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What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.

abeliangrape
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3 Answers3

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Short answer: given a ring $A$ and an element $x$, the notation $A[x]$ denotes the smallest ring that contains both $A$ and $x$. If $A$ happens to be a field, then $A(x)$ denotes the smallest field that contains both $A$ and $x$. That's all.

Longer answer: given a ring $A$, we denote by $A[x]$ the ring of polynomials in the variable $x$ with coefficients in $A$. Starting from the description in the above paragraph, if a ring contains $x$, it must also contain $x^2,x^3$, and so on. And sums of these. And the results of multiplying these with one another and with the elements of $A$. Without any more context, these are just polynomials the way you're used to them.

If $A\subseteq B$ and $x\in B$, then $x$ inherits any relations it may satisfy in $B$ when used in this polynomial ring. One example is the ring of Gaussian integers $\Bbb Z[i]$, where $i^2=-1$. If $x$ satisfies no relations that can be described using only elements of $A$, then it acts like in the previous paragraph, and you get regular polynomials with no special rules. Your $\Bbb Q[\pi]$ is an example.

If $F$ is a field, then $F(x)$ is the smallest field that contains both $F$ and $x$. Starting from the description in the first paragraph, this field must for the same reason contain all polynomials in the variable $x$ with coefficients in $F$. But it must also contain the reciprocal of any non-zero polynomial. Same things as in the above paragraph apply: If $x$ satisfies some algebraic relation that can be phrased using only coefficients from $F$ (we call this "$x$ is algebraic over $F$"), then that's baked into $F(x)$. Your $\Bbb Q(i)$ is an example of that, where $i^2=-1$ is the algebraic relation in question.

When $x$ is not algebraic over $F$, then $F(x)$ becomes the field of rational functions in the variable $x$ with coefficients in $F$. Which is to say, the elements of $F(x)$ are fractions where the numerators and the denominators are polynomials in $F[x]$. One concrete example is $\Bbb Q(\pi)$. But the fact that $\pi$ lies inside a well-known extension of $\Bbb Q$ is unnecessary. $\Bbb Q(x)$ works just as well, where $x$ is just an abstract variable.

If $x$ is algebraic over $F$, then $F[x]$ and $F(x)$ turn out to be isomorphic: fractions in $F(x)$ can always be expanded so that the denominators have no $x$ in them. Thus you will see either one used, which can be a point of confusion. For instance, $\Bbb Q(i)$ and $\Bbb Q[i]$ are, for all practical intents and purposes, the same thing.

