alpha'$?

3) Let $\phi$ and $h$ be like in 1). If $h$ is reducible in $L[X]$, is then also $\phi h$ reducible in $L'[X]$?

In the algebra book I am using there is no proof of these facts. I assume they are rather trivial then, but I still could not come up with the proofs of these facts.

Thanks a lot in advance!

Answer 1

The key thing one needs to understand here is, I opine, that a field isomorphism such as $\phi:L \to L^\prime$, that is, an injective, surjective map 'twixt $L$ and $L'$, which also preserves algebraic stuctures, is merely a re-naming of the elelments of $L$ using the elements of $L'$ as the new names. Furthermore, the map $\phi^{-1}: L^\prime \to L$ is also a field isomorphism and hence may be thought of as returning the new names to their originals; if this view of is borne in mind, problems like those posed here become much more transparent.

Let's start with item (1.); we have $h(X) \in L[X]$ with

$h(X) = \sum_0^n a_i X^i, \tag{1}$

there $a_i \in L$ for $0 \le i \le n$. Given $\alpha \in L$ with $h(\alpha) = 0$, we then may write

$\sum_0^n a_i \alpha^i = h(\alpha) = 0, \tag{2}$

and applying $\phi$ to (2),

$\phi(\sum_0^n a_i \alpha^i) = \phi(0) = 0, \tag{3}$

since $\phi$ preserves algebraic operations, that is

$\phi(s + t) = \phi(s) + \phi(t), \;\phi(st) = \phi(s)\phi(t) \tag{4}$

for $s, t \in L$, it is easily seen that

$\phi(\alpha^i) = (\phi(\alpha))^i, \tag{5}$

whence

$\phi(a_i \alpha^i) = \phi(a_i)\phi(\alpha^i) = \phi(a_i)(\phi(\alpha))^i, \tag{6}$

and thus

$\phi(\sum_0^n a_i \alpha^i) = \sum_0^n \phi(a_i \alpha^i) = \sum_0^n \phi(a_i)(\phi(\alpha))^i; \tag{7}$

we conclude from (3) and (7) that

$\sum_0^n \phi(a_i)(\phi(\alpha))^i = 0, \tag{8}$

which shows that $\phi(\alpha)$ satisfies

$(\phi h)(X) = \sum_0^n \phi(a_i)X^i \in L^\prime[X], \tag{9}$

as anticipated.

Our OP vaoy asked why it is the case that $\phi h(\alpha) = 0$; in a nutshell, because $\phi$ preserves any algebraic relationship which can be derived from the two basic operations of $L$ "$+$" and "$\cdot$", as follows from repeated appliation of (4). In particular, polynomial relationships will be preserved under the action of $\phi$, and hence, the roots of polynomials $h(X) \in L[X]$ will map to the roots of the $\phi h(X) \in L^\prime[X]$.

The preservation of roots under the action of field isomorphisms such as $\phi:L \to L^\prime$ is actually a specific case of a much more general phenomenon. For let $R$ be any commutative ring, let $h(X) \in R[X]$, and suppose $\rho \in R$ satisfies

$h[X] = \sum_0^n a_i X^i \tag{10}$

with $a_i \in R$, $0 \le i \le n$; that is,

$\sum_0^n a_i \rho^i = 0; \tag{11}$

then if $S$ is another commutative ring and $\phi:R \to S$ is a homomorphism, we have

$\phi h(\rho) = \sum_0^n \phi(a_i)(\phi(\rho))^i = 0; \tag{12}$

thus $\phi(\rho)$ is a zero of $\phi h$, just as when $\phi:L \to L^\prime$ is a map 'twixt fields; the proof is essenially identical to that given above.

We turn to item (2): $\phi h[X] \in L^\prime[X]$ is indeed a minimal polynomial of $\phi(\alpha)$. For if this were not the case, $\phi(\alpha)$ would satisfy a polynomial $k[X] \in L^\prime[X]$ with $m = \deg k[X] < n = \deg (\phi h)[X]$. Suppose

$k[X] = \sum_0^m b_i X^i; \tag{13}$

then

$k(\phi(\alpha)) = \sum_0^m b_i (\phi(\alpha))^i = 0, \tag{14}$

and we consider the polynomial $\phi^{-1} k[X] \in L[X]$. Since $\phi^{-1}: L^\prime \to L$ is itself an isomorphism of fields, we have

$\phi^{-1} k[X] = \sum_0^m \phi^{-1}(b_i)X^i \in L[X]; \tag{14}$

thus,

$\phi^{-1}k (\alpha) = \sum_0^m \phi^{-1}(b_i) \alpha^i, \tag{15}$

and

$\phi(\phi^{-1}k (\alpha)) = \sum_0^m \phi(\phi^{-1}(b_i)) \phi(\alpha^i) = \sum_0^m b_i (\phi(\alpha))^i = 0; \tag{16}$

since $\phi$ is an isomorphism, (16) implies that

$\phi^{-1}k (\alpha) = \sum_0^m \phi^{-1}(b_i) \alpha^i = 0; \tag{17}$

since $\deg \phi^{-1}k = m < n = \deg h$, (17) is in contradiction to the minimality of $h(X) \in L[X]$; this contradiction confirms the minimality of $\phi h[X] \in L^\prime[X]$.

