# Polynomial Rings - Lemma 1.13 from Sharp: Steps in Commutative Algebra

#### Peter

##### Well-known member
MHB Site Helper
I am reading R.Y. Sharp: Steps in Commutative Algebra.

Lemma 1.13 on page 7 (see attachment) reads as follows:

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1.13 LEMMA. let R be a commutative ring, and let X be an indeterminate; let T be a commutative R-algebra with structural ring homomorphism $$\displaystyle f \ : \ R \ \to \ T$$; and let $$\displaystyle \alpha \in T$$.

Then there is a unique ring homomorphism $$\displaystyle f_1 \ : \ R[X] \ \to \ T$$ which extends $$\displaystyle f$$ (that is, is such that $$\displaystyle f_{1|R}$$) and satisfies $$\displaystyle f_1(X) = \alpha$$

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I must confess that while I have a vague and general idea of what is going on here (very vague :-( ) I do not fully comprehend this Lemma.

Could someone give me a simple example showing explicitly what is going on here - and if possible, explain the situation?

Is this about the evaluation of polynomials? In the proof (see attachment) Sharp mentions that $$\displaystyle f_1$$ would have to satisfy

$$\displaystyle f_1(rX^i) = f(r) {\alpha}^i$$

This appears to be a generalisation of the evaluation homomorphism (in the evaluation homomorphism we would have f(r) = r) - is that correct?

Would appreciate someone explaining the situation using an example.

Peter

#### Turgul

##### Member
Consider the $\mathbb{Q}$-algebra $\mathbb{Q}(i)$. We have the structure morphism $f: \mathbb{Q} \rightarrow \mathbb{Q}(i)$ taking $a \mapsto a + 0i$.

Now pick some (fixed) element of $\mathbb{Q}(i)$, say $1+2i$. The lemma is saying that we should get a ring homomorphism $f_1: \mathbb{Q}[x] \rightarrow \mathbb{Q}(i)$ which takes $a \mapsto a$ for any $a \in \mathbb{Q}$ and takes $x \mapsto 1 + 2i$.

This is precisely evaluation; for some polynomial $g(x) = a_0 + a_1x + \cdots + a_nx^n$, we should have that $f(g(x)) = g(1+2i) = a_0 + a_1(1+2i) + \cdots + a_n(1+2i)^n$.

The only thing which makes this slightly strange is that I am evaluating at an element not in $\mathbb{Q}$ (ie I am evaluating at a point not in the ring I was defining my polynomial over to begin with). But this is not really a problem, because $\mathbb{Q}$ sits inside of $\mathbb{Q}(i)$, so any polynomial over $\mathbb{Q}$ is a polynomial over $\mathbb{Q}(i)$, where we are doing the evaluating.

The claim that the lemma makes is that there is nothing special about $1+2i$, I could just as well have sent $x \mapsto 7-3i$ (evaluated our polynomials at $7-3i$) and we still would have gotten a ring homomorphism. The other thing to notice is that, since I know where $\mathbb{Q}$ has to be mapped under one of these homomorphisms, the only other choice I have is where to send $x$, and once that has been decided, that determines the map uniquely (any ring homomorphism that sends $x \mapsto 1+2i$ and fixes $\mathbb{Q}$ must be evaluation at $1+2i$).

#### Peter

##### Well-known member
MHB Site Helper
Consider the $\mathbb{Q}$-algebra $\mathbb{Q}(i)$. We have the structure morphism $f: \mathbb{Q} \rightarrow \mathbb{Q}(i)$ taking $a \mapsto a + 0i$.

Now pick some (fixed) element of $\mathbb{Q}(i)$, say $1+2i$. The lemma is saying that we should get a ring homomorphism $f_1: \mathbb{Q}[x] \rightarrow \mathbb{Q}(i)$ which takes $a \mapsto a$ for any $a \in \mathbb{Q}$ and takes $x \mapsto 1 + 2i$.

This is precisely evaluation; for some polynomial $g(x) = a_0 + a_1x + \cdots + a_nx^n$, we should have that $f(g(x)) = g(1+2i) = a_0 + a_1(1+2i) + \cdots + a_n(1+2i)^n$.

The only thing which makes this slightly strange is that I am evaluating at an element not in $\mathbb{Q}$ (ie I am evaluating at a point not in the ring I was defining my polynomial over to begin with). But this is not really a problem, because $\mathbb{Q}$ sits inside of $\mathbb{Q}(i)$, so any polynomial over $\mathbb{Q}$ is a polynomial over $\mathbb{Q}(i)$, where we are doing the evaluating.

