# Rings of polynomials

The hypothesis of theorem 5.5 in page 152 of Hungerford's Algebra requires that the homomorphism maps the identity of R to the identity of S, but the proof "never" use this hypothesis. Is the theorem still true if the stronger hypothesis is removed. I am so sorry that I currently do not have link to the book.

Theorem 5.5 Let $R$ and $S$ be commutative rings with identity and $\varphi: R \to S$ a homomorphism of rings such that $\varphi(1_R) =1_S$. If $s_1, \dots, s_n \in S$, then there is a unique homomorphism of rings $\overline{\varphi}: R[x_1, \dots , x_n] \to S$ such that $\overline{\varphi}|R = \varphi$ and $\overline{\varphi}(x_i) = s_i$ for $i = 1,2, \dots, n$. This property completely determines $R[x_1, \dots, x_n]$ up to isomorphism.

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@Yuan: You should really provide the text of the problem. – Arturo Magidin Jan 24 '11 at 22:18
Isn't it by definition that a ring homomorphism maps identities to identities for a CRWI? – PEV Jan 24 '11 at 22:19
@PEV: it is mutiplicative identity – Yuan Jan 24 '11 at 22:23
@PEV: It's a convention that many, but not all follow; if I recall correctly (not in my office right now), Hungerford does not require a ring homomorphism between rings with identity to restrict to a monoid homomorphism of the multiplicative monoids of the rings, but only a semigroup homomorphism (that is, a fancy way of saying Hungerford does not require ring homomorphisms between rings with identity to map the identity to the identity). – Arturo Magidin Jan 25 '11 at 3:27
@Arturo You're quite right. Hungerford explicitly says so in page 119 of his Algebra book. To quote him: "If $R$ and $S$ both have identities $1_R$ and $1_S$, we do not require that a homomorphism of rings map $1_R$ to $1_S$". – Adrián Barquero Jan 25 '11 at 4:10

The hypothesis is being stealthily used in the assertion that $\overline{\varphi}$ is a homomorphism of rings.

Added and rewritten: More explicitly: the sketch Hungerford provides is to take an element of the polynomial ring $$\sum r_{i}x_1^{k_{1i}}x_2^{k_{2i}}\cdots x_n^{k_{ni}}$$ (suitably finite) and suggests mapping it to $$\sum\varphi(r_i)s_1^{k_{1i}}\cdots x_n^{k_{ni}}.$$ Then Hungerford asserts that this makes $\overline{\varphi}$ into a ring homomorphism.

You don't actually need the full hypothesis that $\varphi(1_R)=1_S$, but you do need something:

For $\varphi$ to be a ring homomorphism, we need $\varphi(1_R)$ to be an identity for $\varphi(R)$; for $\overline{\varphi}$ defined as above to be a ring homomorphism, we also need $\varphi(1_R)s_j = s_j$, because taking $p = 1_R$ and $q=x_j$ we have $\overline{\varphi}(pq) = \overline{\varphi}(x_j) =s_j$, and $\overline{\varphi}(p)\overline{\varphi}(q) = \varphi(1_R)s_j$. Thus, for $\overline{\varphi}$ to be a ring homomorphism, it is necessary that $\varphi(1_R)$ be an identity of the subring of $S$ generated by $\varphi(R)$ and $s_1,\ldots,s_n$ (as $R$ and $S$ are both commutative).

In fact, this is also sufficient for you to be able to get a homomorphism: you must require that $\varphi(1_R)$ be the identity for the subring of $S$ generated by $\varphi(R)$ and $s_1,\ldots,s_n$.

But it is not hard to check that the apparently weaker condition in the theorem (weaker, because it refers to a smaller class of homomorphisms and possible selections of $s_1,\ldots,s_n$) is in fact sufficient to characterize the polynomial ring up to unique isomorphism; the theorem suffices to prove the more general case by considering first the map onto $S'=\langle \varphi(R),s_1,\ldots,s_n\rangle$ and then composing with the embedding $S'\hookrightarrow S$ (since we are not requiring maps of rings with identity to send the identity to the identity).

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Could you please explain "completely determine" means in this theorem? I feel confused about it. – Yuan Jan 27 '11 at 4:43
@Yuan: It means that any commutative ring with identity $T$ and elements $t_1,\ldots,t_n$, together with an embedding $i\colon R\to T$, which satisfies the hypothesis of the theorem (for every homomorphism $\varphi\colon R\to S$ with $\varphi(1_R)=1_S$ and every elements $s_1,\ldots,s_n\in S$ there exists a unique homomorphism $\overline{\varphi}\colon T\to S$ such that $\varphi=\overline{\varphi}\circ i$ and $\overline{\varphi}(t_j)=s_j$, will be isomorphic to the ring of polynomials, and the isomorphism will be canonical. It's a typical category-theory statement about universality. – Arturo Magidin Jan 27 '11 at 5:10