As for item (3.), suppose $h(X)$ is reducible in $L[X]$; then we have

$h(X) = p(X)q(X) \tag{18}$

with $p(X), q(X) \in L[X]$ and $\deg p, \deg q > 0$. Now at this point perhaps the easiest and best way to show the reducibility of $\phi h(X) \in L^\prime[X]$ is establish the fact that the map $\phi$ induces on $L[X]$ via (9) is in fact a ring isomorphism 'twixt $L[X]$ and $L^\prime[X]$; we will call this induced map $\Phi: L[X] \to L'[X]$. It is easy to see that

$\Phi(h(X) + k(X)) = \Phi(h(X)) + \Phi(k(X)) \tag{19}$

and

$\Phi(h(X)k(X)) = \Phi(h(X)) \Phi(k(X)); \tag{20}$

both (19) and (20) follow from direct application of (4) to the coefficients of $h(X)$ and $k(X)$; for instance, with $\deg h = n$ and $\deg k = m$, $h(X)$ and $k(X)$ as in (1) and (13), we have

$h(X)k(X) = \left(\sum_0^n a_i X^i \right) \left(\sum_0^m b_j X^j \right) = \sum_{p = 0}^{m + n} \left( \sum_{q + r = p} a_q b_r\right) X^p, \tag{21}$

and thus, using (4),

$\Phi(h(X)k(X)) = \Phi \left( \sum_{p = 0}^{m + n} \left( \sum_{q + r = p} a_q b_r \right)X^p \right) = \sum_{p = 0}^{m + n} \phi\left(\sum_{q + r = p} a_q b_r \right) X^p$ $ = \sum_{p = 0}^{m + n} \sum_{q + r = p} \phi(a_q) \phi (b_r) X^p = \left(\sum_0^n \phi(a_i) X^i \right) \left(\sum_0^m \phi(b_j) X^j \right)$ $ = \Phi(h(X))\Phi(k(X)); \tag{22}$

the demonstration of (19) is even easier and is left to the reader.

We now immediately see that if $h(X)$ is reducible in $L[X]$, say

$h(X) = f(X)g(X) \in L[X], \tag{23}$

with $\deg f, \deg g > 0$, then

$\Phi(h(X)) = \Phi(f(X)) \Phi(g(X)) \in L^\prime[X], \tag{24}$

with $\deg \Phi(f) = \deg f$ and $\deg \Phi(g) = \deg g$; thus $\Phi(h(X))$ is reducible in $L^\prime[X]$.

Again, the key here is to realize that the isomporphism $\phi: L \to L^\prime$ induces the isomorphism $\Phi: L[X] \to L^\prime[X]$. It is clear from the discussion of item (3) that $\Phi$ is a homomorphism; that it is surjective is in fact tacitly demonstrated in item (2). Clearly, $\ker \Phi = \{ 0 \}$; for if

$\sum_0^n \phi(a_i)X^i = 0 \in L^\prime[X], \tag{25}$

then since $\phi(a_i) = 0$ for each $a_i$, every $a_i = 0$ as well by virtue of the fact that $\phi$ is an isomorphism. Thus $\Phi$, like $\phi$, may be thought of as merely renaming the elements of $L[X]$ whilst preserving algebraic structure; it should thus be no surprise that essential algebraic relationships present in $L[X]$ are preserved under the action of $\Phi$.

These traits of $\phi: L \to L^\prime$ and $\Phi: L[X] \to L^\prime[X]$ are in fact nearly obvious but it helps to see the proofs a least once, especially when one is just starting out.

Answer 2

1) As yanko pointed out in his/her comment: the whole point of field(more generally ring)-morphisms is that they commute with polynomials, i.e. $\phi h(\phi(a)) = \phi(h(a))= \phi(0) = 0$.

The problem with 2) and 3) is that they assume that the isomorphism can be extended to some $\phi : K \to K'$ where $L\subset K, L'\subset K'$ and $\alpha \in K$ is algebraic over $L$.(otherwise the notion of minimal polynomial is trivial and so not interesting). So that's what I will assume.

2) Assume moreover that we have proved 3). Then since $\phi h \neq 0 $ vanishes on $\phi(\alpha) \in K'$, the minimal polynomial of $\phi(\alpha)$ must divide $\phi h$. But since $h$ is irreducible, and since we assumed that we had proved 3), $\phi h$ must be irreducible and therefore it must be the minimal polynomial of $\alpha$.

3) Assume $h$ is irreducible and $\phi h = pq$ where $p,q \in L'[X]$. We write $p= \displaystyle\sum_{k=0}^d p_k X^k$ and similarly for $q$. Then $p = \displaystyle\sum_{k=0}^d \phi(\phi^{-1}(p_k)) X^k = \phi p'$ where $p' = \displaystyle\sum_{k=0}^d \phi^{-1}(p_k) X^k$. Similarly, $q= \phi q'$ for some $q' \in L[X]$. Therefore, $\phi h = \phi p' \phi q' = \phi p'q'$. But it is easy to check that $\phi$ is injective on polynomials and so $h= p'q'$. Now this implies that $p'$ or $q'$ is constant. But $deg(p)= deg(p')$ and $deg(q)=deg(q')$. Therefore $p$ or $q$ is constant.

Since $p,q$ were arbitrary, this implies that $\phi h$ is irreducible.

Of course, this settles this for this question, but I want you to also see why this worked, and why things like this will work more generally.

The thing is, in algebra and more generally actually, isomorphisms "carry all the information". If there is an isomorphism $\phi: L\to L'$ you can virtually consider $L$ and $L'$ to be the same, and so any property verified by $L$ will be verified by $L'$, and conversely. Moreover, an isomorphism preserve information pointwise, that is $a$ and $\phi(a)$ will "look the same" from the respective points of view of $L$ and $L'$. That's why mathematicians often don't even prove that properties are preserved by isomorphisms, simply because that's what isomorphisms are for.

Some people go as far as saying : an interesting property is a property that is preserved under isomorphisms. Almost anything you can think of will always be preserved under isomorphisms.