The claim that the lemma makes is that there is nothing special about $1+2i$, I could just as well have sent $x \mapsto 7-3i$ (evaluated our polynomials at $7-3i$) and we still would have gotten a ring homomorphism. The other thing to notice is that, since I know where $\mathbb{Q}$ has to be mapped under one of these homomorphisms, the only other choice I have is where to send $x$, and once that has been decided, that determines the map uniquely (any ring homomorphism that sends $x \mapsto 1+2i$ and fixes $\mathbb{Q}$ must be evaluation at $1+2i$).
Thanks so much for the example, Turgul

Just working through it carefully now

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
In pictoral form:

$\begin{array}{ccccc}R&\stackrel{1_R}{\rightarrow}&R[x]&\stackrel{x}{\leftarrow}&\{\ast\}\\&\stackrel{f}{\searrow}&\Big\downarrow\rlap{\scriptstyle f^{\ast}}&\stackrel{a}{\swarrow}\\&&T\end{array}$

Where the maps on the right are defined in the only way possible, and are merely functions, while the maps on the left are ring homomorphisms.

The theorem asserts the existence and uniqueness of $f^{\ast}$.

Note that this diagram shows that the theorem is essentially a property of ring homomorphisms. Diagrams like this are typical of "universal objects" such as $R[x]$ ($R[x]$ is the "free $R$-algebra generated by the set $\{x\}$", which goes a long way towards explaining why $R[x]$ has this universal property).

If $R$ is a field, and $f$ is not trivial, then $T$ is typically called an extension ring of $R$ (why does this make sense?). This is also done whenever $f$ is injective.

#### Peter

##### Well-known member
MHB Site Helper
Consider the $\mathbb{Q}$-algebra $\mathbb{Q}(i)$. We have the structure morphism $f: \mathbb{Q} \rightarrow \mathbb{Q}(i)$ taking $a \mapsto a + 0i$.

Now pick some (fixed) element of $\mathbb{Q}(i)$, say $1+2i$. The lemma is saying that we should get a ring homomorphism $f_1: \mathbb{Q}[x] \rightarrow \mathbb{Q}(i)$ which takes $a \mapsto a$ for any $a \in \mathbb{Q}$ and takes $x \mapsto 1 + 2i$.

This is precisely evaluation; for some polynomial $g(x) = a_0 + a_1x + \cdots + a_nx^n$, we should have that $f(g(x)) = g(1+2i) = a_0 + a_1(1+2i) + \cdots + a_n(1+2i)^n$.

The only thing which makes this slightly strange is that I am evaluating at an element not in $\mathbb{Q}$ (ie I am evaluating at a point not in the ring I was defining my polynomial over to begin with). But this is not really a problem, because $\mathbb{Q}$ sits inside of $\mathbb{Q}(i)$, so any polynomial over $\mathbb{Q}$ is a polynomial over $\mathbb{Q}(i)$, where we are doing the evaluating.

The claim that the lemma makes is that there is nothing special about $1+2i$, I could just as well have sent $x \mapsto 7-3i$ (evaluated our polynomials at $7-3i$) and we still would have gotten a ring homomorphism. The other thing to notice is that, since I know where $\mathbb{Q}$ has to be mapped under one of these homomorphisms, the only other choice I have is where to send $x$, and once that has been decided, that determines the map uniquely (any ring homomorphism that sends $x \mapsto 1+2i$ and fixes $\mathbb{Q}$ must be evaluation at $1+2i$).
Thanks Turgul, that really shed some light on LEMMA 13 for me.

I thought I would work through it carefully and formally as follows: (note some minor concerns indicated by ??)

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Let $$\displaystyle R = \mathbb{Q}$$ be the commutative ring in LEMMA 1.13.

Let $$\displaystyle T = Q(i) = \{ a + bi\}$$ be the $$\displaystyle R$$-algebra (in this case a $$\displaystyle \mathbb{Q}$$ algebra.

Then $$\displaystyle f: \ R \to T$$ the structural (????) ring homomorphism is given by $$\displaystyle f: \ \mathbb{Q} \to \mathbb{Q}(i)$$ and we define f to be f(a) = a

(can f be defined differently?? presumably, as long as f is homomorphism??)

Now we choose $$\displaystyle f_1 (x) = \alpha$$ for some $$\displaystyle \alpha \in \mathbb{Q}(i)$$

Let us choose $$\displaystyle \alpha = 1 + 2i$$ so $$\displaystyle f_1 (x) = \alpha = 1 + 2i$$

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Now we should have a unique homomorphism

$$\displaystyle f_1: \ R[X] \to T$$ i.e. $$\displaystyle f_1: \ \mathbb{Q}[X] \to \mathbb{Q}(i)$$ ... ... (1)

Now let us see how $$\displaystyle f_1$$ works. [Indeed we will see that $$\displaystyle f_1$$ depends only on $$\displaystyle f$$ and $$\displaystyle \alpha$$ which are predetermined or given as preconditions in the LEMMA]

Take $$\displaystyle g \in \mathbb{Q}$$ where g is as follows:

$$\displaystyle g = a_0 + a_1x + a_2x^2 + ... \ ... + a_nx^n$$

where $$\displaystyle a_i \in \mathbb{Q}$$ and $$\displaystyle x$$ is indeterminate.