Arthur
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  • You said that if $x$ is algebraic over $F$, then $F[x]$ and $F(x)$ turn out to be isomorphic. I just want to say that they are not only isomorphic, but they are equal! – Ariel Fishbein Mar 28 '24 at 10:57
  • @ArielFishbein Not in my book, because $F(x)$ has elements that look like $\frac 1x$, and $F[x]$ doesn't have those. – Arthur Mar 28 '24 at 11:08
  • remember you said that $x$ is algebraic. E.g $F=\mathbb{Q}$ and $K=\mathbb{R}$. Now, $\frac 14\in F(4)$ but $1/4=\frac{1}{16}\cdot 4 \in F[x]$. – Ariel Fishbein Mar 28 '24 at 12:50
  • If you want, I can prove to you that if $\alpha$ is algebraic over $F$ with a minimal polynomial $p(x)$, then $F(\alpha)=F[\alpha]\cong F[x]/(p(x))$. Just tell me if you want to see this prove, and I'll be happy too – Ariel Fishbein Mar 28 '24 at 12:52
  • The element $\frac{1}{\sqrt2}$ is not in $\Bbb Q[\sqrt2]$. The element $\frac{\sqrt2}{2}$ does exist there. The two rings / fields are isomorphic, but one of them has names for elements that the other one doesn't. – Arthur Mar 28 '24 at 13:41
  • I don't know what to tell you other than that $\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$ – Ariel Fishbein Mar 28 '24 at 14:50
  • @ArielFishbein In $\Bbb Q(\sqrt2)$, yes.. In $\Bbb Q[\sqrt2]$, we only have polynomials, so elements have the names $a+b\sqrt2$ for $a,b\in\Bbb Q$, and no others. Of course, this is being pedantic, but I will stand by my statement that $\Bbb Q[\sqrt 2]$ and $\Bbb Q(\sqrt2)$ are two different rings, because their underlying sets of formal algebraic expressions are different. They are very obviously isomorphic with a very obvious isomorphism, but they are not the same. – Arthur Mar 28 '24 at 14:55
  • (I say no others, but of course we have higher degree expressions, any finite variation of $a+b\sqrt2+c\sqrt2^2+\cdots$. What you DON'T have in $\Bbb Q[\sqrt2]$ are fractions with $\sqrt2$ in the denominator.) – Arthur Mar 28 '24 at 15:12
  • If $\alpha$ is algebraic, then $F[\alpha]$ becomes a field that contains both $\alpha, F$. But by definition, $F(\alpha)$ is the smallest field that contains $\alpha, F$ so we get that $F(\alpha)\subseteq F[\alpha]$. But it is also obvious that $F[\alpha]\subseteq F(\alpha)$. So we can conclude by saying that $F[\alpha]=F(\alpha) \square$ – Ariel Fishbein Mar 28 '24 at 16:31
  • @ArielFishbein We can conclude they are isomorphic. There is no "the" smallest field. There are infinitely many of them, and they are all isomorphic. You can't use universal properties like that to prove that things are actually equal. You can't prove $\subseteq$, you can only prove there exists injective homomorphisms. It is NOT obvious that $F[\alpha]\subseteq F(\alpha)$, because their elements have different names. – Arthur Mar 28 '24 at 16:33
  • $F[\alpha]={\sum\limits_{i=0}^n a_i \alpha^i: a_i\in R }$ and $F(\alpha)={\frac{p(\alpha)}{q(\alpha)}\bigg| p,q\in F[\alpha]}$ – Ariel Fishbein Mar 28 '24 at 16:42
  • Can you please tell me then what is the isomorphism between $F[\alpha]$ and $F(\alpha)$? – Ariel Fishbein Mar 28 '24 at 16:45
  • The isomorphism $F[\alpha]\to F(\alpha)$ is given by $f\mapsto \frac f1$. Note that I didn't say $f\mapsto f$, because by your own definition, the elements of the two fields are written differently. It is even clearer that it isn't an equality if you try to write the inverse isomorphism. If you have a rational expression in $\alpha$, you need to massage it pretty heavily to make it into a polynomial. Rational expressions aren't the same as polynomials, even with an algebraic variable, and therefore a ring of rational expressions is not a ring of polynomials, even if it's isomorphic to one. – Arthur Mar 28 '24 at 17:02
  • Ill think about it. Anyways it was an interesting convo. Cya! – Ariel Fishbein Mar 28 '24 at 17:41
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$F[x]$ is the ring of all polynomials in $x$ with coefficients from $F$. In contrast, $F(x)$ is the smallest field which extends $F$ and contains $x$: in other words, it contains all rational functions (i.e. quotients of polynomials) in $x$ with coefficients from $F$, rather than just all polynomials. Note that sometimes these can be the same, e.g. $\mathbb{Q}\left[\sqrt{2}\right]$ is the same as $\mathbb{Q}\left(\sqrt{2}\right)$, since any polynomial in $\sqrt{2}$ can be written in the form $a+b\sqrt{2}$, and if $a,b,c,d \in \mathbb{Q}$ then $$\frac{a+b\sqrt{2}}{c+d\sqrt{2}} = \frac{ac-2bd}{c^2-2d^2} + \frac{bc-ad}{c^2-2d^2}$$ where the coefficients are still in $\mathbb{Q}$.

Prasiortle
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Well, $F[x]$ means ring adjunction of $x$, while $F(x)$ means field adjunction of $x$. More concretely,

$F(x) = \left\{\frac{f(x)}{g(x)}\mid f(x),g(x)\in F[x], g(x)\ne 0\right\}$ is the field of quotients of $F[x]$.

$\Bbb Q(i) = \{a+ib\mid a,b\in\Bbb Q\}$ with $i^2=-1$ is the field adjunction of $i$ to $\Bbb Q$. It's a field and the same as the ring adjunction $\Bbb Q[i]$ (as its also a field).

$\Bbb Z[i] = \{a+ib\mid a,b\in\Bbb Z\}$ with $i^2=-1$ is the ring adjungation of $i$ to $\Bbb Q$. Its's a ring.

Beware of $\Bbb Q(\pi)$. As $\pi$ is tranzendential over $\Bbb Q$, this field extension of $\Bbb Q$ is infinite.

You need to learn about ring and field adjunction.

Wuestenfux
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    I think in English the word that goes with "adjoin" is more commonly "adjunction" and I have never seen "adjungation". But in mathematics it might also be more common to speak of things like "a ring extension obtained by extending the ring by adjoining an element" than ring or field "adjunctions". And the spelling is usually "transcendental" rather than "tranzendential". – Mark S. Mar 15 '21 at 12:15