Using (1) we have:

$$\displaystyle f_1(g) = f_1(a_0) + f_1(a_1x) + f_1(a_2x^2) + ... \ ... f_1(a_nx^n)$$ ... ... (2)

Now $$\displaystyle f_1(a_0) = a_0$$ [Definition of f and f_1]

and $$\displaystyle f_1(a_1x) = f_1(a_1)f(x) = a_1 \alpha$$

[Definition of f and f_1, f_1 homomorphism - but must demonstrate f_1 is actually a homomorphism ??]

and $$\displaystyle f_1(a_2x^2) = f_1(a_2)f_1(x^2) = f_1(a_2)f_1(x)f_1(x) = a_2 {\alpha}^2$$

[Definition of f and f_1, f_1 homomorphism - but must demonstrate f_1 is actually a homomorphism ??]

... ... etc

Thus (2) becomes

$$\displaystyle f_1(g) = a_0 + a_1 \alpha + a_2 {\alpha}^2 + ... \ ... + a_n {\alpha}^n$$ where $$\displaystyle \alpha = 1 + 2i$$ ... \ ... (3)

Now equation (3) has been established using only (or put differently, depends only on)
(1) the mappings $$\displaystyle f$$ and $$\displaystyle f_1$$
(2) the mapping of x

But both (1) and (2) are fixed as preconditions of the LEMMA

Hoping that all the above makes good sense.

Thanks again for the example - it was REALLY helpful ...

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Well defining $f_1$ as you have shows that we at least have "a" such homomorphism (existence).

Explicitly:

$f_1(g(x)) = f_1(a_0 + a_1x + \cdots + a_nx^n) = a_0 + a_1\alpha + \cdots + a_n\alpha^n = g(\alpha)$.

It is easy to see this indeed is a homomorphism, we have:

$f_1((g+h)(x)) = (g+h)(\alpha) = g(\alpha) + h(\alpha) = f_1(g(x)) + f_1(h(x))$.

Multiplication is shown similarly.

Now suppose you have another, say $f_2$. Show that (you can now USE the fact that $f_2$ is indeed a homomorphism, you do not have to PROVE it) $f_2 = f_1$.

(Hint: consider $f_1(x), f_2(x)$ and the homomorphism properties to show that $f_1 - f_2$ is the 0-map...we are leveraging the fact that we can add homomorphisms $R[x] \to T$ "point-wise":

For two such homomorphisms, $f,g$ we define:

$(f+g)(p(x)) = f(p(x)) + g(p(x))$
$(-f)(p(x)) = -f(p(x))$).

The injective (structural) homomorphism Turgul is talking about is the usual *embedding* of $\Bbb Q$ in $\Bbb Q(i)$, see the last line of my previous post.

#### Turgul

##### Member
An alternate perspective: if you are already happy with the evaluation homomorphism $A[x] \rightarrow A$ taking $x \mapsto \alpha$ for some $\alpha \in A$ for a given ring $A$, you can build this pseudo-evaluation map in a straightforward way.

You have a homomorphism $f: R \rightarrow T$ so you can get a homomorphism $f_0: R[x] \rightarrow T[x]$ which sends $x \mapsto x$ and sends the coefficients of $x$ in $R$ to their images under $f$. Since $T[x]$ has the standard evaluation homomorphism $\pi_\alpha: T[x] \rightarrow T$ defined by taking $x \mapsto \alpha$, we can compose and get $f_1 = \pi_\alpha \circ f_0$. Since $f_0$ restricts to $f$ on $R$ and $\pi_\alpha$ is the identity on coefficients, the composition will restrict to $f$ on $R$. This way of looking at $f_1$ is a surprisingly important idea in modern algebraic geometry.

A note about "structural" homomorphism: you are told that $T$ is an $R$-algebra. But what is an $R$-algebra? It is a ring $T$, along with an action of $R$ (you have a way to multiply anything in $T$ by anything in $R$). This action defines (and is defined by!) a homomorphism $f: R \rightarrow T$ sending $r \mapsto r \cdot 1$. Note this is a homomorphism and that, if instead I started with a homomorphism of rings $f: R \rightarrow T$ of commutative rings, then I can define a product of things in $R$ with things in $T$ by $r \cdot t = f(r)t$. Hence an $R$-algebra is actually the data of both a ring $T$ and a homomorphism $f: R \rightarrow T$. This choice of homomorphism is the "structural" map and is a fixed piece of data coming with $T$ to view it as an $R$-algebra.

In general there may be many ways to view a ring $T$ as an $R$-module (exactly as many ways as there are homomorphisms $R \rightarrow T$ in fact). It just happens that there is always at most 1 way to map $\mathbb{Q}$ homomorphically into any ring since always $1 \mapsto 1$ and knowing where 1 goes uniquely determines where all of $\mathbb{Q}$ goes. The same is true of any finite field.

EDIT: And of course by any finite field I mean any finite field of prime cardinality.

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